Neglect-Zero Effects in Dynamic Semantics

Abstract

The article presents a bilateral update semantics for epistemic modals which captures their discourse dynamics [54] as well as their potential to give rise to fc inferences [58]. The latter are derived as neglect-zero effects as in [3]. Neglect-zero is a tendency in human cognition to disregard structures that verify sentences by virtue of an empty witness set. The upshot of modelling the neglect-zero tendency in a dynamic setting is a notion of dynamic logical consequence which makes interesting predictions concerning possible divergences between everyday and logico-mathematical reasoning.

Appendix

1.1 BiUS, US and PL

Theorem 1

$\alpha _1, \dots , \alpha _n \models _{BiUS} \beta \text { iff } \alpha _1, \dots , \alpha _n \models _{US} \beta \ [\text {if } \alpha _1,... , \alpha _n, \beta \in L_{US}]$

Proof

We only need to check the case of negation, i.e. show that $s[\gamma ]^r=s - s[\gamma ]$ for all s and $\gamma \in L_{PL}$ (recall that $\Diamond$ cannot appear in the scope of negation in $L_{US}$). We prove this by induction on the complexity of $\gamma$.

• (i) $s[p]^r = s \cap \{w \in W \mid V(p,w)=0\} =s - \{w \in W \mid V(p,w)=1\} = s - W[p]=s- s[p]$

• (ii) $s[\alpha \wedge \beta ]^r= s[\alpha ]^r \cup s[\beta ]^r =_{IH} (s - s[\alpha ) \cup (s - s[\beta ])= s -(s[\alpha ] \cap s[\beta ]) = s - s[\alpha \wedge \beta ]$

• (iii) $s[\alpha \vee \beta ]^r= s[\alpha ]^r \cap s[\beta ]^r =_{IH} (s - s[\alpha ]) \cap (s - s[\beta ])= s -(s[\alpha \cup s[\beta ) = s - s[\alpha \vee \beta ]$

• (iv) $s[\lnot \alpha ]^r= s[\alpha ]$. Since ${s[\alpha ]\subseteq s}$ by eliminativity, $s[\alpha ]=s -(s - s[\alpha ]) =_{IH} s - s[\alpha ]^r = s - s[\lnot \alpha ]$.

Theorem 2

$\alpha _1, \dots , \alpha _n \models _{BiUS} \beta \text { iff } \alpha _1, \dots , \alpha _n \models _{PL} \beta \ [\text {if } \alpha _1,\dots , \alpha _n, \beta \in L_{PL}]$

Proof

This follows from the fact that in BiUS (just like in US, see [54], page 231), all $\alpha \in L_{PL}$ are such that for any s, $s[\alpha ]= s \cap W[\alpha ]$.

1.2 Ignorance and Free Choice

The proofs of the facts below use the following lemmas.

Lemma 1

For $\alpha \in L_{BiUS}$ and ne-free, and any state s.

• (i) If $s[|\alpha |^+]$ is defined, then $s[|\alpha |^+] = s[\alpha ]$

• (ii) If $s[|\alpha |^+]^r$ is defined, then $s[|\alpha |^+]^r = s[\alpha ]^r$

Proof

By an easy double induction on the complexity of $\alpha$.

Lemma 2

For $\alpha \in L_{US}$ and any state s.

• (i) Idempotence: $s[\alpha ]=s[\alpha ][\alpha ]$ and $s[\lnot \alpha ]=s[\lnot \alpha ][\lnot \alpha ]$

• (ii) Monotonicity: $s\subseteq t$ implies $s[\alpha ] \subseteq t[\alpha ]$

• (iii) Downward closure of $\lnot \alpha$: $s\subseteq t$ implies $t[\lnot \alpha ]=t$ $\Rightarrow$ $s[\lnot \alpha ]=s$.

Proof

These properties are consequences of the following two facts: (a) in $L_{US}$ all might-formulas have the form $\Diamond \alpha$, where $\alpha$ is $\Diamond$-free; (b) all $\Diamond$-free $\alpha$ (i.e., $\alpha \in L_{PL}$) are such that for all s, $s[\alpha ]= s \cap W[\alpha ]$.

Lemma 3

(Eliminativity). For $\phi \in L_{BiUS}$ and any state s.

• If $s[\phi ]^{(r)}$ is defined, then $s[\phi ]^{(r)}\subseteq s$

Fact 1

(Modal Disjunction). $|\alpha \vee \beta |^{+} \models \Diamond \alpha \wedge \Diamond \beta$              (if $\alpha ,\beta \in L_{US}$)

Proof

Suppose $s[ |\alpha \vee \beta |^{+}]$ is defined. Then $s[ |\alpha \vee \beta |^{+}]$ = $s[|\alpha |^+] \cup s [|\beta |^{+}]$ with both $s[|\alpha |^+]$ and $s[|\beta |^+]$ defined and $\ne \emptyset$. By Lemma 1 we have $s[|\alpha |^+]=s[\alpha ]\ne \emptyset$. From $s[|\alpha |^{+}] \subseteq s[|\alpha \vee \beta |^{+}]$ it follows $s[\alpha ] \subseteq s[|\alpha \vee \beta |^{+}]$. By monotonicity of $\alpha$ (Lemma 2) we conclude $s[\alpha ] [\alpha ]\subseteq s[|\alpha \vee \beta |^{+}][\alpha ]$. Since $s[\alpha ][\alpha ]=s[\alpha ] \ne \emptyset$ (by idempotence of $\alpha$), we conclude $s[ |\alpha \vee \beta |^{+}] [\alpha ] \ne \emptyset$. But then $s[|\alpha \vee \beta |^{+}] \models \Diamond \alpha$. Similarly for $\Diamond \beta$.

