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.
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.
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.
