prove $[¬p\land (p\lor q)]→q ≡ T$ without using the truth table I need to prove $[¬p\land  (p\lor q)]→q ≡ T$ without using the truth table.
Please help me to solve it.
 A: $$\begin{align} [\lnot p \land (p \lor q)] \rightarrow q & \equiv\lnot  [\lnot p  \land (p \lor q)]  \lor q\\\\
& \equiv [p \lor \lnot (p \lor q)]\lor q\\\\
&\equiv [p \lor (\lnot p \land \lnot q)] \lor q\\\\
&\equiv [(p \lor \lnot p) \land (p \lor \lnot q)] \lor q\\\\
&\equiv [T \land (p \lor \lnot q)] \lor q\\\\
&\equiv p \lor \lnot q \lor q\\\\
&\equiv p \lor T\\\\
&\equiv T\end{align}$$
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:
$$\begin{align} [\lnot p \land (p \land q)]\rightarrow q &\equiv [(\lnot p \land p) \land q] \rightarrow q\tag{associativity}\\
&\equiv (F \land q) \rightarrow q\\
&\equiv q\rightarrow q\\
&\equiv \lnot q \lor q\\
&\equiv T
\end{align}$$
A: 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 $$(\neg p \land(p\lor q))\rightarrow q$$
