5.14 A ``short'' one. $A\Longleftrightarrow B\vdash(A\wedge B)\vee(\neg A\wedge\neg B)$

Seems easy: if two expressions are equivalent, it's because they are both true, or both false. I could prove the validity of $A\Longleftrightarrow B\vdash(A\wedge B)\vee(\neg A\wedge\neg B)$ this way:

$$\begin{fitch} \par (A \Rightarrow B) \wedge (B \Rightarrow A)... ...e (\neg A \wedge \neg B) & E$\vee$\ 10,15,23 \par \end{fitch}$$

Firstly: we can't write $A\Longleftrightarrow B$ since we don't have rules for $\Longleftrightarrow$. Since it is seldom used, when a $\Longleftrightarrow$ appears we are allowed to change it to $(A\Rightarrow B)\wedge(B\Rightarrow A)$, which is the same.

Well, this is the only idea I had... I leave as an exercise to find a shorter way to do it (if it does exist). What I did here was to write down that $A\vee\neg A$ is true (we already did this exercise, and here I just copied the same steps). Once I know that $A\vee\neg A$ holds, I see that both the case $A$ and the case $\neg A$ lead to the same formula, which is the solution.