Logic - Propositional calculus

Question

I don't understand how to show the following: (!Q -> P) ∧ !P -> Q

I understand the answer is true as I did it with a truth table but how can I prove this using propositional logic?

Thanks!

Answer 1

If you want to show that $$!Q \rightarrow P \ \equiv \ !P \ \rightarrow Q$$ then use the following argumentation:

In general $A\rightarrow B$ can be expressed as $!A\lor B$. Accordingly $$!Q \rightarrow P \equiv \ !!Q\lor P\ \equiv Q\lor P$$ and $$!P \rightarrow Q \equiv \ !!P\lor Q \ \equiv P \lor Q=Q\lor P.$$

EDITED

If you meant

$$[(!Q\rightarrow P)\land !P]\rightarrow Q \tag 1$$

then use again that in general $A\rightarrow B$ can be expressed as $!A\lor B$.

Regarding $(1)$

$$(!Q\rightarrow P)\land !P \equiv !!Q\lor P\land !P\equiv Q\lor P\land !P\equiv Q.$$