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I have the standard logical equivalence: $(p\rightarrow q)\wedge(q\rightarrow r)\Leftrightarrow p\rightarrow (q\wedge r)$.

Using several distributive laws I was able to get it down to: $(\neg p\wedge\neg q) \vee (\neg p\wedge r) \vee (q\wedge r)$.

I must be missing some manipulation I can do to reduce this.

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Where did you start, where were you trying to get? "Get it down"... get what down? – Arturo Magidin Sep 23 '10 at 19:49
How about a truth table? That would be the simplest thing to do, I think. – Arturo Magidin Sep 23 '10 at 19:58
I haven't worked through the details, but it strikes me that this statement is not correct. Change the bidirectional implication to a simple implication, and it holds, however. – Noldorin Sep 24 '10 at 0:18
@Noldorin: Yes, they are not equivalent. See my answer. – Aryabhata Sep 24 '10 at 1:20
It seems clear to me that it should have been $(p\rightarrow q)\wedge (p\rightarrow r)$ on the right hand side. I checked to make sure it wasn't me who introduced the mistake when I edited the question, and it was incorrect in the original. – Arturo Magidin Sep 24 '10 at 13:53

2 Answers 2

Umm... maybe I am missing something, but

if $p$ is false, $q$ is true and $r$ is false, then we have that

$(p\rightarrow q)\wedge(q\rightarrow r)$ is false

$p\rightarrow (q\wedge r)$ is true.

So I don't see how you can prove the equivalence.

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$$\begin{array}{|c|c|c|c|c|c|c|c|c}\hline p&q&r&p\Rightarrow q&q\Rightarrow r&(p\Rightarrow q)\land (q\Rightarrow r)&q\land r&p\Rightarrow(q\land r)&\cdots\\\hline T&T&T&T&T&T&T&T\\T&T&F&T&F&F&F&F\\T&F&T&F&T&F&F&F\\T&F&F&F&T&F&F&F\\F&T&T&T&T&T&T&T\\F&T&F&T&F&F&F&T\\F&F&T&T&T&T&F&T\\F&F&F&T&T&T&F&T\\\hline \end{array}$$

$$ \begin{array}{c|c|c|}\hline \cdots&\big[(p\Rightarrow q)\land (q\Rightarrow r)\big]\iff \big[ p\Rightarrow(q\land r)\big]\\\hline &T\\&T\\&T\\&T\\&F\\&T\\&T\\\hline \end{array}$$

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