Simplifying on logic Operations

Question

I need simplify the following proposition to 2 logic operations using the laws of the algebra of propositions. Write each step on a separate line with the algebra law you used as a justification.

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

I'm not sure how I could answer this but to start I think maybe using Idempotent law and maybe Associative to solve this. Or maybe I am way off.

Answer 1

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

By identity $A=(A\lor A)$ and associativity: $${\color{Green} {(P' \land Q')}} \lor (P' \land Q) \lor {\color{Green} {(P' \land Q')}} \lor (P \land Q')$$

By distributivity: $$(P' \land (Q' \lor Q)) \lor ((P' \lor P) \land Q')$$

Reduction by the law of excluded middle: $$(P' \land 1) \lor (1 \land Q')$$ and by the neutral element definition: $$P' \lor Q'$$ Finally de Morgan's law: $$(P \land Q)'$$

Answer 2

Consider $S$ as the universal set. Our given expression is this:

$$(P^\prime\land Q^\prime)\lor (P^\prime \land Q)\lor (P\land Q^\prime)$$

Use the converse of distributive laws along with associativity laws to reduce the given expression as:

$$\left\{P^\prime \land (Q^\prime \lor Q)\right\}\lor (P\land Q^\prime)=\{P^\prime \land S\}\lor (P\land Q^\prime)=P^\prime\lor (P\land Q^\prime)$$

$$P^\prime\lor(P\land Q^\prime)=(P^\prime\lor P)\land (P^\prime\lor Q^\prime)=S\land (P^\prime\lor Q^\prime)=P^\prime \lor Q^\prime$$

Using De-Morgan's law, we get,

$$P^\prime \lor Q^\prime = (P\land Q)^\prime$$

We also used the following laws:

• $S=P\lor P^\prime = Q\lor Q^\prime$
• $Q=Q\land S~~,~~P=P\land S$