Give abstraction to formula and use the definition of tautological implication.

Question

1. (a) Give the abstraction of $$ (\lnot((\forall x)(\varnothing(x) \land p)) \rightarrow \lnot q) \rightarrow (\lnot ((\forall x)(\varnothing (x) \land p)) \rightarrow q) \rightarrow ((\forall x)(\varnothing (x) \land p)) $$

(b) Use the definition of tautological implication to show $$ (\lnot((\forall x)(\varnothing(x) \land p)) \rightarrow \lnot q) , (\lnot((\forall x)(\varnothing(x) \land p)) \rightarrow q) \vDash_{taut} ((\forall x)(\varnothing (x) \land p)) $$

My answer to (a) was $ \lnot p \rightarrow \lnot q \rightarrow r $

For (b) $ \lnot p, \lnot q \vDash_{taut} r $

It is not tautological implication as if i draw a truth table I am gonna have one instance in which $ \lnot p $ and $\lnot q$ is true and $ r $ is false.

I use "Mathematical logic" by Tourlakis.

Answer 1

Your answer to (a) overlooks that there are common parts in the three formulas that you abstracted as $p,q,r$. I'd suggest $(\neg p\to\neg q)\to((\neg p\to q)\to p)$ as a more suitable abstraction.