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I'm trying to prove: $(r\leftrightarrow\neg\ p) \wedge p \equiv \neg\ r \wedge p$

I didn't suceed without truth table.

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    Are you supposed to argue informally or with deduction rules? – 2017-01-28
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    I can use only the rules of replacement. – 2017-01-28
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    Replace $r\leftrightarrow \neg p$ with $(r\to \neg p)\land (\neg p\to r)$, then use $\phi\to \psi\equiv \neg \phi \lor \psi$ on both implications, distribute everything, etc. – 2017-01-28
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    I tryed that way but it lead me nowhere... – 2017-01-28
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    This is why it's not a bad idea to show your work. Do it and someone will surely be able to point you in the right direction. – 2017-01-28

2 Answers 2

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We can replace $(r\leftrightarrow\neg\ p)$ by the disjunction $(r \wedge \lnot p) \lor (\lnot r \wedge p).$ Then distribute the $p$ over these two terms. The first one, $(r \wedge \lnot p) \land p,$ is contradictory and so may be dropped, while the remaining one is $(\lnot r \land p) \land p,$ equivalent to $\lnot r \land p.$

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    Can you please explain the disjunction? – 2017-01-28
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    The double arrow is same as $(r \rightarrow \lnot p ) \land (\lnot p \rightarrow r)$ This is in turn the same as $(\lnot r \lor \lnot p) \land ( p \lor r).$ If the and is now distributed over the two ors, two of the four resulting conjunctions are false and the other two are as noted above after the phrase "the disjunction". – 2017-01-29
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The right side proposition is true only when $r$ is false and $p$ is true. The left side proposition is also true only when $p$ is true and $r$ is false. Hence they are equivalent.

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    I know, but I want to get to the left side from the right side. – 2017-01-28