Let $\alpha, \beta $ be some propositions, such that $\alpha\to\beta$ is a tautology. Let $A$ be the set of variables in $\alpha$ and $B$ be the set of variables in $\beta$. Prove that if $A\cap B=\varnothing$, then $\beta$ is a tautology or $\lnot\alpha $ is a tautology.
I've tried solving it but in my solution I didn't use the fact that the intersection is empty so I want to find out what I did wrong..
Here is my attempt:
Assume towards contradiction there exists an assignment $v$, such that $v(\lnot\alpha)=F$ and $v(\beta)=F$.
Then as $\alpha\to\beta $ is a tautology, $v(\alpha\to\beta)=T$. As $v(\lnot\alpha)=F$, then $v(\alpha)=T$ and $v(\beta)$ must be true, which is a contradiction since we assumed $v(\beta)=F$.
Best Answer
Your proof is incorrect, as you did not rule out a possible case in which $v(\neg \alpha) = F$ and $u(\beta) = F$ for different $v, u$. To prove this properly, you must suppose there are some assignments $v, u$ in which $v(\neg \alpha) = F$ and $u(\beta) = F$ without assuming that $v = u$.
More precisely, the proof begins like this:
Suppose that neither $\neg \alpha$ nor $\beta$ are tautologies. The fact that $\neg \alpha$ is not a tautology means that there is some $v$ such that $v(\neg \alpha) = F$. Similarly, the fact that $\beta$ is not a tautology means that there is some $u$ such that $u(\beta) = F$.