# Abduallah's question on FB

#### ZaidAlyafey

##### Well-known member
MHB Math Helper

prove that $$((p\to \neg q) \land q) \to \neg p$$

is a Tautology .

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
By the commutative law we can rewrite as

$$\displaystyle (q \land (p\to \neg q) ) \to \neg p$$

First we need to know that

$$\tag{1}p\to q \equiv \, \neg p \lor q$$

Using this we get

$$q \land (\neg p \lor \neg q)\to \neg p$$

By distributive law

$$(q \land \neg p) \lor (q \land \neg q)\to \neg p$$

Since

$$q \land \neg q \equiv F \,\,\, , \,\,\, q \lor F \equiv q$$

So we have

$$(q \land \neg p)\to \neg p$$

Using (1) again

$$\neg (q \land \neg p) \lor \neg p \equiv \neg q \lor p \lor \neg p\equiv T$$

Using De Morgan law .