Subgame perfect equilibria under the deferred acceptance algorithm

Abstract

We analyze a subgame perfect equilibrium (SPE) of an extensive game with perfect information induced by the firm-oriented deferred acceptance (DA) algorithm in a one-to-one matching market between firms and workers. Our game repeats the following procedure until every firm in the market has a partner: (i) an unmatched firm strategically decides to which worker to make an offer or to exit the market, and (ii) the worker receiving the offer strategically decides whether to tentatively accept or reject it. When no agents are strategic, the resulting outcome is the firm-optimal stable matching. We show that the worker-optimal stable matching is the unique SPE outcome when only workers are strategic. By contrast, multiple SPE outcomes may exist, possibly including unstable matchings when only firms are strategic. We show that every firm weakly prefers any SPE outcome to the worker-optimal stable matching and that the matching induced by Kesten’s efficiency-adjusted DA algorithm can be achieved as an SPE. When both workers and firms are strategic, we also show that the worker-optimal stable matching is still the unique SPE outcome.

Appendices

Properties of stable matchings

In this section, we list well-known properties that we use in our proofs.

Theorem 7

(Side optimality (Gale and Shapley 1962; Knuth 1976)) Let $I=(F,W,{\succ _F},{\succ _W})$ be a matching market.

• There is the firm-optimal (and worker-pessimal) stable matching $\mu ^{F}$ in the sense that for any stable matching $\mu$, $\mu ^{F}(f) \succeq _{f} \mu (f)$ for all $f \in F$ and $\mu (w) \succeq _{w} \mu ^{F}(w)$ for all $w \in W$.

• Similarly, there is the worker-optimal (and firm-pessimal) stable matching $\mu ^{W}$ in the sense that for any stable matching $\mu$, $\mu ^{W}(w) \succeq _{w} \mu (w)$ for all $w \in W$ and $\mu (f) \succeq _{f} \mu ^{W}(f)$ for all $f \in F$.

Moreover, the firm-oriented (resp., worker-oriented) DA algorithm outputs the firm-optimal (resp., worker-optimal) stable matching for any matching market.

The following result is called the lone wolf theorem: the set of unmatched agents is the same over all stable matchings.¹²

Theorem 8

(Lone wolf theorem (McVitie and Wilson 1970)) For a matching market $I=(F,W,{\succ _F},{}{\succ _W})$, if there exists a stable matching under which $r\in F\cup W$ is matched (resp., unmatched), then r is matched (resp., unmatched) under every stable matching.

The following result is a variant of the lone wolf theorem.

Lemma 8

(Abdulkadiroğlu et al. 2009) For a matching market $I=(F,W,{\succ _F},{\succ _W})$, let $\nu$ be an individually rational matching and $\mu$ be a stable matching such that $\nu (f) \succeq _{f} \mu (f)$ for all $f \in F$. For any $r \in F \cup W$, r is matched in $\mu$ if and only if r is matched in $\nu$.

The following theorem is a well-known incentive property under the firm-oriented DA algorithm: no firm (proposer in the DA algorithm) can be better off by misreporting its preferences.

Theorem 9

(One-sided strategy-proofness (Dubins and Freedman 1981; Roth 1982)) For a matching market $I=(F,W,{\succ _F},{}{\succ _W})$, we have

$$\begin{aligned} {\mathrm {DA}}^{F}[{\succ _F},{\succ _W}](f) \succeq _{f} {\mathrm {DA}}^{F}\left[ ({\succ _f'},{\succ _{-f}}), {\succ _{W}}\right] (f) \end{aligned}$$

for any $f \in F$ and any preference relation $\succ '_{f}$ of f.

Proof of Theorem 1

Here, we prove Theorem 1, i.e., for any workers’ sequential game, the outcome matching of any SPE is the worker-optimal stable matching.

To prove this theorem, we observe some of the properties of SPE. We first claim that any SPE outcome from any node is not worse than the current match for each worker.

Lemma 9

For any workers’ sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, we have $\mu [\sigma ;h](w)\succeq _w \mu [h](w)$ for any $w\in W$ and $h\in {\mathcal {H}}$.

Proof

Let $\sigma _w'\in \varSigma _w$ be a w’s strategy such that $\sigma _w'(h')=\mathtt {R}$ for all $h'\in {\mathcal {W}}_w$. Then, we have $\mu [\sigma ;h](w)\succeq _w \mu [(\sigma '_w,\sigma _{-w});h](w)=\mu [h](w)$ since $\sigma$ is an SPE. $\square$

Similarly, any SPE outcome from any node is not worse than the current offering firm for the offered worker.

Lemma 10

For any workers’ sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, we have $\mu [\sigma ;(h,w)](w)\succeq _w f$ for any $(h,w)\in {\mathcal {W}}$ with $f=\iota (h)$.

Proof

Let $\sigma _w'\in \varSigma _w$ be a w’s strategy such that $\sigma _w'((h,w))=\mathtt {A}$ and $\sigma _w'(h')=\mathtt {R}$ for all $h'\in {\mathcal {W}}_w{\setminus }\{(h,w)\}$. Then we have $\mu [\sigma ;(h,w)](w)\succeq \mu [(\sigma '_{w},\sigma _{-w});(h,w)](w)=f$ since $\sigma$ is an SPE. $\square$

We next prove the following lemma, which states the relation between the original preference and associated preference with a node, as defined in Definition 1.

Lemma 11

For a workers’ sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, we have that for any node $h \in {\mathcal {H}}$ and $w \in W$,

• (i) for all $s \in F \cup \{\emptyset \}$, $s \succeq ^h_{w} \mu [\sigma ; h](w)$ implies $s \succeq _{w} \mu [\sigma ; h](w)$,

• (ii) for all $f \in F$, $f \succeq ^h_{w} \mu [\sigma ; h](w)$ (resp., $f \preceq ^h_{w} \mu [\sigma ; h](w)$) if and only if $f \succeq _{w} \mu [\sigma ; h](w)$ (resp., $f \preceq _{w} \mu [\sigma ; h](w))$.

Proof

When $\mu [h](w) = \emptyset$, this statement clearly holds since $\succ ^h_{w}$ is equivalent to $\succ _{w}$. Suppose that $\mu [h](w) \in F$. From Lemma 1, we have $\mu [\sigma ; h](w) \in F$. Thus, $s \succeq ^h_{w} \mu [\sigma ; h](w)$ implies $s \in F$, and hence $s \succeq _{w} \mu [\sigma ; h](w)$. Further, (ii) holds since $\succ ^h_{w}$ and $\succ _{w}$ are equivalent over F. $\square$

Next, we show that, for any given node, an SPE induces a stable matching in the market corresponding to the node.

Lemma 12

For a workers’ sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, $\mu [\sigma ;h]$ is a stable matching in I[h] for any node $h\in {\mathcal {H}}$.

Proof

Note that, for all $w \in W$, if $\mu [h](w)=\emptyset$, then $\mu [\sigma ; h](w) \succeq ^{h}_{w} \emptyset$ by Lemma 9 and the equivalence of $\succeq ^{h}_{w}$ and $\succeq _{w}$, and if $\mu [h](w) \in F$, then $\mu [\sigma ; h](w) \succeq ^{h}_{w} \emptyset$ since $\succeq ^{h}_{w}$ ranks “$\emptyset$” last. Thus, $\mu [\sigma ; h]$ is individually rational in I[h]. Suppose that $\mu [\sigma ;h]$ is unstable. Then, there exists a blocking pair $(f^*,w^*)\in F\times W$ for $\mu [\sigma ;h]$ in I[h]. Hence, $w^* \succ ^h_{f^*} \mu [\sigma ;h](f^*)$ and $f^* \succ ^h_{w^*} \mu [\sigma ;h](w^*)$. From Lemma 11, $f^* \succ _{w^*} \mu [\sigma ;h](w^*)$. Note that $w^* \succ ^h_{f^*} \mu [\sigma ;h](f^*)$ implies $w^*\in \{w'\mid (f^*,w')\in E[h]\}\cup \{\mu [h](f^*)\}$ and $w^*\succ _{f^*}\mu [\sigma ;h](f^*)$ since firms follow straightforward strategies. If $w^*=\mu [h](f^*)$, then $\mu [\sigma ;h](w^*)\succeq _{w^*}\mu [h](w^*)=f^*$ from Lemma 9, which is a contradiction. If, on the contrary, $w^* \ne \mu [h](f^*)$, there exists a node $h^*=(h,a_1,\dots ,a_i)$ $(i<k)$ such that $\iota (h^*)=(f^*,w^*)$, and $\eta [\sigma ;h]=(h,a_1,\dots ,a_k)$ because $w^*\succ _{f^*} \mu [\sigma ;h](f^*)$. Then, we have $\mu [\sigma ;h](w^*)=\mu [\sigma ;h^*](w^*)\succeq _{w^*} f^*$ from Lemma 10, which is also a contradiction. Thus, the lemma is proved. $\square$

Now, we prove Theorem 1. To use backward induction, we need a statement for every node h.

Lemma 13

For a workers’ sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, we have $\mu [\sigma ;h](w)={\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (w)$ for any $w\in W$ and $h\in {\mathcal {H}}$ such that $\mu [h](w)=\emptyset$.

Proof

We prove this lemma by backward induction.

Suppose that h is a terminal node. Let w be a worker such that $\mu [h](w)=\emptyset$. Note that $\mu [h]$ is stable in I[h] from Lemma 12 and $\mu [\sigma ;h]=\mu [h]$. Since $\mu [h](w)=\emptyset$, we obtain ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h](w)=\emptyset$ from the lone wolf theorem (Theorem 8).

Suppose that h is a workers’ node and that the claim holds for every workers’ or terminal node $h'\supsetneq h$. Let us denote $({\hat{f}},{\hat{w}})=\iota (h)$ and ${\hat{h}}=(h,\sigma _{{\hat{w}}}(h))$. We have $\mu [\sigma ;h]=\mu [\sigma ;{\hat{h}}]$. Note that $\mu [\sigma ;h]({\hat{w}})\succeq _{{\hat{w}}}{\hat{f}}$ from Lemma 10 and ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succeq ^{h}_{{\hat{w}}}\mu [\sigma ;h]({\hat{w}})$ from Lemma 12 and the worker optimality (Theorem 7). Thus, it holds that $\mu [\sigma ;h]({\hat{w}})\succeq ^h_{{\hat{w}}}{\hat{f}}$ and ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})\succeq _{{\hat{w}}}\mu [\sigma ;h]({\hat{w}})$ from Lemma 11. We consider the following two cases.

• (i) Suppose that $\sigma _{{\hat{w}}}(h)=\mathtt {R}$. Then, $\mu [{\hat{h}}]$ and $\succ ^{{\hat{h}}}_W$ are equal to $\mu [h]$ and $\succ ^h_W$, respectively. From $\mu [\sigma ;h]({\hat{w}}) \ne {\hat{f}}$, we have $\mu [\sigma ;h]({\hat{w}})\succ ^h_{{\hat{w}}}{\hat{f}}$. From ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succeq ^{h}_{{\hat{w}}}\mu [\sigma ;h]({\hat{w}})$, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succ ^{h}_{{\hat{w}}}{\hat{f}}$. This implies ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W^{{\hat{h}}}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$ from Lemma 2. Thus, we have $\mu [\sigma ;h](w)=\mu [\sigma ;{\hat{h}}](w)={\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W^{{\hat{h}}}}](w)={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}](w)$ for any $w\in W$ with $\mu [h](w)=\emptyset$ according to the induction hypothesis for ${\hat{h}}$.

• (ii) Suppose that $\sigma _{{\hat{w}}}(h)=\mathtt {A}$. Then, we have $\mu [{\hat{h}}]=\{e\in \mu [h]\mid e_W\ne {\hat{w}}\}\cup \{({\hat{f}},{\hat{w}})\}$. We first claim ${\mathrm {DA}}^W[\succ _F^{{\hat{h}}}, \succ ^{{\hat{h}}}_{W}]={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$. Note that $\mu [\sigma ;h]({\hat{w}})\succ _{{\hat{w}}} \mu [h]({\hat{w}})$ from Lemma 9. This implies $\mu [\sigma ;h]({\hat{w}})\succ ^h_{{\hat{w}}} \mu [h]({\hat{w}})$ in each case of $\mu [h]({\hat{w}})= \emptyset$ and $\mu [h]({\hat{w}}) \in F$ from Lemma 11. Thus, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succeq ^h_{{\hat{w}}}\mu [\sigma ;h]({\hat{w}})\succeq ^h_{{\hat{w}}}{\hat{f}}$ and ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succeq ^h_{{\hat{w}}}\mu [\sigma ;h]({\hat{w}})\succ ^h_{{\hat{w}}} \mu [h]({\hat{w}})$.

