0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Your last line simplifies to $P\lor(Q\land R)$. $\endgroup$
    – Michael Hoppe
    Oct 25, 2019 at 8:08
  • $\begingroup$ $(P \lor Q) \land (P \land Q) = P \land Q$ (2nd line from end) $\endgroup$
    – David Diaz
    Oct 25, 2019 at 9:15
  • $\begingroup$ @MichaelHoppe thanks man i got the answer $\endgroup$
    – MaxBrian
    Oct 25, 2019 at 9:18

1 Answer 1

Reset to default
1
$\begingroup$

For your last line,

Use Distributive law (backward) you get

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.