# Simplifying the proposition $\Bigl(\bigl(P\lor Q\bigr)\land \lnot\bigr(\lnot P\land(\lnot Q\lor\lnot R)\bigl)\Bigr)$

Simplify the Proposition $$\Bigl(\bigl(P\lor Q\bigr)\land \lnot\bigr(\lnot P\land(\lnot Q\lor\lnot R)\bigl)\Bigr)\lor(\lnot P\land\lnot Q)\lor(\lnot P\land\lnot R)$$

What I've tried is simplify the $$\Bigl(\bigl(P\lor Q\bigr)\land \lnot\bigr(\lnot P\land(\lnot Q\lor\lnot R)\bigl)\Bigr)$$ First , using the Distributive law i get

$$(P\lor Q) \land\lnot\bigl((\lnot P\land\lnot Q)\lor(\lnot P\land\lnot R)\bigr)$$ Using the DeMorgan Law and Double negation law i get

$$(P\lor Q)\land(P\land Q )\land(P \lor R)$$ Using the Idempotent law i get

$$(P\lor Q)\land (P\lor R)$$

and now i'm stuck

• Your last line simplifies to $P\lor(Q\land R)$. Oct 25, 2019 at 8:08
• $(P \lor Q) \land (P \land Q) = P \land Q$ (2nd line from end) Oct 25, 2019 at 9:15
• @MichaelHoppe thanks man i got the answer Oct 25, 2019 at 9:18

For your last line,

Use Distributive law (backward) you get

$$P\vee(Q\wedge R)$$