  Suppose that $\mu [h]({\hat{w}}) \in F$. The preference profiles $\succ _F^h$ and $\succ _F^{{\hat{h}}}$ are different only for ${\hat{f}}$ and ${\tilde{f}}:=\mu [h]({\hat{w}})$, where $\succ _{{\hat{f}}}^{{\hat{h}}}$ is obtained from $\succ _{{\hat{f}}}^h$ by making the rank of ${\hat{w}}$ first, and $\succ _{{\tilde{f}}}^{{\hat{h}}}$ is obtained from $\succ _{{\tilde{f}}}^h$ by making ${\hat{w}}$ unacceptable. We have already shown that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}}) \succeq ^h_{{\hat{w}}}{\hat{f}}$ and ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}}) \succ ^h_{{\hat{w}}} \mu [h]({\hat{w}})$. This implies that $({\succ _F^h},{\succ _W^h})$ and $({\succ _F^{{\hat{h}}}},{\succ _W^h})$ satisfy the assumption of Lemma 2. Thus, ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ ^{h}_{W}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$ from Lemma 2. In addition, $\succ ^{{\hat{h}}}_{W}$ equals $\succ ^{h}_{W}$ by $\mu [h]({\hat{w}}) \in F$. Hence, ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ ^{{\hat{h}}}_{W}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$. Suppose that $\mu [h]({\hat{w}}) = \emptyset$. The preference profiles $\succ _F^h$ and $\succ _F^{{\hat{h}}}$ are different only for ${\hat{f}}$, where $\succ _{{\hat{f}}}^{{\hat{h}}}$ is obtained from $\succ _{{\hat{f}}}^h$ by making the rank of ${\hat{w}}$ first. We have already shown that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}}) \succeq ^h_{{\hat{w}}}{\hat{f}}$. From Lemma 2, this implies ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}}, {\succ ^{h}_{W}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ ^{h}_{W}}]$. In addition, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}] \succ _{{\hat{w}}} \emptyset$ from ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succeq _{{\hat{w}}}\mu [\sigma ;h]({\hat{w}})$ and $\mu [\sigma ;h]({\hat{w}})\succ _{{\hat{w}}} \mu [h]({\hat{w}}) = \emptyset$. Thus, we get ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W^h}]({\hat{w}}) \succ _{{\hat{w}}} \emptyset$. From Lemma 2, this implies ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}}, {\succ ^{{\hat{h}}}_{W}}]={\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W^h}]$. Hence, ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}}, {\succ ^{{\hat{h}}}_{W}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$ in every case.

  Therefore, we have $\mu [\sigma ;h](w)=\mu [\sigma ;{\hat{h}}](w)={\mathrm {DA}}^W[\succ _F^{{\hat{h}}},\succ _W^{{\hat{h}}}](w)={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}](w)$ for any $w\in W{\setminus }\{{\hat{w}}\}$ with $\mu [h] (w)=\emptyset$.

  It remains to prove that $\mu [\sigma ;h]({\hat{w}})={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})$ for the case in which $\mu [h]({\hat{w}})=\emptyset$. If $({\hat{f}},{\hat{w}})\in {\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$, then we have $\mu [\sigma ;h]({\hat{w}})={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})$ because $\mu [\sigma ;h]({\hat{w}})\succeq _{{\hat{w}}}{\hat{f}}={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})$ from Lemma 10 and $\mu [\sigma ;h]({\hat{w}})\preceq _{{\hat{w}}}{\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})$. Hence, suppose that $({\hat{f}},{\hat{w}})\not \in {\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$. Then, ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})\succ _{{\hat{w}}}{\hat{f}}$ from ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succeq _{{\hat{w}}}{\hat{f}}$. The preference profile $(\succ _F^{(h,\mathtt {R})},\succ _W^{(h,\mathtt {R})})$ is obtained from $(\succ _F^h,\succ _W^h)$ by changing only the preference ordering of ${\hat{f}}$ so that ${\hat{f}}$ makes ${\hat{w}}$ unacceptable. Thus, ${\mathrm {DA}}^W[\succ _F^{(h,\mathtt {R})},\succ _W^{(h,\mathtt {R})}]={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$ from Lemma 2. Note that $\mu [\sigma ;(h,\mathtt {R})]({\hat{w}})={\mathrm {DA}}^W[\succ _F^{(h,\mathtt {R})},\succ _W^{(h,\mathtt {R})}]({\hat{w}})$ according to the induction hypothesis. Thus, we have $\mu [\sigma ;h]({\hat{w}})=\mu [\sigma ;({\hat{h}},\mathtt {A})]({\hat{w}})\succeq _{{\hat{w}}}\mu [\sigma ;({\hat{h}},\mathtt {R})]({\hat{w}})={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})$. Combining this with $\mu [\sigma ;h]({\hat{w}})\preceq _{{\hat{w}}}{\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})$, we obtain $\mu [\sigma ;h]({\hat{w}})={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})$.

Thus, the lemma holds. $\square$

Therefore, Theorem 1 holds because $\mu [h_{\mathrm {init}}](w)=\emptyset$ for all $w\in W$.

Examples of firms’ sequential game

From Example 2, one may expect that any SPE outcome is the firm-optimal stable matching (or a matching that every firm weakly prefers to it). However, this turns out to be incorrect. The following example shows the case in which the worker-optimal (firm-pessimal) stable matching is an SPE outcome when there are multiple stable matchings.

Example 6

Consider a firms’ sequential game G with three firms $F=\{f_1,f_2,f_3\}$ and three workers $W=\{w_1,w_2,w_3\}$. The preferences $\succ _F$ and $\succ _W$ are given as

$$\begin{aligned}&f_1:~ w_1 \succ _{f_1} \emptyset \succ _{f_1} w_2 \succ _{f_1} w_3&w_1:~ f_1 \succ _{w_1} \emptyset \succ _{w_1} f_2 \succ _{w_1} f_3 \\&f_2:~ w_2 \succ _{f_2} w_3 \succ _{f_2} \emptyset \succ _{f_2} w_1&w_2:~ f_3 \succ _{w_2} f_1 \succ _{w_2} f_2 \succ _{w_2} \emptyset \\&f_3:~ w_3 \succ _{f_3} w_2 \succ _{f_3} \emptyset \succ _{f_3} w_1&w_3:~ f_2 \succ _{w_3} f_1 \succ _{w_3} f_3 \succ _{w_3} \emptyset . \end{aligned}$$

Suppose that for each of the firms’ nodes, the offer is made by the active firm with the smallest index.

It is straightforward to see that there are two SPE outcomes in this game. One is the firm-optimal stable matching $\{(f_1,w_1),(f_2,w_2),(f_3,w_3)\}$, which is attained by the profile of straightforward strategies. The other is the worker-optimal stable matching $\{(f_1,w_1),(f_2,w_3),{}(f_3,w_2)\}$, which is attained by a strategy profile $(\sigma _{f_1}, \sigma _{f_2}, \sigma _{f_3})$ where $\sigma _{f_2}$ and $\sigma _{f_3}$ are straightforward strategies, and $\sigma _{f_1}$ is the strategy such that $f_1$ follows the straightforward strategy except for that $f_1$ first makes an offer to $w_2$ (or $w_3$).

Next, we give an example of firms’ sequential game such that the set of unmatched agents is not unique over NE matchings.

Example 7

Let us consider a firms’ sequential game G with three firms $F=\{f_1,f_2,f_3\}$ and two workers $W=\{w_1,w_2\}$. The preferences $\succ _F$ and $\succ _W$ are given as

$$\begin{aligned}&f_1:~ w_1 \succ _{f_1} \emptyset \succ _{f_1} w_2&w_1:~ f_2 \succ _{w_1} f_3 \succ _{w_1} f_1 \succ _{w_1} \emptyset \\&f_2:~ w_2 \succ _{f_2} w_1 \succ _{f_2} \emptyset&w_2:~ f_1 \succ _{w_2} f_2 \succ _{w_2} \emptyset \succ _{w_2} f_3\\&f_3:~ w_1 \succ _{f_3} \emptyset \succ _{f_3} w_2.&\end{aligned}$$

Suppose that for each of the firms’ nodes, the offer is made by the active firm with the smallest index.

In the market, $\{(f_2,w_2),(f_3,w_1)\}$ is the worker-optimal stable matching, while $\mu ^*:=\{(f_1,w_1),(f_2,w_2)\}$ is the outcome of a NE. To see this, let us consider straightforward strategies $\sigma _F'=(\sigma _f)_{f\in F}$ according to $\succ _F'=(\succ _f)_{f\in F}$, where $w_1 \succ _{f_1}' w_2 \succ _{f_1}' \emptyset$, $w_2 \succ _{f_2}' w_1 \succ _{f_2}' \emptyset$, and $\emptyset \succ _{f_3}' w_1 \succ _{f_3}' w_2$. Then, it is easy to check that $\mu [\sigma _F']=\mu ^*$. In addition, $\sigma _F'$ is a NE because $f_1$ and $f_2$ are matched to their most preferred workers and $f_3$ cannot be matched to $w_1$ by changing its strategy. It should be noted that $\sigma _F'$ is not an SPE. The unique SPE outcome is the worker-optimal stable matching.

Finally, we give an example of firms’ sequential game, which has multiple Pareto efficient matchings for firms.

Example 8

Consider a market with four firms $F=\{f_1,f_2,f_3,f_4\}$ and three workers $W=\{w_1,w_2,w_3\}$. The preferences $\succ _F$ and $\succ _W$ are given as

$$\begin{aligned}&f_1:~ w_2 \succ _{f_1} w_3 \succ _{f_1} w_{1} \succ _{f_1} \emptyset&w_1:~ f_1 \succ _{w_1} f_3 \succ _{w_1} f_2 \succ _{w_1} \emptyset \succ _{w_1} f_4\\&f_2:~ w_1 \succ _{f_2} w_2 \succ _{f_2} \emptyset \succ _{f_2} w_3&w_2:~ f_2 \succ _{w_2} f_1 \succ _{w_2} \emptyset \succ _{w_2} f_3 \succ _{w_2} f_4\\&f_3:~ w_1 \succ _{f_3} w_3 \succ _{f_3} \emptyset \succ _{f_3} w_2&w_3:~ f_3 \succ _{w_3} f_4 \succ _{w_3} f_1 \succ _{w_3} \emptyset \succ _{w_3} f_2\\&f_4:~ w_3 \succ _{f_4} \emptyset \succ _{f_4} w_1 \succ _{f_4} w_2.&\end{aligned}$$

Suppose that for each of the firms’ nodes, the offer is made by the active firm with the smallest index.

This firms’ sequential game has the following three SPE outcomes:

$$\begin{aligned} \mu _1&=\{(f_1,w_1),(f_2,w_2),(f_3,w_3) \},\\ \mu _2&=\{(f_1,w_3),(f_2,w_2),(f_3,w_1) \},\\ \mu _3&=\{(f_1,w_2),(f_2,w_1),(f_3,w_3) \}. \end{aligned}$$

Here, $\mu _1$ is the unique stable matching and $\mu _2$ is the EADA matching. Moreover, $\mu _3$ is Pareto efficient for firms, which is achieved by a strategy profile $({\tilde{\sigma }}_{f})_{f \in F}$, where each firm follows the straightforward strategy except for that $f_3$ first makes an offer to $w_3$ and $f_4$ always exits the market.

Proof of Lemma 4

We first claim that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}] = {\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$ for any $h \in {\mathcal {H}}(= {\mathcal {F}} \cup {\mathcal {T}})$. Let $h \in {\mathcal {H}}$. Note that ${\mathrm {DA}}^{F}[{\succ ^h_{F}}, {\succ _{W}}](w) \succeq _{w} \mu [h](w) \succeq _{w} \emptyset$ for all $w \in W$ from the definition of the game and the assumption that all workers follow straightforward strategies. This implies that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}](w) \succ _{w} \emptyset$ for all $w \in W$ with $\mu [h](w) \in F$. Moreover, $\succ ^h_{w}$ equals $\succ _{w}$ for all $w \in W$ with $\mu [h](w) = \emptyset$. Thus, from Lemma 2, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}] = {\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$.

For any SPE $\sigma$ and any worker $w \in W$, $\mu [\sigma ;h](w)\preceq _w^h{\mathrm {DA}}^W[\succ _F^h,\succ _W^h](w)$ if and only if $\mu [\sigma ;h](w)\preceq _w {\mathrm {DA}}^W[\succ _F^h,\succ _W^h](w)$, where the “only if” part holds from Lemma 11 and the “if” part follows from $\mu [\sigma ;h](w) \succeq _{w} \emptyset$. Thus, it is sufficient to show that for any SPE $\sigma$ and $h \in {\mathcal {H}}$,

$$\begin{aligned} \begin{array}{ll} \mu [\sigma ;h](f)\succeq _f^h{\mathrm {DA}}^W\left[ \succ _F^h,\succ _W\right] (f) ~~(\forall f\in F)&{}\text {and}\\ \mu [\sigma ;h](w)\preceq _w {\mathrm {DA}}^W\left[ \succ _F^h,\succ _W\right] (w) ~~(\forall w\in W).&{} \end{array} \end{aligned}$$

(1)

We prove this statement by backward induction. Fix an SPE $\sigma$ arbitrary. Throughout the proof, I[h] denotes $(F, W, \succ ^h_{F}, \succ _{W})$ instead of $(F, W, \succ ^h_{F}, \succ ^h_{W})$ for all $h \in {\mathcal {H}}$.

For each terminal node $h \in {\mathcal {T}}$, $\succ ^h_{f}$ ranks $\mu [\sigma ; h](f) ~(= \mu [h](f))$ first for all $f \in F$ and thus $\mu [\sigma ; h]$ is the firm-optimal stable matching in I[h]. Therefore, (1) holds for any terminal node.

For a non-terminal (firms’) node $h\in {\mathcal {F}}$, suppose that (1) holds for every node $h'\supsetneq h$. Let us denote ${\hat{f}}:=\iota (h)$, ${\hat{w}}:={\mathrm {DA}}^W[\succ _F^h,\succ _W]({\hat{f}})~(\in \{w'\mid (f,w')\in E[h]\}\cup \{\emptyset \})$, ${\hat{s}}:=\sigma _{{\hat{f}}}(h)$, and ${\hat{h}}:=(h,{\hat{s}})$. We remark that $\succ _{{\hat{f}}}$ and $\succ ^h_{{\hat{f}}}$ are equivalent over $\{w'\mid ({\hat{f}},w')\in E[h]\}\cup \{\emptyset \}$ since ${\hat{f}}$ is unmatched at h (i.e., $\mu [h]({\hat{f}})=\emptyset$).

