Semantic Similarity: Foundations

Abstract

This paper investigates measures of semantic similarity between conversations from an axiomatic perspective. We abstract away from real conversations, representing them as sequences of formulas, equipped with a notion of semantic interpretation that maps them into a different space. An example we use to illustrate our approach is the language of propositional logic with its classical semantics. We introduce and study a range of different candidate properties for metrics on such conversations, for the structure of the semantic space, and for the behavior of the interpretation function, and their interactions. We define four different metrics and explore their properties in this setting.

This research was supported by ERC Grant 269427.

A Appendix: Selected Proofs

(Proof of Fact 5.2). Assume that for every $\overrightarrow{{\varvec{\chi _1}}},\overrightarrow{{\varvec{\chi _2}}}$ with $\Vert \overrightarrow{{\varvec{\chi _1}}}\Vert =\Vert \overrightarrow{{\varvec{\chi _2}}}\Vert$ we have $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _1}}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\chi _2}}})$ (i). Now take some $\overrightarrow{{\varvec{\chi }}}$. We have $\Vert \overrightarrow{{\varvec{\chi }}}\Vert =\Vert \overrightarrow{{\varvec{\chi }}}\Vert$. Hence by (i), we have $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi }}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\chi }}})$. But $\overrightarrow{{\varvec{\chi }}}$ was arbitrary, hence $\forall {\chi } \quad d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi }}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\chi }}})$. It follows, by (sem separation), that $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert =\Vert \overrightarrow{{\varvec{\psi }}}\Vert$.    qed

(Proof of Fact 5.3). Assume that $\forall {\overrightarrow{{\varvec{\chi }}}}$, $d(\overrightarrow{{\varvec{\varphi }}}$, $\overrightarrow{{\varvec{\chi }}})$ $=$ $d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\chi }}})$. In particular $d(\overrightarrow{{\varvec{\varphi }}}$, $\overrightarrow{{\varvec{\psi }}})$ $=$ $d(\overrightarrow{{\varvec{\psi }}}$, $\overrightarrow{{\varvec{\psi }}})=0$. Hence by ($\mathsf{{zero~\Rightarrow ~sem\equiv }}$), $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert =\Vert \overrightarrow{{\varvec{\psi }}}\Vert$.    qed

(Proof of Fact 5.4). The right to left direction follows from Fact 5.3. For the left to right direction, assume that $d(\overrightarrow{{\varvec{\varphi }}}, \overrightarrow{{\varvec{\psi }}})=0$ (i). Take any $\overrightarrow{{\varvec{\chi }}}$. By triangle inequality, $d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}}) \le d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}}) + d(\overrightarrow{{\varvec{\psi }}}, \overrightarrow{{\varvec{\varphi }}})$. Hence, by (i) we have $d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}}) \le d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}})$. Similarly we have $d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}}) \le d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}})$. Hence $d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}}) = d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}})$. But $\chi$ was arbitrary, hence for all $\chi$ we have $d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}}) = d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}})$. By (sem separation), it follows that $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert =\Vert \overrightarrow{{\varvec{\psi }}}\Vert$.    qed

(Proof of Fact 5.7). $\Vert p\lnot p\Vert =\Vert q\lnot q\Vert$ but $\delta _{count}(p\lnot p,q\lnot q)=4$.    qed

(Proof of Fact 5.8). $\Vert p\lnot p\Vert =\Vert q\lnot q\Vert$ but $\delta _{synt,count}(p\lnot p,q\lnot q)=6$.    qed

(Proof of Fact 5.9). Take some $\varphi$ such that $\mathsf{{card}}(\Vert \varphi \Vert )\ge 2$.    qed

(Proof of Fact 5.10). Assume that $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert =\Vert \overrightarrow{{\varvec{\psi }}}\Vert$. By (sem preservation) we have $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi }}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\chi }}})$. In particular we have $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\psi }}})=0$    qed

(Proof of Fact 5.11). Assume that $\Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert =\Vert \overrightarrow{{\varvec{\psi _1}}}\Vert$. Take some $\overrightarrow{{\varvec{\chi }}}$. We have $\Vert \overrightarrow{{\varvec{\chi }}}\Vert =\Vert \overrightarrow{{\varvec{\chi }}}\Vert$. Hence by (sem induced) we have $d(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\chi }}})=d(\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\chi }}})$.    qed

