How to proof tautology without truth table in this case?

Question

Hej, i got stucked while finding a solution to proof the following is a tautology. Can someone help me out please with a good tip? Thanks in advance

Answer 1

Back up a step to $$(\overline{\overline{p}\lor q})\lor(\overline{\overline{q}\lor p})\lor(\overline{p}\lor q)\;;$$ you can rewrite it as

$$\left((\overline{\overline{p}\lor q})\lor(\overline{p}\lor q)\right)\lor(\overline{\overline{q}\lor p})\equiv\top\lor(\overline{\overline{q}\lor p})\equiv\top$$

via the law $\overline{x}\lor x\equiv\top$ and the absorption law $\top\lor x\equiv\top$.

Answer 2

You can use the equivalence between $(\lnot p \lor q)$ and $p \rightarrow q$.

With it, from your second line, we have :

$[(p \rightarrow q) \land (q \rightarrow p)] \rightarrow (p \rightarrow q)$

that is an instance of the rule of Simplification : $(P \land Q) \rightarrow P$.