We first prove that $\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})={\hat{w}}$. Note that this implies

$$\begin{aligned} \mu [\sigma ;h]({\hat{f}})=\mu [\sigma ;{\hat{h}}]({\hat{f}})\succeq _{{\hat{f}}}\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})={\hat{w}} \end{aligned}$$

(2)

since $\sigma$ is an SPE. When ${\hat{w}} = \emptyset$, the claim clearly holds. Thus, we assume that ${\hat{w}}\in W$. To show the claim, we observe the difference between $\succ _F^{h}$ and $\succ _F^{(h,{\hat{w}})}$. For ${\hat{f}}$, the preference $\succ _{{\hat{f}}}^{(h,{\hat{w}})}$ can be obtained from $\succ _{{\hat{f}}}^{h}$ by changing the rank of ${\hat{w}}$ to the top rank because ${\hat{w}}$ accepts the offer from ${\hat{f}}$ (i.e., $\mu [(h,{\hat{w}})]({\hat{f}})={\hat{w}}$) by ${\hat{f}}={\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{w}})\succ _{{\hat{w}}}\mu [h]({\hat{w}})$ (since otherwise $(\mu [h]({\hat{w}}),{\hat{w}})$ is a blocking pair for ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$ in I[h]). Further, when ${\tilde{f}} :=\mu [h]({\hat{w}}) \in F$, the preference $\succ _{{\tilde{f}}}^{(h,{\hat{w}})}$ can be obtained from $\succ _{{\tilde{f}}}^{h}$ by changing ${\hat{w}}$ from the top rank to unacceptable. Here, we remark that ${\tilde{f}}=\mu [h]({\hat{w}})\prec _{{\hat{w}}} {\mathrm {DA}}^W[{\succ ^{h}_F},{\succ _W}]({\hat{w}})$ since otherwise $(\mu [h]({\hat{w}}),{\hat{w}})$ is a blocking pair for ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$ in I[h]. Hence, ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}})}},{\succ _{W}}]({\hat{f}})={\mathrm {DA}}^W[{\succ _F^h},{\succ _{W}}]({\hat{f}})$ from Lemma 2 and the definitions of $\succ _F^h$ and $\succ _F^{(h,{\hat{w}})}$. Thus, we obtain

$$\begin{aligned} \mu [\sigma ;(h,{\hat{w}})]({\hat{f}})\succeq _{{\hat{f}}}^{(h,{\hat{w}})} {\mathrm {DA}}^W\left[ \succ _F^{(h,{\hat{w}})},\succ _{W}\right] ({\hat{f}}) ={\mathrm {DA}}^W\left[ \succ _F^{h},\succ _{W}\right] ({\hat{f}})={\hat{w}}, \end{aligned}$$

where the first relation holds by the inductive hypothesis. As ${\hat{w}}$ is the most preferred worker in $\succ _{{\hat{f}}}^{(h,{\hat{w}})}$, we have $\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})={\hat{w}}$.

Second, we show that, when ${\hat{s}} \in W$,

$$\begin{aligned} {\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W}\right] ({\hat{s}})\succeq _{{\hat{s}}}{\hat{f}}. \end{aligned}$$

(3)

To obtain a contradiction, suppose that ${\hat{f}}\succ _{{\hat{s}}} {\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{s}})$. In this case, ${\hat{s}}$ accepts the offer from ${\hat{f}}$ (i.e., ${\hat{f}}\succ _{{\hat{s}}}\mu [h]({\hat{s}})$) since otherwise $(\mu [h]({\hat{s}}),{\hat{s}})$ is a blocking pair for ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$ in I[h] (recall that ${\hat{s}}$ is ranked top in $\succ _{\mu [h]({\hat{s}})}$ if $\mu [h]({\hat{s}})\in F$). Thus, in the change from $\succ _F^h$ to $\succ _F^{{\hat{h}}}$, ${\hat{f}}$ makes the rank of ${\hat{s}}$ first and $\mu [h]({\hat{s}})$ makes ${\hat{s}}$ unacceptable (if $\mu [h]({\hat{s}}) \in F$). Moreover, we can see that ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W}]({\hat{s}})={\hat{f}}$ by considering the worker-oriented DA procedures for I[h] and $I[{\hat{h}}]$ as follows. Note that the procedures are the same until ${\hat{s}}$ makes an offer to ${\hat{f}}$ or $\mu [h]({\hat{s}})$ by the difference of I[h] and $I[{\hat{h}}]$. As ${\hat{f}}\succ _{{\hat{s}}}\mu [h]({\hat{s}})$ (i.e., ${\hat{s}}$ makes an offer to ${\hat{f}}$ before to $\mu [h]({\hat{s}})$) and ${\hat{f}}\succ _{{\hat{s}}}{\mathrm {DA}}^W[\succ _F^{h},\succ _{W}]({\hat{s}})$ (i.e., ${\hat{s}}$ must make an offer to ${\hat{f}}$ in the procedure for I[h]), ${\hat{s}}$ makes an offer to ${\hat{f}}$ in the procedure for $I[{\hat{h}}]$. Then, ${\hat{f}}$ accepts and never rejects the offer from ${\hat{s}}$ since ${\hat{s}}$ is ranked top in $\succ _{{\hat{f}}}^{{\hat{h}}}$, and hence ${\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W}]({\hat{s}})={\hat{f}}$. Now, by the induction hypothesis, we have $\mu [\sigma ;{\hat{h}}]({\hat{f}})\succeq _{{\hat{f}}}^{{\hat{h}}}{\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _{W}}]({\hat{f}})={\hat{s}}$. This means $\mu [\sigma ;{\hat{h}}]({\hat{f}})={\hat{s}}$ since $\succ ^{{\hat{h}}}_{{\hat{f}}}$ ranks ${\hat{s}}$ first. Thus, we obtain ${\hat{s}}=\mu [\sigma ;{\hat{h}}]({\hat{f}})\succeq _{{\hat{f}}}{\hat{w}}$ by (2), which implies that ${\hat{s}}={\hat{w}}$ or $({\hat{f}},{\hat{s}})$ is a blocking pair for ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$ in I[h]. The former case contradicts the assumption ${\hat{f}}\succ _{{\hat{s}}} {\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{s}})$ and the latter case contradicts the stability of ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$.

Next, we prove that

$$\begin{aligned} {\mathrm {DA}}^W\left[ \succ _F^h,\succ _W\right] ={\mathrm {DA}}^W\left[ \succ _F^{{\hat{h}}},\succ _{W}\right] . \end{aligned}$$

(4)

We have three cases to consider:

• (a) Suppose that ${\hat{s}} = \emptyset$. In this case, ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{f}}) = \emptyset$ from

  $$\begin{aligned} \emptyset = \mu [\sigma ; h]({\hat{f}}) \succeq _{{\hat{f}}} {\hat{w}} = {\mathrm {DA}}^W\left[ {\succ _F^{h}},{\succ _{W}}\right] ({\hat{f}}). \end{aligned}$$

  Note that f makes the rank of $\emptyset$ first in the change from $\succ _F^h$ to $\succ _F^{{\hat{h}}}$. This implies that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]={\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _{W}}]$ from Lemma 2 and the definitions of $\succ _F^h$ and $\succ _F^{{\hat{h}}}$.

• (b) Suppose that ${\hat{s}} \in W$ and ${\hat{f}}\succ _{{\hat{s}}}\mu [h]({\hat{s}})$. We here assume that $\mu [h]({\hat{s}}) \in F$ since the following argument is valid in the remaining case of $\mu [h]({\hat{s}}) =\emptyset$. By ${\hat{f}}\succ _{{\hat{s}}}\mu [h]({\hat{s}})$, ${\hat{s}}$ accepts the offer from ${\hat{f}}$. In this case, $E[{\hat{h}}]=E[h]{\setminus }\{({\hat{f}},{\hat{s}})\}$ and $\mu [{\hat{h}}]=\mu [h]{\setminus }\{(\mu [h]({\hat{s}}),{\hat{s}})\}\cup \{({\hat{f}},{\hat{s}})\}$. In the change from $\succ _F^h$ to $\succ _F^{{\hat{h}}}$, ${\hat{f}}$ makes the rank of ${\hat{s}}$ first and $\mu [h]({\hat{s}})$ makes ${\hat{s}}$ unacceptable. Note that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{s}})\succeq _{{\hat{s}}} {\hat{f}}\succ _{{\hat{s}}}\mu [h]({\hat{s}})$ holds by (3). This implies that ${\mathrm {DA}}^W[\succ _F^h,\succ _W]={\mathrm {DA}}^W[\succ _F^{{\hat{h}}},\succ _{W}]$ from Lemma 2 and the definitions of $\succ _F^h$ and $\succ _F^{{\hat{h}}}$.

• (c) Suppose that ${\hat{s}} \in W$ and $\mu [h]({\hat{s}}) \succ _{{\hat{s}}} {\hat{f}}$. Then, ${\hat{s}}$ rejects the offer from ${\hat{f}}$. In this case, $E[{\hat{h}}]=E[h]{\setminus }\{({\hat{f}},{\hat{s}})\}$ and $\mu [{\hat{h}}]=\mu [h]$. When $\mu [h]({\hat{s}}) = \emptyset$, ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{s}}) \succeq _{{\hat{s}}} \mu [h]({\hat{s}})$ holds by individually rationality, and also when $\mu [h]({\hat{s}})\in F$, ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{s}}) \succeq _{{\hat{s}}} \mu [h]({\hat{s}})$ still holds since otherwise $( \mu [h]({\hat{s}}), {\hat{s}})$ is a blocking pair for ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]$ in I[h] (recall that ${\hat{s}}$ is ranked top in $\succ _{\mu [h]({\hat{s}})}^{h}$). Therefore, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]({\hat{s}}) \succeq _{{\hat{s}}} \mu [h]({\hat{s}}) \succ _{{\hat{s}}} {\hat{f}}$. In the change from $\succ _F^h$ to $\succ _F^{{\hat{h}}}$, ${\hat{f}}$ makes ${\hat{s}}$ unacceptable. This implies that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W}]={\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _{W}}]$ from Lemma 2 and the definitions of $\succ _F^h$ and $\succ _F^{{\hat{h}}}$.

Now, we are ready to prove (1). For each $w\in W$, we have

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\hat{h}}](w)\preceq _w {\mathrm {DA}}^W\left[ \succ _F^{{\hat{h}}},\succ _{W}\right] (w)={\mathrm {DA}}^W\left[ \succ _F^h,\succ _{W}\right] (w) \end{aligned}$$

according to the induction hypothesis and (4). Additionally, for each $f\in F{\setminus }\{{\hat{f}}\}$, we have

$$\begin{aligned} \mu [\sigma ;h](f)=\mu [\sigma ;{\hat{h}}](f) \succeq _f^{{\hat{h}}}{\mathrm {DA}}^W\left[ {\succ _F^{{\hat{h}}}},{\succ _{W}}\right] (f)={\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _{W}}\right] (f) \end{aligned}$$

according to the induction hypothesis and (4). Thus, for each $f\in F{\setminus }\{{\hat{f}}\}$, we have

$$\begin{aligned} \mu [\sigma ;h](f)=\mu [\sigma ;{\hat{h}}](f) \succeq _f^{h}{\mathrm {DA}}^W\left[ {\succ _F^{{\hat{h}}}},{\succ _{W}}\right] (f)={\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _{W}}\right] (f), \end{aligned}$$

by the definitions of $\succ _f^h$ and $\succ _f^{{\hat{h}}}$ (i.e., the difference is only the rank of ${\hat{s}}$ for $f=\mu [h]({\hat{s}})$). Finally, we have $\mu [\sigma ;h]({\hat{f}})\succeq _{{\hat{f}}}^h{\mathrm {DA}}^W[{\succ _F^h},{\succ _{W}}]({\hat{f}})$ since $\mu [\sigma ;h]({\hat{f}})\succeq _{{\hat{f}}}{\mathrm {DA}}^W[{\succ _F^h},{\succ _{W}}]({\hat{f}})$ by (2) and $\mu [\sigma ;h]({\hat{f}})\in \{w'\mid ({\hat{f}},w')\in E[h]\}\cup \{\emptyset \}$.

Proof of Lemma 7

Lemma 7

For a matching market $I=(F,W,\succ _F,\succ _W)$, let ${\hat{f}} \in F$ and $\mathrel {{\hat{\succ }}}_{{\hat{f}}}$ be the preference ordering obtained from $\succ _{{\hat{f}}}$ by making ${\mathrm {EADA}}[\succ _{F}]({\hat{f}})$ rank first without changing the others; that is, ${\mathrm {EADA}}[\succ _{F}]({\hat{f}}) \mathrel {{\hat{\succeq }}}_{{\hat{f}}} a$ for all $a \in W \cup \{ \emptyset \}$ and $s \mathrel {{\hat{\succeq }}}_{f} t$ if and only if $s \succeq _{f} t$ for any $s, t \in (W \cup \{\emptyset \}) {\setminus } \{ {\mathrm {EADA}}[\succ _{F}]({\hat{f}})\}$. Then, we have that ${\mathrm {EADA}}[\succ _{F}] = {\mathrm {EADA}}[({\mathrel {{\hat{\succ }}}_{{\hat{f}}}}, {\succ _{-{\hat{f}}}})]$.

We begin by introducing three lemmas. The result of Gale and Sotomayor (Gale and Sotomayor 1985b, Theorem 2) implies the following lemma.

