Proving ((p->~q)∧q)->~p as a Tautology - Abdullah's FB Question

[alyafey22]

Abdullah asked the following question

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

is a Tautology .

 

[alyafey22]

By the commutative law we can rewrite as

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

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 .