# finite satisfiability and truth assignment

The problem goes like this:

Let $$\Delta$$ be a set of wffs such that (i) every finite subset of $$\Delta$$ is satisfiable, and (ii) for every wff $$\alpha$$, either $$\alpha\in\Delta$$ or $$\neg\alpha\in\Delta$$. Define the truth assignment $$v$$: $$v(A)=T$$ iff $$A\in\Delta$$, $$v(A)=F$$ iff $$A\not\in\Delta$$ for each sentence symbol $$A$$. Show that for every wff $$\varphi$$, $$\bar{v}(\bar{\varphi})=T$$ iff $$\varphi\in\Delta$$.

I tried to prove it by contradiction or contrapositive, but I always ended up showing finite satisfiability of finite subset that includes $$\varphi$$ with the given truth assignment which is absurd because finite satisfiability doesn't depend on a special truth assignment; one needs to consider all the possible truth assignments to decide the finite satisfiability. How should one prove this problem?

By induction on the complexity of $$\varphi$$.

Base case: $$\varphi$$ is a sentence symbol $$A$$. Then, by def of $$v$$: $$v(\varphi)=v(A)= \text T$$ iff $$A \in \Delta$$.

Induction step: for simplicity, consider two basic connectives only: $$\lnot, \lor$$.

Let $$\varphi, \psi$$ such that the property holds and let $$\sigma := \lnot \varphi$$.

We have that $$\overline v(\sigma)=\overline v(\lnot \varphi)= \text T$$ iff $$\overline v(\varphi)=\text F$$. And this, by hypothesis: iff $$\varphi \notin \Delta$$.

By (ii) we have that: if $$\varphi \notin \Delta$$, then $$\lnot \varphi \in \Delta$$.

Thus, so far we have: if $$\overline v(\varphi)=\text F$$, then $$\lnot \varphi \in \Delta$$.

Now assume that $$\lnot \varphi \in \Delta$$. By (i) $$\Delta$$ is consistent. Thus $$\varphi \notin \Delta$$.

And by the argument above we have $$\overline v(\varphi)=\text F$$.

Same for $$\sigma := (\varphi \lor \psi)$$.