Show that $\neg(p \Longleftrightarrow q)$ and $p \Longleftrightarrow \neg q$ are logically equivalent Given there are 2 logical variables $p$, $q$ . Show that $\neg(p \Longleftrightarrow q)$ and  $p \Longleftrightarrow \neg q$ are logically equivalent without using the truth table.
And here is my steps: $(p \land q) \lor (\neg p \land \neg q)$
I get this). What should I do in order to get rid of it? ps: I am studing from Discrete mathematics and its application 7th edition rosen thanks
 A: In the question before we have seen that
$$\begin{array}{cl}
& \left(p\land q \right)\lor \left(\neg p \land \neg q\right) \\
\Longleftrightarrow & \left( p\Longleftrightarrow q\right)
\end{array}$$
Thus
$$\begin{array}{cl}
& \left(p\Longleftrightarrow \neg q\right)\\
\Longleftrightarrow & \left(p\land \neg q \right)\lor \left(\neg p \land q\right) \\
\Longleftrightarrow & \left(\left(p\land \neg q \right)\lor \neg p\right) \land \left(\left(p\land \neg q \right)\lor q\right) \\
\Longleftrightarrow &  \left(\neg p\lor \neg q \right) \land \left(p \lor q\right) \\
\Longleftrightarrow &  \neg\left(p\land q \right) \land \left(p \lor q\right) \\
\Longleftrightarrow &  \neg\left(p\land q \right) \land \neg\left(\neg p \land \neg q\right) \\
\Longleftrightarrow &  \neg\left(\left(p\land q \right)\lor \left(\neg p \land \neg q\right)\right) \\
\Longleftrightarrow & \neg\left( p\Longleftrightarrow q\right)
\end{array}$$
A: Suppose p true.  Then (p⟺q) has the same truth value as q.  Thus, ¬(p⟺q) has the same truth value as ¬q.  Also, (p⟺¬q) has the same truth value as ¬q.
Suppose p false.  Then (p⟺q) has the same truth value as ¬q.  Consequently, ¬(p⟺q) has the same truth value as ¬¬q.  Also, (p⟺¬q) has the same truth value as ¬¬q.
A: I would first point out that each statement is equivalent to $(p \land \lnot q) \lor (\lnot p \land q)$, rather than $(p \land q) \lor (\lnot p \land \lnot q)$.
Left to right, we have the hypothesis is equivalent to $(p \land \lnot q) \lor (\lnot p \land q)$. Then in either case, both $p \rightarrow \lnot q$ and $\lnot q \rightarrow p$ hold, so that $p \leftrightarrow \lnot q$ holds.
Right to left, assume that $(\lnot p \land q)$ does not hold, you want to show $(p \land \lnot q)$. If $p$, use the hypothesis to show that $\lnot q$. 
If $\lnot q$, use the hypothesis to show that $p$. In both cases, $(p \land \lnot q)$. I hope that helps.
A: So $(p \iff q)$ is equivalent to $(p \land q) \lor (\lnot p \land \lnot q)$, and $(p \iff \lnot q)$  (where we want to get after symbolic manipulations) is equivalent to $(p \land \lnot q) \lor (\lnot p \land q)$ (so if we get here via valid manipulations, we're done). Then you can take the negation of this, and use De Morgan's laws:
$$
\lnot (A \lor B) \equiv \lnot A \land \lnot B
$$
so
$$
\lnot\left((p \land q) \lor (\lnot p \land \lnot q)\right) \equiv \lnot(p \land q)\land \lnot(\lnot p \land \lnot q)
$$
and 
$$
\lnot (A \land B) \equiv \lnot A \lor \lnot B
$$
so
$$
\lnot(p \land q)\land \lnot(\lnot p \land \lnot q) \equiv (\lnot p \lor \lnot q)\land (p \lor  q)
$$
See if you can finish the proof from here: continue to De Morgan's laws (sometimes you might need to use them "in reverse": $A \lor B = \lnot \lnot (A \lor B) = \lnot (\lnot A \land \lnot B)$ for example)
A: Simply we can find that $p \leftrightarrow q$ is equal to $\neg(p \underline{\vee} p)$. and IF p BE TRUE q MUST BE FALSE & CONVERSELY. it means that we have $(p \rightarrow \neg q) \wedge (\neg q \rightarrow p)$ (we have a symmetry between p and q.)
A: One way to see this is with the method of analytic tableaux. You start with the negation of $$(\neg (p\leftrightarrow q))\leftrightarrow (p\leftrightarrow \neg q)\tag{1}$$ then apply a series of contradiction-hunting rules to get
,
which is closed (i.e., it ends in contradictions), meaning that $(1)$ is a tautology; that is, $\neg (p\leftrightarrow q)$ and $p\leftrightarrow \neg q$ are logically equivalent.
These notes (in pdf form) explain the method in detail.
