Answers: 2
![P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)]](/tpl/images/0284/9635/39c43.png)
is a tautology.![P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)] \equiv](/tpl/images/0284/9635/bb069.png)
![\equiv (P \land [(\lnot P)\rightarrow (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [(\lnot P)\rightarrow (Q \land \lnot Q)])](/tpl/images/0284/9635/db9f0.png)
by the logical equivalences involving bi-conditional statements![\equiv (P \land [\lnot(\lnot P)\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [\lnot(\lnot P)\lor (Q \land \lnot Q)])](/tpl/images/0284/9635/166c4.png)
by the logical equivalences involving conditional statements![\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [ P\lor (Q \land \lnot Q)])](/tpl/images/0284/9635/6e0c9.png)
by the Double negation law![\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q))](/tpl/images/0284/9635/47ba2.png)
by De Morgan's law![\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q))](/tpl/images/0284/9635/540d2.png)
by the Negation law![\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor \lnot(\lnot Q))](/tpl/images/0284/9635/a3ec4.png)
by De Morgan's law![\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor Q)](/tpl/images/0284/9635/0bf03.png)
by the Double negation law
by the Identity law
by the Idempotent law
by the Commutative law
by the Negation law
by De Morgan's law
by the Idempotent law
by the Distributive law
by the Negation law
by the Domination law
![Prove that p (p) = (qa ~ q)] is a tautology.](/tpl/images/0284/9635/e6301.jpg)
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Prove that p (p) = (qa ~ q)] is a tautology....
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