(Proof of Fact 5.12). Assume that $\Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert =\Vert \overrightarrow{{\varvec{\psi _1}}}\Vert$ (i) and $\Vert \overrightarrow{{\varvec{\varphi _2}}}\Vert =\Vert \overrightarrow{{\varvec{\psi _2}}}\Vert$ (ii). By triangle inequality we have:

$$\begin{aligned} d(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})\le \underbrace{d(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\psi _1}}})}_{0, \text { by} \mathrm{(}{\mathsf{{sem\equiv ~\Rightarrow ~zero}}}\mathrm{)}}+d(\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\varphi _2}}})\\ d(\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\varphi _2}}})\le d(\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\psi _2}}})+\underbrace{d(\overrightarrow{{\varvec{\psi _2}}},\overrightarrow{{\varvec{\varphi _2}}})}_{0, \text { by}\mathrm{( }{\mathsf{{sem\equiv ~\Rightarrow ~zero}}}\mathrm{)}} \end{aligned}$$

Hence, $d(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})\le d(\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\psi _2}}})$. Similarly, we have $d(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})\ge d(\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\psi _2}}})$.    qed

(Proof of Corollary 5.13). By Fact 5.12, $d$ satisfies (sem induced), hence for every $\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}},\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\psi _2}}}$ with $\Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert =\Vert \overrightarrow{{\varvec{\psi _1}}}\Vert \text { and }\Vert \overrightarrow{{\varvec{\varphi _2}}}\Vert =\Vert \overrightarrow{{\varvec{\psi _2}}}\Vert$ we have

$$\begin{aligned} d(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})=d(\overrightarrow{{\varvec{\psi _1}}},\overrightarrow{{\varvec{\psi _2}}}) \end{aligned}$$

It follows that $\dot{d}(\Vert \overrightarrow{{\varvec{\varphi }}}\Vert$, $\Vert \overrightarrow{{\varvec{\psi }}}\Vert )$ $:=$ $d(\overrightarrow{{\varvec{\varphi }}}$, $\overrightarrow{{\varvec{\psi }}})$ is well-defined. Moreover for any $\overrightarrow{{\varvec{\varphi }}}$, $d(\Vert \overrightarrow{{\varvec{\varphi }}}\Vert , \Vert \overrightarrow{{\varvec{\varphi }}}\Vert )$ $=$ $d(\overrightarrow{{\varvec{\varphi }}}, \overrightarrow{{\varvec{\varphi }}})=0$. Triangle inequality is proven similarly.    qed

(Proof of Fact 5.14). First observe, that by triangle inequality, we have

$$\begin{aligned} d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}}) \le d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}}) + d(\overrightarrow{{\varvec{\psi }}}, \overrightarrow{{\varvec{\varphi }}}) \end{aligned}$$

Now, assume that $\overrightarrow{{\varvec{\varphi }}}\equiv \overrightarrow{{\varvec{\psi }}}$. By ($\mathsf{{sem\equiv ~\Rightarrow ~zero}}$) we have $d(\overrightarrow{{\varvec{\psi }}}, \overrightarrow{{\varvec{\varphi }}}) = 0$, hence $d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}}) \le d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}})$. Similarly, $d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\psi }}}) \le d(\overrightarrow{{\varvec{\chi }}}, \overrightarrow{{\varvec{\varphi }}})$ which proves (1).    qed

(Proof of Fact 5.15). Assume that $\forall \chi$ we have $d(\varphi ,\chi )=d(\psi ,\chi )$ (i). Take some $\overrightarrow{{\varvec{\chi _1}}},\overrightarrow{{\varvec{\chi _2}}}$ with $\Vert \overrightarrow{{\varvec{\chi _1}}}\Vert =\Vert \overrightarrow{{\varvec{\chi _2}}}\Vert$. By ($\mathsf{{sem\equiv ~\Rightarrow ~zero}}$) we have $d(\overrightarrow{{\varvec{\chi _1}}},\overrightarrow{{\varvec{\chi _2}}})=0$ (ii). By triangle inequality we have:

$$\begin{aligned} d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _1}}})\le {d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _2}}})}+\underbrace{d(\overrightarrow{{\varvec{\chi _2}}},\overrightarrow{{\varvec{\chi _1}}})}_{0, \text { by}(ii)}\\ d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _2}}})\le {d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _1}}})}+\underbrace{d(\overrightarrow{{\varvec{\chi _1}}},\overrightarrow{{\varvec{\chi _2}}})}_{0, \text { by}(ii)}\\ \end{aligned}$$