Lemma 14

(Gale and Sotomayor 1985b) For a matching market $I=(F,W,{\succ _F},{\succ _W})$ and any nonempty $F' \subseteq F$, we have that ${\mathrm {DA}}^F[\succ _{F'}] (f) \succeq _{f} {\mathrm {DA}}^F[\succ _{F}](f)$ for all $f \in F'$.

The next two lemmas provide some of the basic properties of the last proposers.

Lemma 15

(Bando 2014) For a matching market $I=(F,W,{\succ _F},{\succ _W})$, let L be the set of last proposers. For any $f \in F$ and $w \in W$ such that $(f', w) \in {\mathrm {DA}}^F[\succ _{F}]$ for some $f' \in L$ and $w \succ _{f} {\mathrm {DA}}^F[\succ _{F}](f)$, we have $\emptyset \succ _w f$.

Lemma 16

For a matching market $I=(F,W,\succ _F,\succ _W)$, let L be the set of last proposers. For any individually rational matching $\nu$, if $\nu (f) \succeq _{f} {\mathrm {DA}}^F[\succ _{F}](f)$ for all $f\in F$, then $\nu (f') = {\mathrm {DA}}^F[\succ _{F}](f')$ for all $f' \in L$.¹³

Proof

Suppose to the contrary that there exists $f' \in L$ with $\nu (f') \succ _{f'} {\mathrm {DA}}^F[\succ _{F}](f')$. From Lemma 8, $w' :={\mathrm {DA}}^F[\succ _{F}](f') \in W$. Again, from Lemma 8, $f'' :=\nu (w') \in F$. Thus, we have $w'={\mathrm {DA}}^F[\succ _{F}](f') \succ _{f''} {\mathrm {DA}}^F[\succ _{F}](f'')$. From Lemma 15, $\emptyset \succ _{ w' } f'' = \nu (w')$, contradicting the individual rationality of $\nu$. $\square$

We turn to the proof of Lemma 7. Let ${\hat{f}} \in F$ and $\mathrel {{\hat{\succ }}}_{{\hat{f}}}$ be the preference ordering defined in Lemma 7. We denote $(\mathrel {{\hat{\succ }}}_{{\hat{f}} }, \succ _{-{\hat{f}} })$ by $\mathrel {{\hat{\succ }}}_{F}$. We also denote ${\mathrm {EADA}}[\succ _{F}]$ and ${\mathrm {EADA}}[\mathrel {{\hat{\succ }}}_{F}]$ by $\nu$ and $\nu '$, respectively. We show that $\nu = \nu '$. From the definition of $\mathrel {{\hat{\succ }}}_{{\hat{f}}}$, it holds that for any $s, t \in W \cup \{\emptyset \}$ with $\nu ({\hat{f}}) \succeq _{{\hat{f}}} s$, if $t \mathrel {{\hat{\succeq }}}_{{\hat{f}}} s$, then $t \succeq _{{\hat{f}}} s$.

We assume that the EADA algorithm for $(F, W, \succ _{F}, \succ _{W})$ terminates at step $k^*$. For each step $k=1, \dots , k^*$, let $F_k$ be the remaining firms and $L_k$ be the set of last proposers for $(F_k, W, \succ _{F_k}, \succ _{W})$. Note that $L_k$ is removed from $F_k$ for each step and $F=L_1\cup \cdots \cup L_{k^*}$. To obtain $\nu = \nu '$, we show that $\nu (f) = \nu '(f)$ for all $f \in F$ by contradiction. Suppose that $\nu (f) \ne \nu '(f)$ for some $f \in F$. Let $\ell \in \{1, \dots , k^*\}$ be the minimum index such that there exists $f \in L_{\ell }$ with $\nu (f) \ne \nu '(f)$.

We first prove that

$$\begin{aligned} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }} ]({\hat{f}})&= {\mathrm {DA}}^F[\succ _{F_{\ell }}]({\hat{f}}) \text { when }{\hat{f}} \in F_{\ell },\text { and} \end{aligned}$$

(5)

$$\begin{aligned} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }} ](f)&\succeq _{f} {\mathrm {DA}}^F[ \succ _{F_{\ell }}](f) \text { for all }f \in F_{\ell } {\setminus } \{ {\hat{f}} \}. \end{aligned}$$

(6)

Suppose that ${\hat{f}} \notin F_{\ell }$. Then, ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }} ] = {\mathrm {DA}}^F[\succ _{F_{\ell }}]$ by $\mathrel {{\hat{\succ }}}_{F_{\ell }} = \succ _{F_{\ell }}$. Thus, (5) and (6) hold. Hence, we assume that ${\hat{f}} \in F_{\ell }$. From Theorem 9, we have ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }} ]({\hat{f}}) \mathrel {{\hat{\succeq }}}_{{\hat{f}}} {\mathrm {DA}}^F[ \succ _{F_{\ell }} ]({\hat{f}})$. This implies ${\mathrm {DA}}^F[{\mathrel {{\hat{\succ }}}_{F_{\ell }}}]({\hat{f}}) \succeq _{{\hat{f}}} {\mathrm {DA}}^F[ {\succ _{F_{\ell }}} ]({\hat{f}})$ if ${\mathrm {DA}}^F[{\mathrel {{\hat{\succ }}}_{F_{\ell }}}]({\hat{f}})\ne \nu ({\hat{f}})$. In addition, as $\nu ({\hat{f}}) \succeq _{{\hat{f}}} {\mathrm {DA}}^F[{\succ _{F_{\ell }}}] ({\hat{f}})$ holds from Lemma 6, we have ${\mathrm {DA}}^F[{\mathrel {{\hat{\succ }}}_{F_{\ell }}}]({\hat{f}}) \succeq _{{\hat{f}}} {\mathrm {DA}}^F[ {\succ _{F_{\ell }}} ]({\hat{f}})$ in any case. Again, from Theorem 9, we have ${\mathrm {DA}}^F[ {\succ _{F_{\ell }}} ]({\hat{f}}) \succeq _{{\hat{f}}} {\mathrm {DA}}^F[ {\mathrel {{\hat{\succ }}}_{F_{\ell }}} ]({\hat{f}})$. Hence, we obtain (5). It is straightforward to see that ${\mathrm {DA}}^F[ \succ _{F_{\ell }}]$ is also a stable matching for $(F_{\ell }, W, \mathrel {{\hat{\succ }}}_{F_{\ell }}, \succ _{W})$. Therefore, we have ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }}](f) \mathrel {{\hat{\succeq }}}_{f} {\mathrm {DA}}^F[\succ _{F_{\ell }}](f)$ for all $f \in F_{\ell } {\setminus } \{{\hat{f}}\}$ by Theorem 7, and hence we obtain (6).

We next prove that

$$\begin{aligned} \nu '(f) ~(={\mathrm {EADA}}[\mathrel {{\hat{\succ }}}_{F}](f)) \mathrel {{\hat{\succeq }}}_{f} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell } }](f) \text { for all } f \in F_{\ell }. \end{aligned}$$

(7)

To obtain a contradiction, suppose that there exists $j \in F_{\ell }$ such that ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell } }](j) \mathrel {{\hat{\succ }}}_{j} \nu '(j)$. Let us consider the EADA algorithm for $(F, W, \mathrel {{\hat{\succ }}}_{F}, \succ _{W})$. Let p be the step at which j is removed and $F'_{p}$ be the set of the remaining firms at the beginning of step p. Note that j is included in the set of last proposers for $(F'_{p}, W, \mathrel {{\hat{\succ }}}_{F'_p}, \succ _{W})$. We define $F^* = F_{\ell } \cap F'_{p}$ which is nonempty by $j \in F^*$.

Let us consider

$$\begin{aligned} \nu '' :=\{ (f, w)\in {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F^*}]\mid f \in F^*\} \cup \{ (f, w)\in \nu ' \mid f \in F'_{p} {\setminus } F^*\}. \end{aligned}$$

There are two cases: $\nu ''$ is a matching (between $F'_{p}$ and W) and is not a matching.

Suppose that $\nu ''$ is not a matching. Then, there exist $w^* \in W$, $f_1 \in F^*$, and $f_2 \in F'_{p} {\setminus } F^*$ such that ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F^*}](f_1) = \nu '(f_2) = w^*$. From $f_1 \in F^* \subseteq F_{\ell }$ and Lemma 14, we have $w^*={\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F^*}](f_1) \mathrel {{\hat{\succeq }}}_{f_1} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }}](f_1)$. When $f_1 \ne {\hat{f}}$, we have $w^* \succeq _{f_1} {\mathrm {DA}}^F[\succ _{F_{\ell }}](f_1)$ since ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }}](f_1) \succeq _{f_1} {\mathrm {DA}}^F[\succ _{F_{\ell }}](f_1)$ holds by (6). Further, suppose that $f_1 ={\hat{f}}$. Then, $\nu ({\hat{f}}) \succeq _{{\hat{f}}} {\mathrm {DA}}^F[\succ _{F_{\ell } }]({\hat{f}}) = {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell } }]({\hat{f}})$ by Lemma 6 and (5). Together with $w^* \mathrel {{\hat{\succeq }}}_{{\hat{f}}} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell }}](f_1)$, we have $w^* \succeq _{{\hat{f}}} {\mathrm {DA}}^F[\succ _{F_{\ell }}](f_1)$ from the definition of ${\hat{\succeq }}_{{\hat{f}}}$. Therefore, $w^* \succeq _{f_1} {\mathrm {DA}}^F[\succ _{F_{\ell }}](f_1)$ holds in every case. Owing to $f_2 \in F'_{p} {\setminus } F^*~(=F'_{p}{\setminus } F_{\ell })$, we have $f_2 \in L_1 \cup \cdots \cup L_{\ell -1}$. Because of the minimality of $\ell$, we have $(w^*=)~\nu '(f_2) = \nu (f_2)$. By $\nu (f_2)=w^*$, $f_2$ is matched with $w^*$ before step $\ell -1$ at the EADA algorithm for $(F, W, \succ _{F}, \succ _{W})$. Thus, ${\mathrm {DA}}^F[\succ _{F_{\ell }}](f_1)\ne w^*$ since $f_1$ is remained at step $\ell$ by $f_1 \in F^* \subseteq F_\ell$. This implies that $w^* \succ _{f_1} {\mathrm {DA}}^F[\succ _{F_{\ell }}](f_1)$ since $w^* \succeq _{f_1} {\mathrm {DA}}^F[\succ _{F_{\ell }}](f_1)$. From Lemmas 6 and 15, we have $\emptyset \succ _{w^*} f_1$ which contradicts ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F^*}](f_1) = w^*$. Hence, (7) holds in this case.

Suppose that $\nu ''$ is a matching between $F'_{p}$ and W. By the definition of $\nu ''$, $\nu ''(f) = {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F^*}](f)$ for all $f \in F^*$, and $\nu ''(f) = \nu '(f)= {\mathrm {EADA}}[\mathrel {{\hat{\succ }}}_{F}](f)$ for all $f \in F'_p {\setminus } F^*$. Thus, $\nu ''(f) \mathrel {{\hat{\succeq }}}_{f} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F'_p}](f)$ for all $f \in F^*$ by Lemma 14 and $F^* \subseteq F'_p$. Recall that $F'_{p}$ is the set of the remaining firms at step p in the EADA algorithm for $(F, W, \mathrel {{\hat{\succ }}}_{F}, \succ _{W})$. By Lemma 6, $\nu ''(f) = \nu '(f) \mathrel {{\hat{\succeq }}}_{f} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F'_p}](f)$ for all $f \in F'_{p}{\setminus } F^*$. Therefore, $\nu ''$ is an individually rational matching for $(F'_{p}, W, \mathrel {{\hat{\succ }}}_{F'_{p}}, \succ _{W})$ such that $\nu ''(f) \mathrel {{\hat{\succeq }}}_{f} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F'_p}](f)$ for all $f \in F'_{p}$. Since j is a last proposer for $(F'_{p}, W, \mathrel {{\hat{\succ }}}_{F'_{p}}, \succ _{W})$, we have $\nu ''(j) = {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F'_p}](j)$ from Lemma 16. Recall that $j \in F^*$ and ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell } }](j) \mathrel {{\hat{\succ }}}_{j} \nu '(j) ={\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F'_p}](j)$ are assumed. Thus, we have ${\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F_{\ell } }](j) \mathrel {{\hat{\succ }}}_{j} {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F^*}](j)$ from $\nu ''(j) = {\mathrm {DA}}^F[\mathrel {{\hat{\succ }}}_{F^*}](j)= {\mathrm {DA}}^F[{\mathrel {{\hat{\succ }}}_{F'_p}}](j)$, contradicting Lemma 14. Hence, (7) also holds in this case.

We are now in a position to prove Lemma 7. Note that $\nu '(f) \succeq _{f} {\mathrm {DA}}[\succ _{F_{\ell } }](f)$ for all $f \in F_{\ell }$. This is because, for $f={\hat{f}}$, we have $\nu '({\hat{f}}) \succeq _{{\hat{f}}} {\mathrm {DA}}[\mathrel {{\hat{\succ }}}_{F_{\ell } }]({\hat{f}}) = {\mathrm {DA}}[\succ _{F_{\ell } }]({\hat{f}})$ (by (5) and (7)). In addition, for $f\in F_\ell {\setminus }\{{\hat{f}}\}$, we have $\nu '(f) \mathrel {{\hat{\succeq }}}_{f} {\mathrm {DA}}[\mathrel {{\hat{\succ }}}_{F_{\ell } }](f)\succeq _f {\mathrm {DA}}[\succ _{F_{\ell } }](f)$ (by (6) and (7)), which implies $\nu '(f) \succeq _{f} {\mathrm {DA}}[\succ _{F_{\ell } }](f)$. Recall that from the definition of $\ell$, there exists $f \in L_\ell$ such that $\nu '(f) \ne \nu (f) ~(={\mathrm {DA}}[{\succ _{F_{\ell }} }](f))$. Hence, $\nu '(f) \succ _{f} {\mathrm {DA}}[\succ _{F_{\ell } }](f)$ for some $f \in L_{\ell }$. However, this contradicts Lemma 16. Therefore, we have $\nu '=\nu$, which completes the proof.

