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I want to prove that $[(P \lor A) \land ( \neg P \lor B)] \rightarrow (A \lor B)$, using distributions or reductions (even though I am aware that simpler proofs exist). The issue is that I keep wandering around like a fool. Here is what I'e come up with so far:

$(P \lor A) \land ( \neg P \lor B) \Leftrightarrow$

$(P \land ( \neg P \lor B)) \lor (A \land (\neg P \lor B)) \Leftrightarrow$

$((P \land \neg P) \lor (P \land B)) \lor (A \land (\neg P \lor B)) \Leftrightarrow$

$(\bot \lor (P \land B)) \lor ( A \land (\neg P \lor B)) \Leftrightarrow$

$(P \land B) \lor (A \land ( \neg P \lor B)) \Leftrightarrow$

$(P \land B) \ \lor A) \land ((P \land B) \lor ( \neg P \lor B)) \Leftrightarrow$

$((P \land B) \ \lor A) \land ((P \land B) \lor \neg P) \lor B) \Leftrightarrow$

$((P \land B) \ \lor A) \land (((P \lor \neg P) \land (\neg P \lor B)) \lor B) \Leftrightarrow$

$((P \land B) \ \lor A) \land (( \top \land (\neg P \lor B)) \lor B) \Leftrightarrow$

$((P \land B) \ \lor A) \land ( (\neg P \lor B)) \lor B) \Leftrightarrow$

$((P \land B) \ \lor A) \land (\neg P \lor B) \Leftrightarrow$

$((P \land B) \land (\neg P \lor B)) \lor ((\neg P \lor B) \land A) \Leftrightarrow $

The point where I give up

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  • $\begingroup$ You want to prove that $((P \lor A) \land ( \neg P \lor B)) \rightarrow (A \lor B)$. It would be nice if you could get $((P \lor A) \land ( \neg P \lor B)) \leftrightarrow ((A\lor B)\land \text{Something})$. And it seems to be you can get exactly this from $(P \land B) \lor (A \land ( \neg P \lor B))$. I did it my head, so I might be wrong. $\endgroup$ – Git Gud May 23 '16 at 22:54
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First start with turning $[(P \lor A) \land ( \neg P \lor B)] \rightarrow (A \lor B)$

Into this $\neg [(P \lor A) \land ( \neg P \lor B)] \lor (A \lor B)$

This is where you might have been going in circles:

$$ \neg [(P \lor A) \land ( \neg P \lor B)] \lor (A \lor B) \\ \neg(P \lor A) \lor \neg( \neg P \lor B) \lor (A \lor B) \\ (\neg P \land \neg A) \lor ( P \land \neg B) \lor (A \lor B) \\ (\neg P \land \neg A) \lor ( P \land \neg B) \lor (A \lor B) \\ (\neg P \land \neg A) \lor ( P \land \neg B) \lor A \lor B $$

Can you continue from here?

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[Note that a common convention is that $\land$ has higher precedence over $\lor$, which lessens brackets.]

The systematic way is:

  1. Convert everything into using only $\neg,\land,\lor$.

  2. Use distributivity of $\land$ over $\lor$ and De Morgan's laws to expand into disjunctive normal form.

  3. Use the law of excluded middle to simplify. Another useful law is $P \lor \neg P \land Q \equiv P \lor Q$.

Following this gives: $\def\imp{\rightarrow}$

  $(P \lor A ) \land ( \neg P \lor B ) \imp A \lor B$

  $\ \equiv \neg ( (P \lor A ) \land ( \neg P \lor B ) ) \lor ( A \lor B )$

  $\ \equiv ( \neg (P \lor A ) \lor \neg ( \neg P \lor B ) ) \lor ( A \lor B )$

  $\ \equiv ( ( \neg P \land \neg A ) \lor ( P \land \neg B ) ) \lor ( A \lor B )$

  $\ \equiv ( \neg P \land \neg A \lor A ) \lor ( P \land \neg B \lor B )$

  $\ \equiv ( \neg P \lor A ) \lor ( P \lor B )$

  ... [which I'm sure you can easily finish using excluded middle].

Here is a proof of the useful law which I used above:

  $P \lor \neg P \land Q$

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

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

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

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

  $\ \equiv P \lor Q$.

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  • $\begingroup$ @CFLAGS: Does my answer help you? $\endgroup$ – user21820 May 30 '16 at 4:24

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