Hence $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _1}}})=d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _2}}})$. Moreover by (i) we have $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _2}}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\chi _2}}})$. Hence $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\chi _1}}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\chi _2}}})$. Since $\overrightarrow{{\varvec{\chi _1}}},\overrightarrow{{\varvec{\chi _2}}}$ were arbitrary, it follows by (min sem separation), that $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert =\Vert \overrightarrow{{\varvec{\psi }}}\Vert$.    qed

(Proof of Fact 6.12). Let $v$ be an isotone upper valuation and $z \le x,y$. Assume without loss of generality that $v(y) \ge v(x)$. We can write, for $0 \le \alpha \le 1$:

$$-\alpha \cdot 2^{-v(y)} \ge -2^{-v(y)} \ge -2^{-v(x)}, $$

which ensures that

$$ -\alpha \cdot 2^{-v(y)} - \frac{1}{\alpha } 2^{-v(x)} \le -2^{-v(y)} - 2^{-v(x)}.$$

Instantiating this result for $\alpha = 2^{v(z)-v(x)}$ yields after development

$$-2^{v(z) - v(x) - v(y)} -2^{-v(z)} \le -2^{-v(y)} - 2^{-v(x)}. $$

Since $v$ is an upper-valuation, we have $-v(x \vee y) \ge v(z) - v(x) - v(y)$ and thus

$$-2^{-v(x \vee y)} - 2^{-v(z)} \le -2^{v(z) - v(x) - v(y)} -2^{-v(z)} \le -2^{-v(y)} -2^{-v(x)}$$

i.e., $w(x \vee y) + w(z) \le w(x) + w(y)$ which concludes the proof.

The case $v(x) \ge v(y)$ is symmetrically dealt with.    qed

(Proof of Fact 6.16). Let $\overrightarrow{{\varvec{\varphi }}}$ $:=$ $(p_1\wedge p_2)\vee (\lnot p_1\wedge \lnot p_2 \wedge \lnot p_3)$ and $\overrightarrow{{\varvec{\psi }}}$ $:=$ $(p_1\wedge (p_2\rightarrow p_3))$, and assume some intersective interpretation of concatenation. We have $\Vert \overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}}\Vert$ $=$ $\Vert p_1\wedge p_2 \wedge p_3\Vert$. $d_H^{ham}(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}})=3$, but $d_H^{ham}(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}})=1$.    qed

(Proof of Fact 6.18). For any $\varphi$, $d_H^{ham}$ is neither well-defined for $(\varphi ,\bot )$ nor for $(\bot ,\varphi )$.    qed

(Proof of Fact 6.21). Adding a new propositional letter that does not occur in either sequence will not affect the symmetric difference of the range of formulas, nor the symmetric difference of the respective signature. Allowing for the negation of the propositional letter that was previously forbidden will not change the sets either.    qed

(Proof of Fact 7.1). Assume that $d(\varphi ,\chi )\le d(\psi ,\chi )$ then $d(\varphi \psi \chi ,\chi \psi \chi )\le d(\psi \psi \chi ,\chi \psi \chi )$. By $\mathsf{{exchange}}$, $\mathsf{{contraction}}$, $\mathsf{{expansion}}$, and ($\mathsf{{sem\equiv ~\Rightarrow ~zero}}$), we have $d(\psi \psi \chi ,\chi \psi \chi )=0$. Hence $d(\varphi \psi \chi ,\chi \psi \chi )=0$. By $\mathsf{{exchange}}$, $\mathsf{{contraction}}$, $\mathsf{{expansion}}$, and ($\mathsf{{sem\equiv ~\Rightarrow ~zero}}$), we have $d(\varphi \psi \chi ,\chi \psi )=0$. Concluding our proof.    qed