Proof of Theorem 5

Let ${\tilde{h}} :=(a_1, \dots , a_{{\tilde{k}}})$ be the terminal node such that, for each $i=1, \dots , {\tilde{k}}$, $a_{i} = {\mathrm {EADA}}[\succ _{F}](\iota ((a_1, \dots , a_{i-1})))$.¹⁴ That is, every firm makes an offer directly to the partner of the EADA matching in its node. For any $f \in F$, there uniquely exists ${\tilde{h}}_{f} \subsetneq {\tilde{h}}$ such that ${\tilde{h}}_{f} \in {\mathcal {F}}_{f}$, i.e., ${\tilde{h}}_{f}=(a_1,\dots ,a_{i-1})$, where $f=\iota ((a_1, \dots , a_{i-1}))$. For each $f \in F$, we define f’s strategy ${\tilde{\sigma }}_{f}$ as follows:

$$\begin{aligned} {\tilde{\sigma }}_{f}(h)={\left\{ \begin{array}{ll} {\mathrm {EADA}}[\succ _{F}](f) &{}(\text {if } h= {\tilde{h}}_{f} ),\\ \sigma ^*_{f}(h) &{}(\text {if } h \ne {\tilde{h}}_{f}), \end{array}\right. } \qquad (h\in {\mathcal {F}}_f). \end{aligned}$$

In other words, f follows the straightforward strategy except for that f first makes an offer to ${\mathrm {EADA}}[\succ _{F}](f)$. We denote $({\tilde{\sigma }}_{f})_{f \in F}$ by ${\tilde{\sigma }}$. By definition, $\mu [{\tilde{\sigma }}] = {\mathrm {EADA}}[\succ _{F}]$ holds.

We prove that ${\tilde{\sigma }} = ({\tilde{\sigma }}_{f})_{f \in F}$ constitutes an SPE. Let $f \in F$, $h \in {\mathcal {F}}_f$, and $\sigma '_{f} \in \varSigma _{f}$. There are two cases to consider.

• (1) Suppose that $h \ne {\tilde{h}}_f$. Then, we have $\mu [{\tilde{\sigma }}; h] =\mu [\sigma ^*; h]$ and $\mu [(\sigma '_{f}, {\tilde{\sigma }}_{-f}); h ] = \mu [(\sigma '_{f}, \sigma ^*_{-f}); h]$ from the definition of ${\tilde{\sigma }}$. Thus, from Lemma 5, we obtain $\mu [{\tilde{\sigma }}; h](f)=\mu [\sigma ^*; h](f) \succeq _{f} \mu [(\sigma '_{f}, \sigma ^*_{-f}); h](f)=\mu [(\sigma '_{f}, {\tilde{\sigma }}_{-f}); h](f)$.

• (2) Suppose that $h = {\tilde{h}}_{f}$. Let ${\hat{F}} :=\{ \iota (h') \mid h' \subsetneq {\tilde{h}}_f\}$ be the set of firms that moves before f at ${\tilde{h}}_f$, which may be empty. Let $\succ '_{f}$ be a preference ordering that is consistent with $\eta [(\sigma '_{f}, {\tilde{\sigma }}_{-f}); {\tilde{h}}_{f}]~ (=\eta [(\sigma '_{f}, {\tilde{\sigma }}_{-f})])$. From the definition of ${\tilde{\sigma }}$, $\mu [(\sigma '_{f}, {\tilde{\sigma }}_{-f}); {\tilde{h}}_{f} ] ={\mathrm {DA}}^F[(\succ '_{f}, \mathrel {{\hat{\succ }}}_{{\hat{F}}}, \succ _{F {\setminus } ({\hat{F}} \cup \{f\}) })]$, where $\mathrel {{\hat{\succ }}}_{{\hat{f}}}$ is the preference ordering defined in Lemma 7 for each ${\hat{f}} \in {\hat{F}}$. Then, we have that

  $$\begin{aligned} {\mathrm {EADA}}[\succ _{F}](f)&= {\mathrm {EADA}}[(\succ _{f}, \mathrel {{\hat{\succ }}}_{{\hat{F}}}, \succ _{F {\setminus } ({\hat{F}} \cup \{f\}) })](f)\\&\succeq _{f} {\mathrm {DA}}^F[(\succ _{f}, \mathrel {{\hat{\succ }}}_{{\hat{F}}}, \succ _{F {\setminus } ({\hat{F}} \cup \{f\}) })](f)\\&\succeq _{f} {\mathrm {DA}}^F[(\succ '_{f}, \mathrel {{\hat{\succ }}}_{{\hat{F}}}, \succ _{F {\setminus } ({\hat{F}} \cup \{f\}) })](f), \end{aligned}$$

  where the first line follows from repeated applications of Lemma 7, the second line follows from Lemma 6, and the third line follows from the strategy-proofness of the DA (Theorem 9). Thus, we obtain $\mu [{\tilde{\sigma }}; {\tilde{h}}_{f}] (f) ={\mathrm {EADA}}[\succ _{F}](f) \succeq _{f} \mu [(\sigma '_{f}, {\tilde{\sigma }}_{-f}); {\tilde{h}}_{f}](f)$.

Therefore, $\mu [{\tilde{\sigma }}; h] (f) \succeq _{f} \mu [(\sigma '_{f}, {\tilde{\sigma }}_{-f});h]$ in every case. This completes the proof.

Proof of Theorem 6

Here, we prove Theorem 6, i.e., for any sequential game, the outcome matching of any SPE is the worker-optimal stable matching.

We first prove basic properties of SPE. Then, we show a lemma (Lemma 19) which is analogous to Lemma 4 for firms’ sequential game. The proof is similar to the one of Lemma 4, but strategic behavior of workers has to be taken into account. By using Lemma 19, we prove Theorem 6 by backward induction, whose proof is similar to Lemma 13.

We begin by proving a property for SPE for workers that corresponds to Lemmas 9 and 10.

Lemma 17

For a sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {F}}\cup {\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, we have $\mu [\sigma ;h](w)\succeq _w \mu [h](w)$ and $\mu [\sigma ;h](w)\succeq _w f$ for any $h\in {\mathcal {W}}$ with $(f,w)=\iota (h)$.

Proof

Fix any $h\in {\mathcal {W}}$ with $(f,w)=\iota (h)$. Let $\sigma _w'\in \varSigma _w$ be a strategy such that w always rejects an offer, i.e., $\sigma _{w}'(h')=\mathtt {R}$ for all $h'\in {\mathcal {W}}_{w}$. Then, since $\sigma$ is an SPE, we have $\mu [\sigma ;h](w)\succeq _w \mu [(\sigma _F,(\sigma '_w,\sigma _{-w}));h](w)=\mu [h](w)$.

Similarly, let $\sigma _{w}''$ be a strategy such that w only accepts an offer at h, i.e.,

$$\begin{aligned} \sigma _{w}''(h')= {\left\{ \begin{array}{ll} \mathtt {A}&{}(\text {if }h'= h),\\ \mathtt {R}&{}(\text {if }h'\ne h), \end{array}\right. } \qquad (h'\in {\mathcal {W}}_{w}). \end{aligned}$$

Then, since $\sigma$ is an SPE, we have $\mu [\sigma ;h](w)\succeq _w \mu [(\sigma _F,(\sigma ''_w,\sigma _{-w}));h](w)=f$. $\square$

For a workers’ node $h\in {\mathcal {W}}$, the next lemma connects the worker-optimal stable matching for I[h] with the one for $I[(h,\mathtt {A})]$ or $I[(h,\mathtt {R})]$.

Lemma 18

Let $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {F}}\cup {\mathcal {W}}\cup {\mathcal {T}},\iota )$ be a sequential game and $h\in {\mathcal {W}}$ be a workers’ node with $({\hat{f}},{\hat{w}})=\iota (h)$. Then, we have (i) ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succ _{{\hat{w}}} {\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]({\hat{w}})\succeq _{{\hat{w}}}{\hat{f}}$ if ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})=\mu [h]({\hat{w}})\succ _{{\hat{w}}}{\hat{f}}$, and (ii) ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succ _{{\hat{w}}} {\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})\succeq _{{\hat{w}}}\mu [h]({\hat{w}})$ if ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})={\hat{f}}$.

Proof

We first prove (i). Suppose that ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})=\mu [h]({\hat{w}})\succ _{{\hat{w}}}{\hat{f}}$. Note that when $({\succ _F^h},{\succ _W^h})$ changes to $({\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}})$, (a) ${\hat{f}}$ makes the rank of ${\hat{w}}$ first, (b) $\mu [h]({\hat{w}})$ makes ${\hat{w}}$ unacceptable (if $\mu [h]({\hat{w}}) \in F$), and (c) ${\hat{w}}$ makes the rank of $\emptyset$ the last (if $\mu [h]({\hat{w}})= \emptyset$). We also remark that $E[(h,\mathtt {A})]=E[h]{\setminus }\{({\hat{f}},{\hat{w}})\}$ and $\mu [(h,\mathtt {A})]=(\mu [h]{\setminus }\{(\mu [h]({\hat{w}}),{\hat{w}})\})\cup \{({\hat{f}},{\hat{w}})\}$. Since ${\hat{w}}$ is the most preferred worker in $\succ _{{\hat{f}}}^{(h,\mathtt {A})}$, we have ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]({\hat{w}})\succeq ^{(h, \mathtt {A})}_{{\hat{w}}} {\hat{f}}$ from the stability of ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]$. This implies that ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]({\hat{w}})\in F$ (since every firm is acceptable in $\succ _{{\hat{w}}}^{(h,\mathtt {A})}$) and ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]({\hat{w}})\succeq _{{\hat{w}}}{\hat{f}}$ (since $\succ _{{\hat{w}}}$ and $\succ _{{\hat{w}}}^{(h,\mathtt {A})}$ are equivalent over F). By ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})=\mu [h]({\hat{w}})\succ _{{\hat{w}}}{\hat{f}}$, the worker ${\hat{w}}$ never makes an offer to ${\hat{f}}$ in the worker-oriented DA algorithm for I[h]. This implies that the behaviors of the worker-oriented DA algorithm for I[h] and $I[(h,\mathtt {A})]$ are the same until just before ${\hat{w}}$ makes an offer to ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})$ (possibly $\emptyset$). If ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}}) \in F$, ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})~(=\mu [h]({\hat{w}}))$ rejects ${\hat{w}}$ for $I[(h,\mathtt {A})]$ (since ${\hat{w}}$ is unacceptable in $\succ _{{\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})}^{(h,\mathtt {A})}$), and hence we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succ _{{\hat{w}}}{\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]({\hat{w}})$ from ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]({\hat{w}})\in F$ (recall that $\succ _{{\hat{w}}}$ and $\succ _{{\hat{w}}}^{(h,\mathtt {A})}$ are equivalent over F). If ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}}) = \emptyset$, then ${\hat{w}}$ makes an offer to an unacceptable firm (w.r.t. $\succ _{{\hat{w}}}$) for $I[(h,\mathtt {A})]$ since $\emptyset \succ _{w} {\hat{f}}$ and $\succ ^{(h,\mathtt {A})}_{w}$ ranks $\emptyset$ the last. Thus, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succ _{{\hat{w}}}{\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {A})}},{\succ _W^{(h,\mathtt {A})}}]({\hat{w}})$ in every case.