For a counterexample to Modal Disjunction with $\alpha \not \in L_{US}$, let $\alpha$ be $(p \wedge \Diamond \lnot p)$. Then $\{w_p, w_{\emptyset }\} [|(p \wedge \Diamond \lnot p) \vee p| ^+] =\{w_p\}$ is defined but does not support $\Diamond (p \wedge \Diamond \lnot p)$. Thus $|(p \wedge \Diamond \lnot p) \vee p| ^+ \not \models \Diamond (p \wedge \Diamond \lnot p)$.

Fact 2

(Narrow Scope fc ). $|\Diamond (\alpha \vee \beta )|^+ \models \Diamond \alpha \wedge \Diamond \beta$

Proof

Suppose $s [|\Diamond (\alpha \vee \beta )|^+ ]$ is defined. Then $s [|\Diamond (\alpha \vee \beta )|^+ ]= s [\Diamond |(\alpha \vee \beta )|^+]= s\ne \emptyset$ and $s [| \alpha \vee \beta |^+] = s[| \alpha |^+] \cup s[|\beta |^+] \ne \emptyset$. It follows that $s[|\alpha |^+] \ne \emptyset \ne s[|\beta |^+]$. By Lemma 1 we conclude $s[\alpha ] \ne \emptyset$. Hence $s[|\Diamond (\alpha \vee \beta )|^+][\Diamond \alpha ]=s$ and thus $s[|\Diamond (\alpha \vee \beta )|^+] \models \Diamond \alpha$. Similarly for $\Diamond \beta$.

Fact 3

(Wide Scope fc ). $|\Diamond \alpha \vee \Diamond \beta |^+ \models \Diamond \alpha \wedge \Diamond \beta$

Proof

Suppose $s [|\Diamond \alpha \vee \Diamond \beta |^+ ]$ is defined. Then $s [|\Diamond \alpha \vee \Diamond \beta |^+]=s[|\Diamond \alpha |^+ \vee |\Diamond \beta |^+] = s[|\Diamond \alpha |^+] \cup s[|\Diamond \beta |^+]=s\ne \emptyset$. Hence both $s[|\Diamond \alpha |^+]$ and $s[|\Diamond \beta |^+]$ are defined which means $s[\Diamond |\alpha |^+]= s[\Diamond |\beta |^+] =s \ne \emptyset$. It follows that $s[|\alpha |^+] \ne \emptyset \ne s[|\beta |^+]$. By Lemma 1 we conclude $s[\alpha ] \ne \emptyset$. Hence $s[|\Diamond \alpha \vee \Diamond \beta |^+ ][\Diamond \alpha ]=s$ and thus $s[|\Diamond \alpha \vee \Diamond \beta |^+ ] \models \Diamond \alpha$. Similarly for $\Diamond \beta$.

Fact 4

(Negation 1). $|\lnot (\alpha \vee \beta )|^+ \models \lnot \alpha \wedge \lnot \beta$              (if $\alpha ,\beta \in L_{US}$)

Proof

Suppose $s[ |\lnot (\alpha \vee \beta )|^{+}]$ is defined. Then $s[| \lnot (\alpha \vee \beta )|^{+}]$ = $s[|\alpha |^+ \vee |\beta |^+]^r =$ $s[|\alpha |^+]^r \cap s[|\beta |^+]^r$. By Lemma 1 we have $s[|\alpha |^+]^r=s[\alpha ]^r =s[\lnot \alpha ]\ne \emptyset$. From $s[|\lnot (\alpha \vee \beta )|^{+}] \subseteq s[|\alpha |^+]^r$ we have then $s[|\lnot (\alpha \vee \beta )|^{+}] \subseteq s[\lnot \alpha ]$. By idempotence, $s[\lnot \alpha ] =s[\lnot \alpha ] [\lnot \alpha ]$, and by downword closure, $s[|\lnot (\alpha \vee \beta )|^{+}] = s[|\lnot (\alpha \vee \beta )|^{+}] [\lnot \alpha ]$. Hence $s[|\lnot (\alpha \vee \beta )|^{+}] \models \lnot \alpha$. Similarly for $\lnot \beta$.

Fact 5

(Negation 2 ). $|\lnot \Diamond (\alpha \vee \beta )|^+ \models \lnot \Diamond \alpha \wedge \lnot \Diamond \beta$

Proof

Suppose $s[ |\lnot \Diamond (\alpha \vee \beta )|^{+}]$ is defined. Then $s[| \lnot \Diamond (\alpha \vee \beta )|^{+}] = s[\Diamond |(\alpha \vee \beta )|^{+}]^r \ne \emptyset$. This means that $s[\Diamond |(\alpha \vee \beta )|^{+}]^r =s$ and so also $s[|(\alpha \vee \beta )|^{+}]^r = s[|\alpha |^+ \vee |\beta |^+]^r = s[|\alpha |^+]^r \cap s[|\beta |^+]^r =s$. By Lemma 1, $s[|\alpha |^+]^r = s[\alpha ]^r$ and so $s\subseteq s[\alpha ]^r$. By eliminativity, $s[\alpha ]^r=s$ and so $s[\Diamond \alpha ]^r=s$ and $s[\lnot \Diamond \alpha ] =s$. Hence $s[|\lnot \Diamond (\alpha \vee \beta )|^{+}] \models \lnot \Diamond \alpha$. Similarly for $\lnot \Diamond \beta$.