(Proof of Corollary 7.2). We only give the idea of the proof. The idea of the proof is to define a linear order on $\mathsf{{L}}^*$ compatible with $\le _o$. By induction, using Fact 7.1 we first show the claim for formulas in the same $o$-equivalence class, then we show that the claim propagate downward, that is for every $\overrightarrow{{\varvec{\psi }}}\in \downarrow \![\overrightarrow{{\varvec{\varphi }}}]$. Finally we show that the claim propagates with transitive closure.    qed

(Proof of Corollary 7.3). Direct from Fact 5.15 and Corollary 7.2.    qed

(Proof of Fact 7.4). Let $k\ge 2$, $n=2k$. Now let

$$\begin{aligned} \varphi&:=p_1\rightarrow (\lnot p_2\wedge \ldots \wedge \lnot p_n)\wedge \lnot p_1\rightarrow (p_2\wedge \ldots \wedge p_n),\\ \psi&:=p_1 \wedge \lnot p_2 \wedge \ldots \wedge p_{2k-1} \wedge \lnot p_{2k}, \end{aligned}$$

$\chi :=p_1\wedge \ldots \wedge p_n$ and $\varphi _0:=p_1$. Since $k\ge 2$ we have

$$\begin{aligned}1=d_H^{ham}(\varphi ,\chi )&< d_H^{ham}(\varphi ,\chi )=k,\text { and,}\\n-1=d_H^{ham}(\varphi \varphi _0,\chi \varphi _0)&> d_H^{ham}(\varphi \varphi _0,\chi \varphi _0)=k=n/2\end{aligned}$$

Concluding our proof.    qed

(Proof of Fact 7.6). $(1\Rightarrow 2)$. Take some $\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}}$. By $\mathsf{{contraction}}$, $\mathsf{{expansion}}$ and $\mathsf{{exchange}}$ we have

$$\begin{aligned} \Vert \overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}}\Vert =\Vert \overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}}\Vert \end{aligned}$$

Hence by ($\mathsf{{sem\equiv ~\Rightarrow ~zero}}$), $d(\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}}$,$\overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}})=0$ (i). Hence $d(\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}})\le d(\overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}},{\overrightarrow{{\varvec{\psi }}}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}})$ (i). Now, let $\overrightarrow{{\varvec{\chi }}}=\overrightarrow{{\varvec{\psi }}}$ and $\overrightarrow{{\varvec{\varphi _0}}}=\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}}$. By (i) and (uniform anti- preservation) we have

$$\text {If }d(\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}})\le d(\overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}},{\overrightarrow{{\varvec{\psi }}}}\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}}) \text { then }d(\overrightarrow{{\varvec{\varphi }}},{\overrightarrow{{\varvec{\psi }}}})\le d(\overrightarrow{{\varvec{\psi }}},{\overrightarrow{{\varvec{\psi }}}})=0$$

Hence by (i), $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}})=0$. Concluding the proof for this direction. The other direction is trivial.    qed

(Proof of Fact 7.8). Let $n\in \omega$ be such that $n>5$. Moreover let: $\varphi :=(p_3\wedge \ldots \wedge p_n)\wedge \lnot (p_1\wedge p_2)$, $\psi _1:=(p_1\vee \ldots \vee p_n)\wedge \lnot (p_2\wedge \ldots \wedge p_n)$ and $\psi _2:=(p_2\wedge \ldots \wedge p_n)$. We have $d_{\alpha }(\varphi ,\psi _1)=1-\frac{2}{n+3}=\frac{n+1}{n+3}>d_{\alpha }(\varphi ,\psi _2)=1-\frac{1}{4}=\frac{3}{4}$. But we have $d_{\alpha }(\varphi ,\varphi \psi _1)=1-\frac{2}{3}=\frac{1}{3}<d_{\alpha }(\varphi ,\varphi \psi _2)=1-\frac{1}{3}=\frac{2}{3}$.    qed