Next, we prove (ii). Suppose that ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})={\hat{f}}$. Note that when $({\succ _F^h},{\succ _W^h})$ changes to $({\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}})$, ${\hat{f}}$ makes ${\hat{w}}$ unacceptable. We also remark that $E[(h,\mathtt {R})]=E[h]{\setminus }\{({\hat{f}},{\hat{w}})\}$ and $\mu [(h,\mathtt {R})]=\mu [h]$. If $\mu [h]({\hat{w}})\in F$, then ${\hat{w}}$ is the most preferred worker in $\succ _{\mu [h]({\hat{w}})}^{(h,\mathtt {R})}$, and ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})\succeq ^{(h,\mathtt {R})}_{{\hat{w}}}\mu [h]({\hat{w}})$ from the stability of ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]$. Hence, we have ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})\in F$ (since every firm is acceptable in $\succ _{{\hat{w}}}^{(h,\mathtt {R})}$ and $\mu [h]({\hat{w}})\in F$) and ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})\succeq _{{\hat{w}}}\mu [h]({\hat{w}})$ (since $\succ _{{\hat{w}}}$ and $\succ _{{\hat{w}}}^{(h,\mathtt {R})}$ are equivalent over F). If $\mu [h]({\hat{w}}) =\emptyset$, then $\succ ^{(h, \mathtt {R})}_{{\hat{w}}}$ equals $\succ _{{\hat{w}}}$ and ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})\succeq _{{\hat{w}}}\mu [h]({\hat{w}})$ by the individually rationality of ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]$. Hence, we have ${\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})\succeq _{{\hat{w}}}\mu [h]({\hat{w}})$ in every case. Note that the behavior of the worker-oriented DA algorithm for I[h] and $I[(h,\mathtt {R})]$ is the same until just before ${\hat{w}}$ makes an offer to ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})={\hat{f}}$ while ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})= {\hat{f}}$ rejects ${\hat{w}}$ for $I[h, \mathtt {R}]$ (recall that ${\hat{w}}$ is unacceptable in $\succ _{{\hat{f}}}^{(h,\mathtt {R})}$). This implies that

$$\begin{aligned} {\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] ({\hat{w}})= {\hat{f}} \succ ^{(h,\mathtt {R})}_{{\hat{w}}} {\mathrm {DA}}^W\left[ {\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}\right] ({\hat{w}})\succeq _{{\hat{w}}}\mu [h]({\hat{w}}). \end{aligned}$$

If $\mu [h]({\hat{w}}) \in F$, we have

$$\begin{aligned} {\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] ({\hat{w}})\succ _{{\hat{w}}}{\mathrm {DA}}^W\left[ {\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}\right] ({\hat{w}}) \end{aligned}$$

since ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}}), {\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})\in F$ and $\succ _{{\hat{w}}}$ and $\succ _{{\hat{w}}}^{(h,\mathtt {R})}$ are equivalent over F. If $\mu [h]({\hat{w}}) = \emptyset$, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{w}})\succ _{{\hat{w}}}{\mathrm {DA}}^W[{\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}]({\hat{w}})$ since $\succ _{{\hat{w}}}$ equals $\succ _{{\hat{w}}}^{(h,\mathtt {R})}$. Thus, we have

$$\begin{aligned} {\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] ({\hat{w}})\succ _{{\hat{w}}} {\mathrm {DA}}^W\left[ {\succ _F^{(h,\mathtt {R})}},{\succ _W^{(h,\mathtt {R})}}\right] ({\hat{w}})\succeq _{{\hat{w}}}\mu [h]({\hat{w}}) \end{aligned}$$

in either case. $\square$

Next, we prove that all the firms prefer the outcome matching of any SPE at least as well as the worker-optimal stable matching. The following lemma is analogous to Lemma 4, which is for the case when only firms act strategically.

Lemma 19

For a sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {F}}\cup {\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, we have

$$\begin{aligned} \begin{matrix} \mu [\sigma ;h](f)\succeq _f^h{\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (f) ~~(\forall f\in F)&{} \text {and}\\ \mu [\sigma ;h](w)\preceq _w^h{\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (w) ~~(\forall w\in W) \end{matrix} \end{aligned}$$

(8)

for any firms’ or terminal node $h\in {\mathcal {F}}\cup {\mathcal {T}}$.

Proof

We prove this lemma by backward induction. For each terminal node $h \in {\mathcal {T}}$, $\succ ^h_{f}$ ranks $\mu [\sigma ; h](f)~(=\mu [h](f))$ first for all $f \in F$ and $\mu [\sigma ; h](w)~(=\mu [h](w))\succeq _w^h \emptyset$ for all $w\in W$. Thus, $\mu [\sigma ; h]$ is the firm-optimal stable matching in I[h]. Therefore, we have $\mu [\sigma ;h](f)\succeq _f^h{\mathrm {DA}}^W[\succ _F^h,\succ _W^h](f)$ for all $f\in F$ and $\mu [\sigma ;h](w)\preceq _w^h{\mathrm {DA}}^W[\succ _F^h,\succ _W^h](w)$ for all $w\in W$ by Theorem 7.

For a firms’ node $h\in {\mathcal {F}}$, suppose that the claim holds for every firms’ or terminal node $h'\supsetneq h$. Let us denote ${\hat{f}}=\iota (h)$, ${\hat{w}}={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{f}})~(\in \{w'\mid ({\hat{f}},w')\in E[h]\}\cup \{\emptyset \})$, ${\hat{s}}=\sigma _{{\hat{f}}}(h)$, and ${\hat{h}}=(h,{\hat{s}})$. If ${\hat{s}}\in W$, let ${\tilde{h}}=({\hat{h}},\sigma _{{\hat{s}}}({\hat{h}}))$. We remark that $\succ _{{\hat{f}}}$ and $\succ ^h_{{\hat{f}}}$ are equivalent over $\{w'\mid ({\hat{f}},w')\in E[h]\}\cup \{\emptyset \}$ since ${\hat{f}}$ is unmatched at h (i.e., $\mu [h]({\hat{f}})=\emptyset$). We also remark that when ${\hat{s}} \in W$, we have

$$\begin{aligned} {\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] ({\hat{s}}) \succeq _{{\hat{s}}}^h \mu [h]({\hat{s}}) \end{aligned}$$

(9)

by the individually rationality of ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})$ if $\mu [h]({\hat{s}})=\emptyset$ and by the stability of ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$ if $\mu [h]({\hat{s}})\in F$ (since ${\hat{s}}$ is ranked top in $\succ _{\mu [h]({\hat{s}})}^{h}$).

We first prove that $\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})={\hat{w}}$. This implies

$$\begin{aligned} \mu [\sigma ;h]({\hat{f}})\succeq _{{\hat{f}}}\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})={\hat{w}}~\left( ={\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] ({\hat{f}})\right) \end{aligned}$$

(10)

since $\sigma$ is an SPE. When ${\hat{w}} = \emptyset$, this statement clearly holds. Thus, we assume that ${\hat{w}} \in W$.

Suppose that $\sigma _{{\hat{w}}}((h,{\hat{w}}))=\mathtt {R}$. Note that $\succ _{W}^h$ is equal to $\succ _{W}^{(h,{\hat{w}},\mathtt {R})}$. From Lemma 18, we have ${\hat{f}}={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})\succ _{{\hat{w}}}{\mathrm {DA}}^W[\succ _F^{(h,{\hat{w}},\mathtt {R})},\succ _W^{(h,{\hat{w}},\mathtt {R})}]({\hat{w}})$, and hence

$$\begin{aligned} {\hat{f}}\succ _{{\hat{w}}}^h {\mathrm {DA}}^W\left[ {\succ _F^{(h,{\hat{w}},\mathtt {R})}},{\succ _W^{(h,{\hat{w}},\mathtt {R})}}\right] ({\hat{w}}) \end{aligned}$$

by ${\hat{f}}\in F$. Then, according to the induction hypothesis, we have

$$\begin{aligned} {\hat{f}}\succ _{{\hat{w}}}^h{\mathrm {DA}}^W\left[ \succ _F^{(h,{\hat{w}},\mathtt {R})},\succ _W^{(h,{\hat{w}},\mathtt {R})}\right] ({\hat{w}})\succeq _{{\hat{w}}}^h\mu [\sigma ;(h,{\hat{w}},\mathtt {R})]({\hat{w}}). \end{aligned}$$

In addition, we have

$$\begin{aligned} \mu [\sigma ;(h,{\hat{w}},\mathtt {R})]({\hat{w}})=\mu [\sigma ;(h,{\hat{w}})]({\hat{w}})\succeq _{{\hat{w}}}{\hat{f}} \end{aligned}$$

from Lemma 17. From ${\hat{f}} \succ ^h_{{\hat{w}}} \mu [\sigma ;(h,{\hat{w}},\mathtt {R})]({\hat{w}})$ and $\mu [\sigma ;(h,{\hat{w}},\mathtt {R})]({\hat{w}}) \succeq _{{\hat{w}}} {\hat{f}}$, we have $\mu [h]({\hat{w}})\in F$. Moreover, since $\succ _{{\hat{w}}}$ and $\succ _{{\hat{w}}}^h$ are equivalent over F, we have $\mu [\sigma ;(h,{\hat{w}},\mathtt {R})]({\hat{w}})=\emptyset$. However, this is impossible from Lemma 1.

Suppose that $\sigma _{{\hat{w}}}((h,{\hat{w}}))=\mathtt {A}$. Note that ${\hat{f}}\succ _{{\hat{w}}}^h\mu [h]({\hat{w}})$ since otherwise $\mu [h]({\hat{w}})\in F$ by $\mu [h]({\hat{w}})\succeq _{{\hat{w}}}^h{\hat{f}}\succ _{{\hat{w}}}^h\emptyset$ and $(\mu [h]({\hat{w}}),{\hat{w}})$ is a blocking pair for ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$ in I[h] (recall that ${\hat{w}}$ is ranked top in $\succ _{\mu [h]({\hat{w}})}^h$). We claim that ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^{(h,{\hat{w}},\mathtt {A})}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$. When $\succ _F^h$ changes to $\succ _F^{(h,{\hat{w}},\mathtt {A})}$, ${\hat{f}}$ makes the rank of ${\hat{w}}={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{f}})$ first and $\mu [h]({\hat{w}})$ makes ${\hat{w}}$ unacceptable (if $\mu [h]({\hat{w}}) \in F$). By Lemma 2 and ${\hat{f}}={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}}) \succ _{{\hat{w}}}^h\mu [h]({\hat{w}})$, we have ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^h}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$. If $\mu [h]({\hat{w}}) \in F$, then $\succ _W^h$ is equal to $\succ _W^{(h,{\hat{w}},\mathtt {A})}$ and hence ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^{(h,{\hat{w}},\mathtt {A})}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$. Suppose that $\mu [h]({\hat{w}}) = \emptyset$. Then, ${\hat{w}}$ makes the rank of $\emptyset$ the last in the change from $\succ _W^h$ to $\succ _W^{(h,{\hat{w}},\mathtt {A})}$. Recall that ${\hat{f}} = {\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}}) \succ _{{\hat{w}}}^h\mu [h]({\hat{w}}) = \emptyset$ holds. By ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^h}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$, we have ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^h}] ({\hat{w}}) \succ _{{\hat{w}}}^h \emptyset$. From Lemma 2, we have ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^{(h,{\hat{w}},\mathtt {A})}}]={\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^h}]$. Therefore, ${\mathrm {DA}}^W[{\succ _F^{(h,{\hat{w}},\mathtt {A})}},{\succ _W^{(h,{\hat{w}},\mathtt {A})}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$ in every case. Since $\mu [\sigma ;(h,{\hat{w}},\mathtt {A})]({\hat{f}})\succeq _{{\hat{f}}}^{(h,{\hat{w}},\mathtt {A})} {\mathrm {DA}}^W[\succ _F^{(h,{\hat{w}},\mathtt {A})},\succ _W^{(h,{\hat{w}},\mathtt {A})}]({\hat{f}})~(={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{f}})={\hat{w}})$ according to the induction hypothesis and ${\hat{w}}$ is the most preferred worker in $\succ _{{\hat{f}}}^{(h,{\hat{w}},\mathtt {A})}$, we have $\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})=\mu [\sigma ;(h,{\hat{w}},\mathtt {A})]({\hat{f}})={\hat{w}}$.

Second, we show that

$$\begin{aligned} {\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] ({\hat{s}})\succeq _{{\hat{s}}}^h{\hat{f}} \quad \text {when}\quad {\hat{s}} \in W. \end{aligned}$$

(11)

Suppose to the contrary that ${\hat{f}}\succ _{{\hat{s}}}^h {\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$. We separately consider the following two cases: (i) $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {R}$ and (ii) $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$.

(i) Suppose that $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {R}$. Note that, in this case, $\succ _{W}^h$ is equal to $\succ _{W}^{{\tilde{h}}}$ since the set of temporarily matched workers is unchanged. Note also that only the difference between $\succ _F^h$ and $\succ _F^{{\tilde{h}}}$ is that ${\hat{s}}$ is acceptable to ${\hat{f}}$ at $\succ _F^h$ while ${\hat{s}}$ is unacceptable to ${\hat{f}}$ at $\succ _F^{{\tilde{h}}}$. We claim that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})={\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]({\hat{s}})$. We assume that ${\hat{s}} \succ ^{h}_{{\hat{f}}} \emptyset$ since otherwise this claim clearly holds owing to the equivalence of $\succ _F^{{\tilde{h}}}$ and $\succ _F^{h}$ for the acceptable workers. From ${\hat{f}} \ne {\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})$ and the difference of $\succ _F^h$ and $\succ _F^{{\tilde{h}}}$, the matching ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$ is also stable in $I[{\tilde{h}}]$. Thus, ${\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]({\hat{s}}) \succeq ^{h}_{{\hat{s}}} {\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})$ from the equivalence of $\succ _{W}^h$ and $\succ _{W}^{{\tilde{h}}}$. Suppose that ${\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]({\hat{s}}) \succ ^{h}_{{\hat{s}}} {\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$. Let $\succ '_{{\hat{s}}}$ be a preference ordering constructed from $\succ ^{h}_{{\hat{s}}}$ by making ${\hat{f}}$ unacceptable without changing the others. Then, ${\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]({\hat{s}}) = {\mathrm {DA}}^W[{\succ _F^h},({\succ '_{{\hat{s}}}}, {\succ ^{h}_{-{\hat{s}}})}]({\hat{s}}) \succ ^{h}_{{\hat{s}}} {\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$ holds, which contradicts the strategy-proofness of the DA (Theorem 9). Thus, ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})={\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]({\hat{s}})$. Therefore, we have

$$\begin{aligned} {\hat{f}}&\succ _{{\hat{s}}}^h {\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] ({\hat{s}}) ={\mathrm {DA}}^W\left[ {\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}\right] ({\hat{s}}) \\&\succeq _{{\hat{s}}}^h\mu [\sigma ;{\tilde{h}}]({\hat{s}}) =\mu [\sigma ;h]({\hat{s}}) \succeq _{{\hat{s}}}{\hat{f}}, \end{aligned}$$

where the last relation holds by Lemma 17. In particular, ${\hat{f}} \succ _{{\hat{s}}}^h \mu [\sigma ;h]({\hat{s}}) \succeq _{{\hat{s}}}{\hat{f}}$. These relations are possible only when $\mu [\sigma ;h]({\hat{s}})=\emptyset$, and hence $\mu [h]({\hat{s}})=\emptyset$ by Lemma 1. We also have that $\succ _{{\hat{s}}}^h$ and $\succ _{{\hat{s}}}$ are equal. However, this implies ${\hat{f}}\succ _{{\hat{s}}}{\hat{f}}$, which is a contradiction.

