Boolean law simplification

Question

How would someone go about simplifying the expression:

$((p \vee r)\wedge((p \wedge q) \vee (p \wedge ¬q)))\vee (p \wedge r) \vee \neg(p \vee q) \vee r$

Any help would be appreciated.

EDITED: Is this a correct truth table?

Answer 1

Hints:

• $(p \land q) \lor (p \land \lnot q) = p \land(q \lor \lnot q) = p\;$ by distributivity and complements;

• $(p \land r) \lor r = r\;$ by absorption;

• $\lnot(p \lor q) = \lnot p \land \lnot q$ by De Morgan's.

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[ EDIT ]  After effecting the simplifications above, the expression becomes:

$$((p \lor r) \land p) \lor r \lor (\lnot p \land \lnot q)$$

More hints:

• $(p \lor r) \land p = p\;$ again by absorption;

• $p \lor (\lnot p \land \lnot q) = (p \lor \lnot p) \land (p \lor \lnot q)=p \lor \lnot q\;$ by distributivity and complements.