(Proof of Fact 7.9). Take some $\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}}$. By (coherent deviation) we have

$$\begin{aligned} \text {If }d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})<d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\epsilon }}})\text { then }d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\epsilon }}})<d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\epsilon }}}) \end{aligned}$$

$$\begin{aligned} \text {If }d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\epsilon }}})<d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})\text { then }d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\psi }}}\overrightarrow{{\varvec{\epsilon }}})<d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\epsilon }}}) \end{aligned}$$

Since $\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}}$ were arbitrary, it follows that for any $\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}}$, $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\epsilon }}})$. In particular $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})=d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\epsilon }}})=d(\overrightarrow{{\varvec{\epsilon }}},\overrightarrow{{\varvec{\epsilon }}})=0$. Hence by (triangle inequality) we have $\forall {\overrightarrow{{\varvec{\varphi }}}, \overrightarrow{{\varvec{\psi }}}} d(\overrightarrow{{\varvec{\varphi }}}, \overrightarrow{{\varvec{\psi }}}) \le d(\overrightarrow{{\varvec{\varphi }}}, \overrightarrow{{\varvec{\epsilon }}}) + d(\overrightarrow{{\varvec{\epsilon }}}, \overrightarrow{{\varvec{\psi }}}) = 0$.    qed

(Proof of Fact 7.10). $(1\Rightarrow 2)$. Take some $\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}}$. We have

$$0=d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\epsilon }}})\le d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\psi }}})$$

Hence by (converse strong action pref) $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})\le d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}})$. But $\psi$ was arbitrary, hence, in particular $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})\le d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}})=0$. But $\varphi$ was arbitrary as well, hence $\forall {\chi } d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})=0$. Hence by triangle inequality for any formula $\overrightarrow{{\varvec{\varphi }}}, \overrightarrow{{\varvec{\psi }}}$ we have $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}})\le d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})+d(\overrightarrow{{\varvec{\epsilon }}},\overrightarrow{{\varvec{\psi }}})=0$. Concluding our proof.    qed

(Proof of Fact 7.11). $(1\Rightarrow 2)$. Take some $\overrightarrow{{\varvec{\varphi }}}$ and $\overrightarrow{{\varvec{\psi }}}$. We have

$$0=d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}}\overrightarrow{{\varvec{\epsilon }}})\le d(\overrightarrow{{\varvec{\epsilon }}},\overrightarrow{{\varvec{\epsilon }}}\overrightarrow{{\varvec{\epsilon }}})=0$$

By (converse coherent deviation) $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})\le d(\overrightarrow{{\varvec{\epsilon }}},\overrightarrow{{\varvec{\epsilon }}})=0$ (i). Similarly, we have $d(\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\epsilon }}})=0$ (ii). By (i), (ii) and triangle inequality we have $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}})\le$   $d(\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\epsilon }}})+$ $d(\overrightarrow{{\varvec{\epsilon }}},\overrightarrow{{\varvec{\psi }}})=0$. Concluding our proof.    qed

(Proof of Fact 8.1). Take some $\overrightarrow{{\varvec{\chi }}}$ and assume that $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert \le \Vert \overrightarrow{{\varvec{\psi }}}\Vert$. Since $\mathsf{{target}}(\Vert \cdot \Vert )$ is a lattice. We have $\Vert \overrightarrow{{\varvec{\chi }}}\Vert \wedge \Vert \overrightarrow{{\varvec{\varphi }}}\Vert \le \Vert \overrightarrow{{\varvec{\chi }}}\Vert \wedge \Vert \overrightarrow{{\varvec{\psi }}}\Vert$. Hence by (${\mathsf{{strong~\wedge ~rule}}}$) $\delta (\overrightarrow{{\varvec{\chi }}},\overrightarrow{{\varvec{\varphi }}})\ge \delta (\overrightarrow{{\varvec{\chi }}},\overrightarrow{{\varvec{\psi }}})$.    qed

(Proof of Corollary 8.2). Assume that $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert \le \Vert \overrightarrow{{\varvec{\psi }}}\Vert$. Since $d$ satisfies (${\mathsf{{strong~\wedge ~rule}}}$), we have by Fact 8.1 we have in particular $0=\delta (\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\varphi }}})\ge \delta (\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}})=\delta (\overrightarrow{{\varvec{\psi }}},\overrightarrow{{\varvec{\varphi }}})$.    qed

