Show that two statements are logically equivalent using different laws

Question

I have to show that

$(\backsim(P\vee Q))\vee(\backsim P\wedge Q)\cong \backsim P.$

I tried like this but do not know how to get the desired result using different laws

L.H.S$=(\backsim(P\vee Q))\vee(\backsim P\wedge Q)$

$=(\backsim(P\vee Q))\vee\backsim(\backsim((\backsim P\wedge Q))$

$=(\backsim(P\vee Q))\vee\backsim(\backsim\backsim P\vee\backsim Q))$ by De-Morgan' law

$=\backsim[(P\vee Q))\wedge( P\vee\backsim Q))]$ because $\backsim\backsim P=P$

I do not know how to proceed further to get the desired result $\backsim P.$

I can prove this using truth table but do not know how to prove using different laws.

I need in detail answer.

Thank You!

Answer 1

$\neg [(P∨Q))∧(P∨\neg Q)]$ is equivalent to $\neg [P\wedge (Q\vee\neg Q)]$ by distributivity. But $Q\vee \neg Q$ is logically true and so the formula is equivalent to $\neg [P\wedge true]$ which is $\neg P$ as required.