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How does one get rid of $ p $ in $(p\Leftrightarrow q) \wedge (p\Leftrightarrow r) $? I have already tried to simplify the formula, applied DeMorgan's laws, etc, but nothing helps. Does anyone know how to deal with this?

Constantine
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  • Do you want to prove that $[(p\iff q)\land (p \iff r)]\implies(q\iff r)$? If so, what exactly is at your disposal? Would a truth table suffice? – Git Gud Sep 07 '13 at 22:22
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    What do you mean by "get rid of $p$"? You can certainly prove that the statement you gave implies a certain relationship between $r$ and $q$... – dfeuer Sep 07 '13 at 22:22

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You can't get rid of $p$ without losing some information. However, this does imply that $$q \iff r,$$ so if you're just looking to remove $p$, that's probably the most informative result.

qaphla
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  • But they are not equivalent, their truth tables differ, don't they? – Constantine Sep 07 '13 at 22:23
  • They aren't equivalent, but the one that you gave implies this one, and I can't see any way to get rid of $p$ that gives you more information about the relationship of $q$ and $r$. – qaphla Sep 07 '13 at 22:24