Axioms for signatures with domain and demonic composition

Abstract

Demonic composition $*$ is an associative operation on binary relations, and demonic refinement ${\sqsubseteq }$ is a partial order on binary relations. Other operations on binary relations considered here include the unary domain operation D and the left restrictive multiplication operation $\circ$ given by $s\circ t=D(s)*t$. We show that the class of relation algebras of signature $\{\, \sqsubseteq , D, *\, \}$, or equivalently $\{\, \subseteq , \circ , *\, \}$, has no finite axiomatisation. A large number of other non-finite axiomatisability consequences of this result are also given, along with some further negative results for related signatures. On the positive side, a finite set of axioms is obtained for relation algebras with signature $\{\, \sqsubseteq , \circ , *\, \}$, hence also for $\{\, \subseteq , \circ , *\, \}$.

Appendix

Proof of Lemma 2.5. Recall from Definition 2.3 the $(\le , ;)$-structure ${{\mathcal {A}}}_n$.

Definition 5.1

An element $s\in {{\mathcal {A}}}_n$ is i-short if there are $\alpha _0, \alpha _1, \dots , \alpha _{i-1}\in \{\, {\bar{g}}, f\, \}^*$, $\beta _0, \beta _1, \dots , \beta _{i-1}\in \{\, g, {\bar{f}}\, \}^*$ such that $s\le \alpha _0\beta _0\dots \alpha _{i-1} \beta _{i-1}$. (For example, $b\le (fg)^n$ is n-short but not $n-1$-short.)

A label $s\in N(x, y)$ is witnessed in N if for all u, t where $s=ut$, $ut={{\,\mathrm{\mathsf{nf}}\,}}(ut)$ and u, t are minimal subject to $u; v\in N(x, y)$, there is $z\in \mathsf{nodes}(N)$ such that $u\in N(x, z)$, $t\in N(z, y)$. For $i<n$, an ${{\mathcal {A}}}_n$-network N is i-good if every i-short, minimal label is witnessed.

Lemma 5.2

Let $i, j<\omega$. If $s\in I^*$ and s is i-short then ${\bar{s}}$ is also i-short. If s is i-short and t is j-short and $u\le s; t$ then u is $(i+j)$-short. If s; t is i-short then either $s; t=\Lambda$ or s is i-short.

For the last statement, note that if $s; t=\Lambda$ then $s\in \{\, {\bar{g}}, f\, \}^*$ and $t={\bar{s}}$. Then s is 1-short and if it is non-empty it is not 0-short, but $s; t=\Lambda$ is 0-short. Suppose $s; t\ne \Lambda$. Since $s\in I^*$ we may write $s=\alpha _0\beta _0\dots \alpha _{j-1}\beta _{j-1}$ where $\alpha _k\in \{\, {\bar{f}}, g\, \}^*$, $\beta _k\in \{\, f, {\bar{g}}\, \}^*$ for some j and we may assume that none of the $\alpha _k$ or $\beta _k$ in the decomposition of s is empty except possibly $\alpha _0$ or $\beta _{j-1}$. If s is not i-short then $j>i$. When we multiply s on the right by $t\in I^*$ we might possibly cancel the final $\beta _{j-1}$ (in the case where $t=\overline{\beta _{j-1}}t_0$), but all the other parts of the decomposition of s will be unaffected, hence s; t is not i-short.

In the following, whenever we refer to i-short elements we have $j\le n$, so an i-short element will not involve b.

Lemma 5.3

In a play of $\Gamma _k({{\mathcal {A}}}_n)$, if the current network N is $2^i$-good, then for any move by $\forall$, there is an $2^{i-1}$-good extension $N^+$ of N that $\exists$ can play.

Proof

The proof differs from the proof of [7, Theorem 19] here, because domain moves and range moves are not treated there. Suppose $\forall$ makes a domain move (d, x, a). First suppose $a\in \Sigma$ is a one letter word. If there is $w\in \mathsf{nodes}(N)$ where $a\in N(x, w)$ then she lets $N^+=N$, otherwise she adds a single new node z to N, lets $N^+(z, z)=\Lambda ^\uparrow$ and

$$\begin{aligned} N^+(v, z)&=(N(v, x); a)^\uparrow ,&N^+(z, v)&={\left\{ \begin{array}{ll}\emptyset &{} \text{ if } a=b\\ ({\bar{a}}; N(x, v))^\uparrow &{} \text{ if } a\in I,\end{array}\right. } \end{aligned}$$

for $v\in \mathsf{nodes}(N)$. If $a=b$ then no new irreflexive label is $2^{i}$-short. If $a\in I$ (so a is 1-short), $s\in N(v, x)$ and $s; a\in N^+(v, z)$ is minimal and $2^{i}$-short then either $s; a=\Lambda$ or s is $2^i$-short, by Lemma 5.2. The former case is ruled out by our assumption that no suitable $w\in \mathsf{nodes}(N)$ exists. In the latter case, inductively, we have ${\bar{s}}\in N(x, v)$. Also, we must have $s; a =sa$ (as otherwise a suitable witness already exists in N). If $sa\le ut={{\,\mathrm{\mathsf{nf}}\,}}(ut)$ where $ut\in N(v, z)$ is minimal then $sa=ut$, $t=t_0a$ (some $t_0$) and $s= u t_0$ where $ut_0\in N(v, w)$ is minimal, hence inductively there is $w\in \mathsf{nodes}(N)$ where $u\in N(v, w)$ and $t_0\in N(w, x)$. It follows that $u\in N(v, w)$ and $t=t_0a\in N(w, z)$, so $N^+$ is a consistent, $2^i$-good extension of N. More generally for $a\in \Sigma ^*$ she computes her extension $N^+$ of N by iterating the previous extension |a| times, still $2^i$-good. Her response to a range move (r, x, a) is defined similarly.