(Proof of Corollary 8.3). $(1\Rightarrow 2)$. Take two arbitrary $\overrightarrow{{\varvec{\varphi }}},\overrightarrow{{\varvec{\psi }}}$. By $(\overrightarrow{{\varvec{\epsilon }}}$–$\top )$ we have $\Vert \overrightarrow{{\varvec{\varphi }}}\Vert \le \Vert \overrightarrow{{\varvec{\epsilon }}}\Vert$ and $\Vert \overrightarrow{{\varvec{\psi }}}\Vert \le \Vert \overrightarrow{{\varvec{\epsilon }}}\Vert$. Since $d$ satisfies (${\mathsf{{strong~\wedge ~rule}}}$), it follows by Corollary 8.2 that $d(\overrightarrow{{\varvec{\varphi }}},{\overrightarrow{{\varvec{\epsilon }}}})=d(\overrightarrow{{\varvec{\epsilon }}},{\overrightarrow{{\varvec{\psi }}}})=0$. By triangle inequality it follows that $d(\overrightarrow{{\varvec{\varphi }}},{\overrightarrow{{\varvec{\psi }}}})=0$. The $(2\Rightarrow 1)$ direction is trivial.    qed

(Proof of Fact 8.5). Take $\overrightarrow{{\varvec{\varphi }}}$, $\overrightarrow{{\varvec{\psi _1}}}$, $\overrightarrow{{\varvec{\psi _2}}}$ with $\overrightarrow{{\varvec{\varphi }}} \cap \overrightarrow{{\varvec{\psi _1}}} \subsetneq \overrightarrow{{\varvec{\varphi }}} \cap \overrightarrow{{\varvec{\psi _2}}}$. And consider the cardinalities assigned in Fig. 2. Let $X_1:=\alpha _1+\eta _0$ and let $X_2:=\alpha _2+\eta _0$.

Fig. 2.

Assigning cardinalities to the respective intersections.

Using these cardinalities and inserting them in the expression of the distance, gives us:

$$\begin{aligned} d_{\mathcal {C}}(\varphi , \psi _1)&= \frac{2^{\mathsf{{card}}(\Vert \overrightarrow{{\varvec{\varphi }}})\Vert }+2^{\mathsf{{card}}(\Vert \overrightarrow{{\varvec{\psi _1}}}\Vert )}-2\cdot 2^{\mathsf{{card}}(\Vert \overrightarrow{{\varvec{\varphi }}}\Vert \cap \Vert \overrightarrow{{\varvec{\psi _1}}}\Vert )}}{2^{\mathsf{{card}}(\Vert \overrightarrow{{\varvec{\varphi }}}\Vert ) + \mathsf{{card}}(\Vert \overrightarrow{{\varvec{\psi _1}}}\Vert )}}\\ {}&= \frac{2^{X_\varphi +\eta _1+\eta _2}+2^{X_1 + \eta _1}-2\cdot 2^{\eta _1}}{2^{X_{\varphi }+X_1+2\eta _1+\eta _2}} \end{aligned}$$

Similarly:

$$\begin{aligned} d_{\mathcal {C}}(\varphi , \psi _2) = \frac{2^{X_{\varphi } + \eta _1 + \eta _2}+2^{X_2 + \eta _1 + \eta _2}-2\cdot 2^{\eta _1 + \eta _2}}{2^{X_{\varphi } + X_2 +2\eta _1 + 2\eta _2}} \end{aligned}$$

From the two previous expression, after simplifications we find:

$$\begin{aligned} d_{\mathcal {C}}(\varphi , \psi _1) - d_{\mathcal {C}}(\varphi , \psi _2) = \frac{2^{X_{\varphi }}(2^{X_{2}+\eta _2} - 2^{X_1}) + 2(2^{X_1}- 2^{X_2})}{2^{X_{\varphi }+X_1+X_2+\eta _1+\eta _2}} \end{aligned}$$

