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Show $\lnot (A\iff B)$ is logically equivalent to $A\iff \lnot B$.

Making some calculations I got this

$$\lnot (A\iff B)=(A\land\lnot B)\lor(B\land\lnot A) \tag{1}$$ and

$$(A\iff \lnot B)=(\lnot A\lor\lnot B)\land(\lnot B\lor\lnot A)\tag{2}$$

I don't know how to make a relation between (1) and (2). I believe there is a trick in the solution.

Could someone lighten my brain?

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  • $\begingroup$ Have you tried using truth tables? These will not only show the equivalence, they will also show that your second equation is not correct (but your first is). $\endgroup$ – skyking Oct 18 '17 at 5:58
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Your calculations are not correct. Recall that $(A\implies B)=\lnot A\lor B$.

Now note that for the distributive law, $$(A\iff \lnot B)=(\lnot A\lor\lnot B)\land(B\lor A)= (\lnot A\land B)\lor( \lnot A\land A)\lor (\lnot B\land B)\lor(\lnot B\land A). $$ Is the above expression equivalent to $$\lnot (A\iff B)=(A\land\lnot B)\lor(B\land\lnot A) \quad ?$$

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Logically $A\Leftrightarrow B$ means $A$ and $B$ have equal truth values (either both true or both false). If this is not the case, they have different truth values, which then means $A$ and $\neg B$ have the same truth value. This is a consequence of the two-valuedness of classical logic. You can also use a truth table to see this more clearly.

\begin{align} \begin{array}{cll} A\Leftrightarrow B & 0 &1\\ 0 & 1 & 0\\ 1 & 0 & 1 \end{array}\qquad \begin{array}{cll} A\Leftrightarrow\neg B & 0 &1\\ 0 & 0 & 1\\ 1 & 1 & 0 \end{array} \end{align}

From the above truth tables, the truth value of $\neg(A\Leftrightarrow B)$ is always the same as that of $A\Leftrightarrow\neg B$.

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By logic table the first tells $A$ and $B$ have not the same logic values, the second that they have different logic values (so they are equivalent under tertium non datur assumption). I know this is not the searched answer, but was funny and worth to note.

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