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Please help. I've missed some lectures, and now I'm stuck (my fault!). The lectures notes don't explain elaborately, and I can't find good tutorials online. I've somehow managed to arrive at $(Q \lor P) \land P$. If this is correct, can this be simplified further? Thanks heaps.

Levi
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3 Answers3

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Replacing $\lor$ by $+$ and $\land$ by $\cdot$ makes it easy to see the further possible simplification
$$(Q+P)\cdot P = Q\cdot P+P = (Q+1)\cdot P = P$$

$P\cdot P = P\land P = P$ because "$\land$" of two inputs is true when "both" its inputs are true. Saying both $P, P$ are true is same as saying $P$ is true.

$Q+1=1$ because "$\lor$" of two inputs is true when "at least" one input is true. Since one input is already true, $Q+1=1$.

AgentS
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Yes, it can still be simplified further, the idea is using Commutative law and Absorption law,
I suppose you need to simplify this with Logical equivalence, and the proof is simple: $$¬((¬Q \land ¬P) \lor ¬P)\tag*{Start from}$$ $$\neg(¬Q \land ¬P)\land P\tag*{De Morgan's law}$$ $$(Q\lor P)\land P\tag*{De Morgan's law}$$ $$P\land(P\lor Q)\tag*{ Commutative law$\times2$}$$ $$P\tag*{Absorption law}$$

Therefore this simplified the expression.

Ethan
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Using de Morgan's laws and the double negation law:

$$\neg ((\neg Q \land \neg P) \lor \neg P) \equiv \neg (\neg Q \land \neg P) \land \neg \neg P$$ $$\equiv \neg (\neg Q \land \neg P) \land P$$ $$\equiv (\neg \neg Q \lor \neg \neg P)\land P$$ $$\equiv (Q \lor P)\land P$$

Next, you could use the absorption law and commutative law to conclude that

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

If you haven't proved the absorption law in class then you could directly prove the result through the same strategy in this question and here.

Axion004
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