From the assumption that $\overrightarrow{{\varvec{\varphi }}} \cap \overrightarrow{{\varvec{\psi _1}}} \subsetneq \overrightarrow{{\varvec{\varphi }}} \cap \overrightarrow{{\varvec{\psi _2}}}$, it follows that $\eta _2 \ge 1$, hence:

$$\begin{aligned} 2^{X_{\varphi }}(2^{X_{2}+\eta _2} - 2^{X_1}) + 2(2^{X_1}- 2^{X_2}) \ge 2^{X_{2}+\eta _2} - 2^{X_2+1} + 2^{X_1+1} - 2^{X_1} \ge 0 \end{aligned}$$

which concludes the proof.    qed

(Proof of Fact 8.7). Take $\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\varphi }}}_2$. Assume that $\Vert \overrightarrow{{\varvec{\psi }}}\Vert =\Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert \vee \Vert \overrightarrow{{\varvec{\varphi _2}}}\Vert$ (i). By definition, we have

$$\begin{aligned} d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\varphi }}}_2)=\max \{\max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\min _{y\in \Vert \overrightarrow{{\varvec{\varphi _2}}}\Vert } \delta (x,y),\\ {}&\max _{y\in \Vert \overrightarrow{{\varvec{\varphi _2}}}\Vert } \min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert } \delta (x,y)\} \end{aligned}$$

Hence, we are in one of two cases.

• (1) $\max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert } \min _{y\in \Vert \overrightarrow{{\varvec{\varphi _2}}}\Vert } \delta (x,y)=d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$, or,

• (2) $\max _{y\in \Vert \overrightarrow{{\varvec{\varphi _2}}}\Vert } \min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert } \delta (x,y)=d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$.

Case 1. There are two subcases.

Subcase 1a. Assume that $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\psi }}})=\max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }$ $\min _{y\in \Vert \overrightarrow{{\varvec{\psi }}}\Vert }$ (a). By (i),

$$\max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\min _{y\in \Vert \overrightarrow{{\varvec{\psi }}}\Vert }\le \max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\min _{y\in \Vert \overrightarrow{{\varvec{\varphi }}}_2\Vert }$$

By (a) and (1) we have $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\psi }}})\le d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$.

Subcase 1b. $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\psi }}})= \max _{y\in \Vert \overrightarrow{{\varvec{\psi }}}\Vert }\min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\delta (x,y)$ (b). Since $\delta$ is a metric we have $\max _{y\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\delta (x,y)=0$. Hence by (i) and (b) we have $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\psi }}})= \max _{y\in \Vert \overrightarrow{{\varvec{\varphi }}}_2\Vert }\min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\delta (x,y)$ (ii). But by (1) and definition of $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$ we have $\max _{y\in \Vert \overrightarrow{{\varvec{\varphi }}}_2\Vert }\min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\delta (x,y)\le d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$ (iii). By (ii) and (iii) we have $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\psi }}})\le d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$.

Case 2. Since $\delta$ is a metric we have

$$\max _{y\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\delta (x,y)=0$$

Hence by (2), we have

We now consider two subcases.

Subcase 2a. $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\psi }}})=\max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert } \min _{y\in \Vert \overrightarrow{{\varvec{\psi }}}\Vert }$ (a). But since $\delta$ is a metric we have $\max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert } \min _{y\in \Vert \overrightarrow{{\varvec{\psi }}}\Vert }\le \max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert } \min _{y\in \Vert \overrightarrow{{\varvec{\varphi }}}_2\Vert }$ (v). But by definition $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$ and (2) we have $\max _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert } \min _{y\in \Vert \overrightarrow{{\varvec{\varphi }}}_2\Vert }\le d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\varphi }}}_2)$ (vi). By (v), (vi) and (a), it follows that $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\psi }}})\le d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$.

Subcase 2b. Assume that. $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\psi }}})= \max _{y\in \Vert \overrightarrow{{\varvec{\psi }}}\Vert }\min _{x\in \Vert \overrightarrow{{\varvec{\varphi _1}}}\Vert }\delta (x,y)$ (b). By (iv), it follows that $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\psi }}})$ $=$ $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$.

Hence in all cases $d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi }}}_1,\overrightarrow{{\varvec{\psi }}})\le d_{H}^{\delta }(\overrightarrow{{\varvec{\varphi _1}}},\overrightarrow{{\varvec{\varphi _2}}})$. Concluding our proof.    qed