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With the absorption law, I know I can do

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

Can the same be applied like this?

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

Because, technically, I could move $Q$ to associate with the left parenthesis, and then switch its position with the first $P$, making this:

$$[(\lnot R \land Q)\land P] \lor P$$

Which looks a lot like the first scenario.

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3 Answers

up vote 2 down vote accepted

No, that doesn't work. Each of $\land$ and $\lor$ are associative, but they don't associate together.

That's the same as ordinary arithmetic: $\times$ and $+$ are individually associative, but it's not true that $a\times(b+c)=(a\times b)+c$.

In your concrete example, note that $(\neg R\land P)\land(Q\lor P)$ is false when $R$ is true, no matter what $P$ is. So it can't be equivalent to $P$.

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Aaaand now I see why I failed the other exercises. Thank you. –  Omega Sep 21 '13 at 2:21
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A very good way to attack problems such as this to determine logical equivalence (or lack thereof) is to do it by assuming $P = T$ and $P = F$ and see if both sides are the same.

Suppose $P = T$ then let's evaluate $(\neg R\wedge P) \wedge (Q \vee P)$. This becomes

$$(\neg R \wedge T) \wedge (Q\vee T) = \neg R \wedge T = \neg R.$$

So as you can see it is dependent upon the truth of $R$. Thus it cannot possibly hold that $(\neg R\wedge P) \wedge (Q \vee P) \equiv P$. Just out of curiousity, let's work with the scenario that $P = F$ just to see what happens. Then

$$(\neg R\wedge P) \wedge (Q \vee P) = (\neg R \wedge F) \wedge (Q\vee F) = F \wedge Q = F = P$$

Even though it didn't work out in the case where $P = T$, it does work for $P = F$.

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To simplify $\;(\lnot R \land P) \land (Q \lor P)\;$, you can use the absorption law, but not in the way you asked about. Here are the detailed steps: \begin{align} & (\lnot R \land P) \land (Q \lor P) \\ \equiv & \;\;\;\;\;\text{"$\;\land\;$ is associative -- to prepare for absorption"} \\ & \lnot R \land (P \land (Q \lor P)) \\ \equiv & \;\;\;\;\;\text{"$\;\lor\;$ is symmetric"} \\ & \lnot R \land (P \land (P \lor Q)) \\ \equiv & \;\;\;\;\;\text{"absorption law -- exactly as you formulated it"} \\ & \lnot R \land P \\ \end{align} This cannot be further simplified to $\;P\;$, obviously.

Personally, when using logic instead of studying it, I leave the associativity and symmetry implicit, and just write something like \begin{align} & \lnot R \land P \land (Q \lor P) \\ \equiv & \;\;\;\;\;\text{"logic: use $\;P \equiv \text{true}\;$ on other side of $\;\land\;$; simplify"} \\ & \lnot R \land P \\ \end{align}

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