Contradiction Detection

If there is a direct contradiction in the reasoning process, the contradictands appear in $\dis(i)$ for some $i$. This is given by Theorem 5.7.1. The theorem states that every direct contradiction that occurs in the reasoning gets detected.

Proof. That is, prove that for any model $\Dh$ of $\mal$, if

$$\Dh \models (\mu \in (AF(t) \cup \bel(t)) \wedge \false(\mu) \in (\AF(t) \cup \bel(t)))$$

then

$$\Dh \models (\mu \in \dis(t) \wedge \false(\mu) \in \dis(t)))$$

Assuming the antecedant,

$$\Dh \models (\mu \in (AF(t) \cup \bel(t))$$ (25)

$$\Dh \models (\false(\mu) \in (\AF(t) \cup \bel(t)))$$ (26)

By (5.16) and Axiom (21),

$$\Dh \models (\mu \notin AF(t) \rightarrow \mu \in \bel(t))$$ (27)

and by (5.17) and Axiom (21)

$$\Dh \models (\false(\mu) \notin AF(t) \rightarrow \false(\mu) \in \bel(t))$$ (28)

By (5.18), if $\Dh \models \mu \notin AF(t)$ then $\Dh \models \mu \in \bel(t)$. That is, either $\Dh \nvDash \mu \notin AF(t)$ or $\Dh \models \mu \in \bel(t)$. And, if $\Dh \nvDash \mu \notin AF(t)$ then $\Dh \models \mu \in \AF(t)$ $\therefore$ $\Dh \models \mu \in \AF(t)$ or $\Dh \models \mu \in \bel(t)$ Similarly, by (5.19), $\Dh \models \false(\mu) \in \AF(t)$ or $\Dh \models \false(\mu) \in \bel(t)$ $\therefore$ the possible scenarios that arise when (5.16) and (5.17) hold are as follows:

i.

$\Dh \models \mu \in \AF(t)$ and $\Dh \models \false(\mu) \in \AF(t)$

ii.

$\Dh \models \mu \in \AF(t)$ and $\Dh \models \false(\mu) \in \bel(t)$

iii.

$\Dh \models \mu \in \bel(t)$ and $\Dh \models \false(\mu) \in \AF(t)$

iv.

$\Dh \models \mu \in \bel(t)$ and $\Dh \models \false(\mu) \in \bel(t)$

Theorem 5.6.1 states that [iv] never occurs. For [i], [ii] and [iii], Axioms (43), (44) and (45) respectively, ensure that $\Dh \models (\mu \in \dis(t) \wedge \false(\mu) \in \dis(t))$ $\qedsymbol$