Tautology problem

Question

Show that $((p \vee q) \wedge \neg (\neg p \wedge (\neg q \vee \neg r))) \vee ( \neg p \wedge \neg q) \vee (\neg p \vee r )$

is a tautology (without using truth table).

After simplification I got $ ((p \vee q) \wedge (p \vee r)) \vee \neg (p \vee q) \vee \neg (p \wedge \neg r)$, which could easily become $T$ if there was $\neg (p \vee r) $ instead of last term.

But this expression is also a tautology using truth table.

Answer 1

Taking your simplification one step farther:

$((p∨q)∧(p∨r))∨¬(p∨q)∨¬(p∧¬r)\equiv (p\vee (q\wedge r))\vee \neg(p\vee q)\vee(\neg p\vee r)$

If $p$ is true, the first parentheses, $(p\vee (q\wedge r))$, is true. If $p$ is false, the last parentheses, $(\neg p \vee r)$ is true.

So the whole expression is true since it is a disjunction of three parentheses, at least one of which must be true.

Answer 2

OR distributes, so after you have simplified the expression to:

$$(p\lor (q\land r))\lor \lnot(p\lor q)\lor(\lnot p\lor r)$$

you can then write:

$$p\lor (q\land r)\lor \lnot(p\lor q)\lor\lnot p\lor r$$

and OR also commutes:

$$p\lor\lnot p\lor (q\land r)\lor \lnot(p\lor q)\lor r$$

and $p\lor\lnot p\equiv\top$, and $\top\lor X=\top$, so the expression is a tautology.