(ii) Suppose that $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$. In this case, ${\tilde{h}} = (h, {\hat{s}}, \mathtt {A})$. Note that when $({\succ _F^h},{\succ _W^h})$ changes to $({\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}})$, (i) ${\hat{f}}$ makes the rank of ${\hat{s}}$ first, (ii) $\mu [h]({\hat{s}})$ makes ${\hat{s}}$ unacceptable (if $\mu [h]({\hat{s}}) \in F$), and (iii) ${\hat{s}}$ makes the rank of $\emptyset$ the last (if $\mu [h]({\hat{s}})= \emptyset$). Let us consider the worker-oriented DA algorithm for I[h] and $I[{\tilde{h}}]$. In I[h], ${\hat{s}}$ makes an offer to ${\hat{f}}$ before $\mu [h]({\hat{s}})$ because ${\hat{f}} \succ _{{\hat{s}}}^h {\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}}) \succeq _{{\hat{s}}}^h \mu [h]({\hat{s}})$ by (9). This implies that ${\hat{s}}$ also makes an offer to ${\hat{f}}$ in $I[{\tilde{h}}]$ according to the definition of $I[{\tilde{h}}]$. Thus, ${\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}] ({\hat{f}}) = {\hat{s}}$ since $\succ ^{{\tilde{h}}}_{{\hat{f}}}$ ranks ${\hat{s}}$ first. According to the induction hypothesis, we have $\mu [\sigma ;{\tilde{h}}]({\hat{f}})\succeq _{{\hat{f}}}^{{\tilde{h}}}{\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}]({\hat{f}})={\hat{s}}$. Since $\succ ^{{\tilde{h}}}_{{\hat{f}}}$ ranks ${\hat{s}}$ first, we have $\mu [\sigma ;{\tilde{h}}]({\hat{f}})={\hat{s}}$. Thus, we obtain ${\hat{s}}=\mu [\sigma ;{\tilde{h}}]({\hat{f}})\succeq _{{\hat{f}}}\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})={\hat{w}}$. Then, $({\hat{f}},{\hat{s}})$ is a blocking pair for ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$ in I[h] by ${\hat{f}}\succ _{{\hat{s}}}^h {\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$, which is a contradiction.

Finally, we prove (8). For $f={\hat{f}}$, we have $\mu [\sigma ; h]({\hat{f}}) \succeq ^h _{{\hat{f}}} {\mathrm {DA}}^W[\succ _F^h,\succ _W^h] ({\hat{f}})$ from (10) and equality of $\succeq _{{\hat{f}}}$ and $\succeq _{{\hat{f}}}^h$ by $\mu [h]({\hat{f}})=\emptyset$. Hence, it is sufficient to prove this for the other firms $F{\setminus }\{{\hat{f}}\}$ and the workers W. We discuss the following four cases separately: when ${\hat{s}}=\emptyset$ (Case 1), when ${\hat{s}} \in W$ and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {R}$ (Case 2), when ${\hat{s}}\in W$, $\mu [h]({\hat{s}})\in F$, and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$ (Case 3), and when ${\hat{s}}\in W$, $\mu [h]({\hat{s}})=\emptyset$, and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$ (Case 4).

Case 1: Suppose that ${\hat{s}} = \emptyset$. In this case, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{f}}) = \emptyset$ from

$$\begin{aligned} \emptyset = \mu [\sigma ; h]({\hat{f}}) \succeq ^h_{{\hat{f}}} {\hat{w}} = {\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] ({\hat{f}}). \end{aligned}$$

In the change from $\succ _F^h$ to $\succ _F^{{\hat{h}}}$, ${\hat{f}}$ makes the rank of $\emptyset$ first. Further, $\succ _{W}^h$ is equal to $\succ _{W}^{{\hat{h}}}$. Thus, ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]={\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W^{{\hat{h}}}}]$ from Lemma 2 and the definitions of I[h] and $I[{\hat{h}}]$. Hence, for each $w\in W$, we have

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\hat{h}}](w)\preceq _w^h{\mathrm {DA}}^W\left[ \succ _F^{{\hat{h}}},\succ _W^{{\hat{h}}}\right] (w)={\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (w) \end{aligned}$$

according to the induction hypothesis and the equivalence of $\succ _{W}^h$ and $\succ _{W}^{{\hat{h}}}$. In addition, for each $f\in F{\setminus }\{{\hat{f}}\}$, we have

$$\begin{aligned}\mu [\sigma ;h](f)=\mu [\sigma ;{\hat{h}}](f)\succeq _f^h{\mathrm {DA}}^W\left[ {\succ _F^{{\hat{h}}}},{\succ _W^{{\hat{h}}}}\right] (f)={\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] (f)\end{aligned}$$

according to the induction hypothesis and the equivalence of $\succ ^{h}_{f}$ and $\succ ^{{\hat{h}}}_{f}$.

Case 2: Suppose that ${\hat{s}} \in W$ and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {R}$. Thus, ${\tilde{h}}= (h, {\hat{s}}, \mathtt {R})$. Then, $E[{\tilde{h}}]=E[h]{\setminus }\{({\hat{f}},{\hat{s}})\}$ and $\mu [{\tilde{h}}]=\mu [h]$. Note that $\succ _{W}^h$ is equal to $\succ _{W}^{{\tilde{h}}}$. Further, ${\hat{f}}$ makes ${\hat{s}}$ unacceptable in the change from $\succ _F^h$ to $\succ _F^{{\tilde{h}}}$. We remark that ${\hat{s}}\ne {\hat{w}}$ from $\mu [\sigma ;(h,{\hat{w}})]({\hat{f}})={\hat{w}}$, and hence ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\succ _{{\hat{s}}}^h{\hat{f}}$ by (11). Thus, ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]={\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]$ from Lemma 2 and the definitions of I[h] and $I[{\tilde{h}}]$. Hence, for each $w\in W$, we have

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\tilde{h}}](w)\preceq _w^h{\mathrm {DA}}^W\left[ \succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}\right] (w)={\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (w) \end{aligned}$$

according to the induction hypothesis and the equivalence of $\succ _{W}^h$ and $\succ _{W}^{{\tilde{h}}}$. Further, for each $f\in F{\setminus }\{{\hat{f}}\}$, we have

$$\begin{aligned} \mu [\sigma ;h](f)=\mu [\sigma ;{\tilde{h}}](f)\succeq _f^{h}{\mathrm {DA}}^W\left[ \succ _F^{{\tilde{h}}}, \succ _W^{{\tilde{h}}}\right] (f)={\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (f) \end{aligned}$$

according to the induction hypothesis and the equivalence of $\succ _f^h$ and $\succ _f^{{\tilde{h}}}$.

Case 3: Suppose that ${\hat{s}}\in W$, $\mu [h]({\hat{s}})\in F$, and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$. Thus, ${\tilde{h}}= (h, {\hat{s}}, \mathtt {A})$. Then, $E[{\tilde{h}}]=E[h]{\setminus }\{({\hat{f}},{\hat{s}})\}$ and $\mu [{\tilde{h}}]=\mu [h]{\setminus }\{(\mu [h]({\hat{s}}),{\hat{s}})\}\cup \{({\hat{f}},{\hat{s}})\}$. Note that $\succ _{W}^h$ is equal to $\succ _{W}^{{\tilde{h}}}$. Further, ${\hat{f}}$ makes the rank of ${\hat{s}}$ first and $\mu [h]({\hat{s}})$ makes ${\hat{s}}$ unacceptable in the change from $\succ _F^h$ to $\succ _F^{{\tilde{h}}}$. As ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\succeq _{{\hat{s}}}^h\mu [h]({\hat{s}})$ by (9), we consider the following two cases: ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})=\mu [h]({\hat{s}})$ and ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succ _{{\hat{s}}}^h\mu [h]({\hat{s}})$.

Suppose that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})=\mu [h]({\hat{s}})$. By (11), ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})=\mu [h]({\hat{s}}) \succ _{{\hat{s}}} {\hat{f}}$ since $\succ _{{\hat{s}}}$ and $\succ ^{h}_{{\hat{s}}}$ are equivalent over F. Then, ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\succ _{{\hat{s}}}{\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]({\hat{s}})$ from Lemma 18. According to the induction hypothesis, we have

$$\begin{aligned} {\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] ({\hat{s}})\succ _{{\hat{s}}} {\mathrm {DA}}^W\left[ {\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}\right] ({\hat{s}}) \succeq _{{\hat{s}}}^{{\tilde{h}}}\mu [\sigma ;{\tilde{h}}]({\hat{s}}) =\mu [\sigma ;h]({\hat{s}}). \end{aligned}$$

Here, since $\mu [\sigma ;h]({\hat{s}})$ and ${\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}]({\hat{s}})$ are in F by $\mu [h]({\hat{s}})\in F$ and Lemma 1, we obtain ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\succ _{{\hat{s}}}\mu [\sigma ;h]({\hat{s}})$. However, we have $\mu [\sigma ;h]({\hat{s}})\succeq _{{\hat{s}}}\mu [h]({\hat{s}})={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})$ from Lemma 17, which is a contradiction.

Suppose that ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succ _{{\hat{s}}}^h\mu [h]({\hat{s}})$. Then, ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]={\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}]$ from Lemma 2 and the definitions of I[h] and $I[{\tilde{h}}]$ (recall that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\succeq _{{\hat{s}}}^h{\hat{f}}$ by (11)). Hence, for each $w\in W$, we have

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\tilde{h}}](w)\preceq _w^h{\mathrm {DA}}^W\left[ \succ _F^{{\tilde{h}}}, \succ _W^{{\tilde{h}}}\right] (w)={\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (w) \end{aligned}$$

according to the induction hypothesis and the equivalence of $\succ _W^h$ and $\succ _W^{{\tilde{h}}}$. In addition, for each $f\in F$, we have

$$\begin{aligned} \mu [\sigma ;h](f)=\mu [\sigma ;{\tilde{h}}](f)\succeq _f^{{\tilde{h}}} {\mathrm {DA}}^W\left[ \succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}\right] (f)={\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] (f) \end{aligned}$$

according to the induction hypothesis. Hence, for $f\in F{\setminus }\{{\hat{f}},\mu [h]({\hat{s}})\}$, we obtain $\mu [\sigma ;h](f)\succeq _f^{h}{\mathrm {DA}}^W[{\succ _F^h}, {\succ _W^h}](f)$ since $\succ _f^h$ equals $\succ _f^{{\tilde{h}}}$. Finally, for $f=\mu [h]({\hat{s}})$, we obtain $\mu [\sigma ;h](f)\succeq _f^{h}{\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}](f)$ since $\succ _f^h$ and $\succ _f^{{\tilde{h}}}$ are equivalent over $(W\cup \{\emptyset \}){\setminus }\{{\hat{s}}\}$ and $\mu [\sigma ;h]({\hat{s}})=\mu [\sigma ;{\tilde{h}}]({\hat{s}})\ne \mu [h]({\hat{s}})\ne {\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]({\hat{s}})={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$.

Case 4: Suppose that ${\hat{s}}\in W$, $\mu [h]({\hat{s}})=\emptyset$, and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$. Thus, ${\tilde{h}}= (h, {\hat{s}}, \mathtt {A})$. Then, $E[{\tilde{h}}]=E[h]{\setminus }\{({\hat{f}},{\hat{s}})\}$ and $\mu [{\tilde{h}}]=\mu [h]\cup \{({\hat{f}},{\hat{s}})\}$. Note that ${\hat{s}}$ makes the rank of $\emptyset$ the last in the change from $\succ _W^h$ to $\succ _W^{{\tilde{h}}}$. Further, ${\hat{f}}$ makes the rank of ${\hat{s}}$ first in the change from $\succ _F^h$ to $\succ _F^{{\tilde{h}}}$.

As ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succeq _{{\hat{s}}}\emptyset$, we consider the following two cases: ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})=\emptyset$ and ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succ _{{\hat{s}}}\emptyset$.

Suppose that ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})=\emptyset$. By (11), ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})=\mu [h]({\hat{s}})=\emptyset \succ _{{\hat{s}}} {\hat{f}}$ since $\succ ^{h}_{{\hat{s}}}$ is equal to $\succ _{{\hat{s}}}$. Then, we have $\emptyset \succ _{{\hat{s}}}{\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}]({\hat{s}})$ from Lemma 18. In addition, we have

$$\begin{aligned} \emptyset \succ _{{\hat{s}}}{\mathrm {DA}}^W\left[ \succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}\right] ({\hat{s}}) \succeq _{{\hat{s}}}\mu [\sigma ;{\tilde{h}}]({\hat{s}})=\mu [\sigma ;h]({\hat{s}}) \end{aligned}$$

according to the induction hypothesis and ${\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}]({\hat{s}}),\mu [\sigma ;{\tilde{h}}]({\hat{s}})\in F$ from $\mu [{\tilde{h}}]({\hat{s}})={\hat{f}}\in F$ and Lemma 1. On the contrary, we have $\mu [\sigma ;h]({\hat{s}})\succeq _{{\hat{s}}}\mu [h]({\hat{s}})=\emptyset$ from Lemma 17, which is a contradiction.

