An Inconsistent Set of Formulas Syntactically Entails Anything

logicpropositional-calculussolution-verification

Here's what I'm trying to prove:

Let $\Delta$ be an inconsistent set of formulas. Then, for any formula $\psi$, $\Delta \vdash \psi$.

Proof Attempt:

We have previously shown that $\{(\alpha \land \beta) \} \vdash \alpha$. Let $\psi$ be any formula. Since $\Delta$ is inconsistent, it follows that:

$$\Delta \vdash (\lnot(\psi \to \psi))$$

However, $(\lnot(\psi \to \psi))$ is just $(\psi \land (\lnot \psi))$. Then, there is a deduction from $(\psi \land (\lnot \psi))$ to $\psi$. Hence, there is a deduction from $\Delta$ to $\psi$ and that is what we wanted to show. $\Box$

Does the argument above work? If it doesn't, then why? How can I fix it?

Would there be any alternative argument that doesn't actually rely on the fact that I proved previously?

Best Answer

We assume as axioms those of the linked proof system.

With Ax.1 and Ax.2 we can prove (many copy of the proof on this site): $\vdash (\alpha \to \alpha)$.

Using the following instance of Ax.1:

$(\psi \to \psi) \to (\lnot \psi \to (\psi \to \psi))$.

and the previous result we have:

$\text { (1) } \vdash (\lnot \psi \to (\psi \to \psi))$.

Using the following instance of Ax.1:

$\lnot (\psi \to \psi) \to (\lnot \psi \to \lnot (\psi \to \psi))$

we have:

$\text { (2) } \lnot (\psi \to \psi) \vdash (\lnot \psi \to \lnot (\psi \to \psi))$.

Using the following instance of Ax.3:

$[(\lnot \psi \to \lnot (\psi \to \psi)) \to ((\lnot \psi \to (\psi \to \psi)) \to \psi)]$

we have, form (1) and (2) by MP:

$\text { (3) } \lnot (\psi \to \psi) \vdash \psi$.

Thus, if $\text {Incons}(\Delta)$ is defined as: $\Delta \vdash ¬(α → α)$ for any formula $α$, using (3) above we have:

if $Δ$ is an inconsistent set of formulas, then $Δ \vdash ψ$, for any formula $\psi$.

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