Suppose $\forall$ plays a composition move (c, x, y, s, t), so we can assume that s, t are minimal subject to $s; t\in N(x, y)$, that is,

$$\begin{aligned} (s'\le s)\wedge (t'\le t)\wedge (s'; t'\in N(x, y))\implies (s'=s)\wedge (t'=t). \end{aligned}$$

If s, t are both $2^{i-1}$-short then $s; t\in N(x, y)$ is $2^i$-short by Lemma 5.2. By (2.1) and minimality of s, t, there are $s_0, t_0, u\in \{\, f, {\bar{f}}, g, {\bar{g}}\, \}^*$ such that $s=s_0u$, $t={\bar{u}} t_0$ and $s_0t_0\in N(x, y)$ is minimal. Since N is $2^i$-good there is $v\in \mathsf{nodes}(N)$ such that $N(x, v)=s_0$, $N(v, y)=t_0$. $\exists$ pretends (to herself) that $\forall$ played (d, v, u) to compute her extension $N^+$, as above, so $s=s_0; u\in N^+(x, z)$, $t={\bar{u}}; t_0\in N^+(z, y)$. Again, $N^+$ is a legal $2^i$-good (hence $2^{i-1}$-good) extension of N.

Suppose s is $2^{i-1}$-short but t is not. First suppose $s=c\in I$, a one letter word. If there is $v\in \mathsf{nodes}(N)$ such that $N(x, v)=c$, $N(v, y)\le t$ then $\exists$ may let $N^+=N$, if not $\exists$ adds a single new node z to the network and lets

$$\begin{aligned} N^+(v, z)&=(N(v, x); c)^\uparrow ,\\ N^+(z, z)&=\Lambda ^\uparrow ,\\ N^+(z, v)&=({\bar{c}}; N(x, v)\cup t; N(y, v))^\uparrow , \end{aligned}$$

for $v\in \mathsf{nodes}(N)$.

Since t is not $2^{i-1}$-short, the only new $2^{i-1}$-short labels have the form $u; c\in N^+(v, z)$ and ${\bar{c}}{\bar{u}}\in N^+(z,v)$, for $2^{i-1}$-short $u\in N(v, x)$, and since we are assuming that no witness for the composition exists in N we must have $uc={{\,\mathrm{\mathsf{nf}}\,}}(uc)$. As in the proof that her response to domain moves is $2^i$-good, a minimal label $uc={{\,\mathrm{\mathsf{nf}}\,}}(uc)\in N^+(v, z)$ is witnessed, where $u\in N(v, x)$ is $2^{i}$-short. Hence $N^+$ is $2^{i-1}$-good. More generally, in response to a move (c, x, y, s, t) where s is $2^{i-1}$-short but t is not, she computes her $2^{i-1}$-good response $N^+$ by iterating the preceeding 1-character case |s| times. The case where t is $2^{i-1}$-short but s is not is similar.

Finally, if neither s nor t is $2^{i-1}$-short then $\exists$ adds a single new node z and lets $N^+(v, z)=(N(v, x); s)^\uparrow$, $N^+(z, v)=(t; N(y, v))^\uparrow$, for $v\in \mathsf{nodes}(N)$ and of course $N^+(z, z)=\Lambda ^\uparrow$. No irreflexive edges incident with the new node z have $2^{i-1}$-short labels by Lemma 5.2, so $N^+$ is $2^{i-1}$-good. $\square$

We are now in a position to prove Lemma 2.5.

Proof

In the initial round, if $\forall$ plays $a\not \le c$ and a is not $2^k$-short then $\exists$ plays $N_0$ where $\mathsf{nodes}(N_0)=\{\, x, y\, \}$, $N_0(x, y)=a^\uparrow$, $N_0(x, x)=N_0(y, y)=\Lambda ^\uparrow$ and $N_0(y, x)=\emptyset$. If a is $2^k$-short then it is in $I^*$, say $a=a_0a_1\dots a_{|a|-1}$, then $N_0$ has $|a|+1$ nodes $x=x_0, x_1, \dots , x_{|a|}=y$ and $N_0(x_i, x_j)=a[i, j]^\uparrow$, for all $i, j\le |a|$. Observe that $N_0$ is $2^k$-good. By the previous lemma, $\exists$ can play a $2^{k-i}$-good extension network $N_i$ in each round. Since $N_i$ is an extension of $N_0$ we have $c\not \in N_i(x, y)=N_0(x, y)$, so $\exists$ wins the play. $\square$