Suppose that ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succ _{{\hat{s}}}\emptyset$. Then, we have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]={\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]$ from Lemma 2 and the definitions of I[h] and $I[{\tilde{h}}]$ (recall that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\succeq _{{\hat{s}}}^h{\hat{f}}$ by (11)). Hence, for each $w\in W$, we have

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\tilde{h}}](w)\preceq _w^h{\mathrm {DA}}^W\left[ \succ _F^{{\tilde{h}}}, \succ _W^{{\tilde{h}}}\right] (w)={\mathrm {DA}}^W[\succ _F^h,\succ _W^h](w) \end{aligned}$$

according to the induction hypothesis, the equivalence of $\succ _w^h$ and $\succ _w^{{\tilde{h}}}$ when $w\ne {\hat{s}}$, and ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\succ _{{\hat{s}}}\emptyset$ when $w={\hat{s}}$. In addition, for each $f\in F{\setminus }\{{\hat{f}}\}$, we have

$$\begin{aligned} \mu [\sigma ;h](f)=\mu [\sigma ;{\tilde{h}}](f)\succeq _f^h{\mathrm {DA}}^W\left[ {\succ _F^{{\tilde{h}}}}, {\succ _W^{{\tilde{h}}}}\right] (f)={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}](f) \end{aligned}$$

according to the induction hypothesis and the equivalence of $\succ _f^h$ and $\succ _f^{{\tilde{h}}}$. $\square$

We are now ready to prove Theorem 6. To use backward induction, we prove a stronger claim, which is similar to Lemma 13 in the case of strategic workers.

Lemma 20

For a sequential game $G=(I=(F,W,\succ _F,\succ _W),{\mathcal {H}}={\mathcal {F}}\cup {\mathcal {W}}\cup {\mathcal {T}},\iota )$ with an SPE $\sigma$, we have $\mu [\sigma ;h](w)={\mathrm {DA}}^W[\succ _F^h,\succ _W^h](w)$ for any firms’ or terminal node $h\in {\mathcal {F}}\cup {\mathcal {T}}$ and any worker $w\in W$ such that $\mu [h](w)=\emptyset$.

Proof

We prove this lemma by backward induction.

Suppose that h is a terminal node. Let ${\hat{w}}$ be a worker such that $\mu [h]({\hat{w}})=\emptyset$. Then, $\mu [\sigma ;h]({\hat{w}})=\mu [h]({\hat{w}})=\emptyset$. Moreover, as $\succ ^h_{f}$ ranks $\mu [h](f)$ first for all $f \in F$ and $\mu [h](w)\succeq _{w}^h\emptyset$ for all $w\in W$, $\mu [h]$ is the firm-optimal stable matching in I[h], i.e., $\mu [h]={\mathrm {DA}}^F[\succ _F^h,\succ _W^h]$. Since $\mu [h]({\hat{w}})~(={\mathrm {DA}}^F[\succ _F^h,\succ _W^h])=\emptyset$, we obtain ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{w}})=\emptyset$ by the lone wolf theorem (Theorem 8).

Suppose that h is a firms’ node and that the claim holds for every firms’ or terminal node $h'\supsetneq h$. Let us denote ${\hat{f}}=\iota (h)$, ${\hat{w}}={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{f}})~(\in \{w'\mid ({\hat{f}},w')\in E[h]\}\cup \{\emptyset \})$, ${\hat{s}}=\sigma _{{\hat{f}}}(h)$, and ${\hat{h}}=(h,{\hat{s}})$. If ${\hat{s}}\in W$, let ${\tilde{h}}=({\hat{h}},\sigma _{{\hat{s}}}({\hat{h}}))$. Here, we have $\mu [\sigma ;h]=\mu [\sigma ;{\hat{h}}]$. In addition, if ${\hat{s}}\in W$, we have $\mu [\sigma ;h]=\mu [\sigma ;{\tilde{h}}]$. In what follows, we discuss the following four cases separately: when ${\hat{s}}={\hat{w}}=\emptyset$ (Case 1), when ${\hat{s}}=\emptyset$ and ${\hat{w}}\in W$ (Case 2), when ${\hat{s}}\in W$ and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {R}$ (Case 3), and when ${\hat{s}}\in W$ and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$ (Case 4).

Case 1: Suppose that ${\hat{s}}={\hat{w}}=\emptyset$. In this case, we have $\mu [h]=\mu [{\hat{h}}]$. Note that ${\succ _W^h}$ is equal to ${\succ _W^{{\hat{h}}}}$. In addition, ${\hat{f}}$ makes the rank of $\emptyset = {\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{f}})$ first in the change from ${\succ _F^h}$ to ${\succ _F^{{\hat{h}}}}$. We also have ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]={\mathrm {DA}}^W[{\succ _F^{{\hat{h}}}},{\succ _W^{{\hat{h}}}}]$ from Lemma 2. Hence,

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\hat{h}}](w)={\mathrm {DA}}^W\left[ {\succ _F^{{\hat{h}}}},{\succ _W^{{\hat{h}}}}\right] (w)={\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] (w) \end{aligned}$$

holds for any $w\in W$ with $\mu [h](w)~(=\mu [{\hat{h}}](w))=\emptyset$ according to the induction hypothesis for ${\hat{h}}$.

Case 2: Suppose that ${\hat{s}}=\emptyset$ and ${\hat{w}}\in W$. Then, we have $\mu [\sigma ;h]({\hat{f}})=\emptyset$. On the contrary, from Lemma 19, we have $\mu [\sigma ;h]({\hat{f}})\succeq _{{\hat{f}}}^h {\hat{w}}\succ _{{\hat{f}}}^h\emptyset$. This is a contradiction.

Case 3: Suppose that ${\hat{s}}\in W$ and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {R}$. Then, we have ${\tilde{h}}=(h, {\hat{s}}, \mathtt {R})$ and $\mu [{\tilde{h}}]=\mu [h]$. Note that $\mu [\sigma ;h]({\hat{s}})\succ _{{\hat{s}}}{\hat{f}}$ from Lemma 17. In addition, ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succeq _{{\hat{s}}}^h\mu [\sigma ;h]({\hat{s}})$ from Lemma 19. If $\mu [h]({\hat{s}})=\emptyset$, we have ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succ _{{\hat{s}}}^h{\hat{f}}$ since $\succ _{{\hat{s}}}$ is equal to $\succ _{{\hat{s}}}^h$. Otherwise (i.e., $\mu [h]({\hat{s}})\in F$), we have ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succ _{{\hat{s}}}^h{\hat{f}}$ since $\mu [\sigma ;h]({\hat{s}}),{\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\in F$ from Lemma 1. Hence, we have ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succ _{{\hat{s}}}^h{\hat{f}}$. Note that ${\succ _W^h}$ is equal to ${\succ _W^{{\tilde{h}}}}$, and ${\hat{f}}$ makes ${\hat{s}}$ unacceptable in the change from ${\succ _F^h}$ to ${\succ _F^{{\tilde{h}}}}$. From Lemma 2, ${\mathrm {DA}}^W[{\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}]={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]$. Thus, we have

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\tilde{h}}](w)={\mathrm {DA}}^W\left[ {\succ _F^{{\tilde{h}}}},{\succ _W^{{\tilde{h}}}}\right] (w)={\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] (w) \end{aligned}$$

for any $w\in W$ with $\mu [h](w)=\emptyset$ according to the induction hypothesis for ${\tilde{h}}$.

Case 4: Suppose that ${\hat{s}}\in W$ and $\sigma _{{\hat{s}}}({\hat{h}})=\mathtt {A}$. Then, we have ${\tilde{h}}= (h, {\hat{s}}, \mathtt {A})$ and $\mu [{\tilde{h}}]=\{e\in \mu [h]\mid e_W\ne {\hat{s}}\}\cup \{({\hat{f}},{\hat{s}})\}$.

We claim that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\in F$. To obtain a contradiction, suppose that ${\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})=\emptyset$. Then, we obtain $\mu [h]({\hat{s}})=\emptyset$ from Lemma 1. Moreover, we have $\emptyset ={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}}) \succeq _{{\hat{s}}} \mu [\sigma ;h]({\hat{s}}) \succeq _{{\hat{s}}} \mu [h]({\hat{s}})=\emptyset$ from Lemmas 17 and 19. Thus, $\mu [\sigma ;h]({\hat{s}}) = \emptyset$. On the contrary, $\sigma _{{\hat{s}}}({\hat{h}}) = \mathtt {A}$ implies $\mu [\sigma ;h]({\hat{s}}) = \mu [\sigma ;{\hat{h}}]({\hat{s}}) \in F$ from Lemma 1, which is a contradiction.

Since $\mu [\sigma ;h]({\hat{s}}),{\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})\in F$, we have $\mu [\sigma ;h]({\hat{s}})\preceq _{{\hat{s}}}{\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$ from Lemma 19. Thus, from Lemma 17, we obtain

$$\begin{aligned} \begin{array}{l} {\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] ({\hat{s}})\succeq _{{\hat{s}}}\mu [\sigma ;h]({\hat{s}})\succeq _{{\hat{s}}}{\hat{f}} \quad \text {and}\\ {\mathrm {DA}}^W\left[ \succ _F^h,\succ _W^h\right] ({\hat{s}})\succeq _{{\hat{s}}}\mu [\sigma ;h]({\hat{s}})\succ _{{\hat{s}}} \mu [h]({\hat{s}}). \end{array} \end{aligned}$$

(12)

We claim that ${\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}]={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$. In the change from I[h] to $I[{\tilde{h}}]$, (i) ${\hat{f}}$ makes the rank of ${\hat{s}}$ first, (ii) $\mu [h]({\hat{s}})$ makes ${\hat{s}}$ unacceptable (if $\mu [h]({\hat{s}}) \in F$), and (iii) ${\hat{s}}$ makes the rank of $\emptyset$ the last (if $\mu [h]({\hat{s}})= \emptyset$). Note that ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}}) \in F$, ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}})\succeq _{{\hat{s}}} {\hat{f}}$ and ${\mathrm {DA}}^W[\succ _F^h,\succ _W^h]({\hat{s}}) \succ _{{\hat{s}}} \mu [h]({\hat{s}})$ by (12). By Lemma 2, this implies ${\mathrm {DA}}^W[\succ _F^{{\tilde{h}}},\succ _W^{{\tilde{h}}}]={\mathrm {DA}}^W[\succ _F^h,\succ _W^h]$ in any cases of $\mu [h]({\hat{s}})= \emptyset$ and $\mu [h]({\hat{s}}) \in F$. Hence, we obtain

$$\begin{aligned} \mu [\sigma ;h](w)=\mu [\sigma ;{\tilde{h}}](w)={\mathrm {DA}}^W\left[ {\succ _F^{{\tilde{h}}}}, {\succ _W^{{\tilde{h}}}}\right] (w)={\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] (w) \end{aligned}$$

for any $w\in W$ with $\mu [{\tilde{h}}](w)=\emptyset$ according to the induction hypothesis for ${\tilde{h}}$. As $\mu [{\tilde{h}}](w)=\mu [h](w)$ for any $w\in W{\setminus }\{{\hat{s}}\}$, we get $\mu [\sigma ;h](w)={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}](w)$ for any $w\in W{\setminus }\{{\hat{s}}\}$ with $\mu [h](w)=\emptyset$.

Finally, prove that $\mu [\sigma ;h]({\hat{s}})={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$ for the case when $\mu [h]({\hat{s}})=\emptyset$. If ${\hat{s}}={\hat{w}}$, then we have $\mu [\sigma ;h]({\hat{s}})={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$ because $\mu [\sigma ;h]({\hat{s}})\succeq _{{\hat{s}}}{\hat{f}}={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$ from Lemma 17 and $\mu [\sigma ;h]({\hat{s}})\preceq _{{\hat{s}}}{\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$ from Lemma 19. Hence, suppose that ${\hat{s}}\ne {\hat{w}}$. Then, we have ${\mathrm {DA}}^W[{\succ _F^h}, {\succ _W^h}]({\hat{s}})\succ _{{\hat{s}}}{\hat{f}}$ from (12). Note that ${\succ _W^{{\hat{h}}}} (\succ _W^{h})$ is equal to ${\succ _W^{({\hat{h}}, \mathtt {R})}}$, and ${\hat{f}}$ makes ${\hat{s}}$ unacceptable when ${\succ _F^{{\hat{h}}}} (\succ _F^{h})$ changes to ${\succ _F^{({\hat{h}}, \mathtt {R})}}$. From Lemma 2, ${\mathrm {DA}}^W[{\succ ^{({\hat{h}},\mathtt {R})}_F},{\succ _W^{({\hat{h}},\mathtt {R})}}]={\mathrm {DA}}^W[{\succ _F^h}, {\succ _W^h}]$. According to the induction hypothesis, we have $\mu [\sigma ;({\hat{h}},\mathtt {R})]({\hat{s}})={\mathrm {DA}}^W[{\succ ^{({\hat{h}},\mathtt {R})}_F}, {\succ _W^{({\hat{h}},\mathtt {R})}}]({\hat{s}})$. Thus, we have

$$\begin{aligned} \mu [\sigma ;h]({\hat{s}})=\mu [\sigma ;({\hat{h}},\mathtt {A})]({\hat{s}}) \succeq _{{\hat{s}}}\mu [\sigma ;({\hat{h}},\mathtt {R})]({\hat{s}}) ={\mathrm {DA}}^W\left[ {\succ _F^h},{\succ _W^h}\right] ({\hat{s}}), \end{aligned}$$

and hence we obtain $\mu [\sigma ;h]({\hat{s}})={\mathrm {DA}}^W[{\succ _F^h},{\succ _W^h}]({\hat{s}})$ from (12). $\square$

Therefore, Theorem 6 is proved because $\mu [h_{\mathrm {init}}](w)=\emptyset$ for all $w\in W$.