prove $[¬p\land (p\lor q)]→q ≡ T$ without using the truth table

Question

I need to prove $[¬p\land (p\lor q)]→q ≡ T$ without using the truth table. Please help me to solve it.

Answer 1

The exact details of the proof will differ, depending on your proof system. It will go something like this though:

Suppose $(\neg p \land (p\lor q))$, therefore $\neg p$ and $(p\lor q)$.

If $p$, then $p$ and $\neg p$, a contradiction, hence $q$.

Else, $q$, hence $q$.

In either case of the disjunction, we have $q$. Therefore $$(\neg p \land(p\lor q))\rightarrow q$$

Answer 2

$$\begin{align} [\lnot p \land (p \lor q)] \rightarrow q & \equiv\lnot [\lnot p \land (p \lor q)] \lor q\\\\ & \equiv [p \lor \lnot (p \lor q)]\lor q\\\\ &\equiv [p \lor (\lnot p \land \lnot q)] \lor q\\\\ &\equiv [(p \lor \lnot p) \land (p \lor \lnot q)] \lor q\\\\ &\equiv [T \land (p \lor \lnot q)] \lor q\\\\ &\equiv p \lor \lnot q \lor q\\\\ &\equiv p \lor T\\\\ &\equiv T\end{align}$$

ADDED:

There seems to be some confusion in the editing of this question. If in fact you are needing to prove that $$[\lnot p \land (p \land q)]\rightarrow q \equiv T$$

Then the proof is much simpler:

$$\begin{align} [\lnot p \land (p \land q)]\rightarrow q &\equiv [(\lnot p \land p) \land q] \rightarrow q\tag{associativity}\\ &\equiv (F \land q) \rightarrow q\\ &\equiv q\rightarrow q\\ &\equiv \lnot q \lor q\\ &\equiv T \end{align}$$