Proving a Tautology Formally

Question

I wish to prove: $(\neg p\leftrightarrow q)\leftrightarrow\neg(p\leftrightarrow q)$

Answer 1

If you're willing to prove by truth table:

~ = NOT, <=> = iff
p  q  ~p<=>q  ~(p<=>q)
T  T     F        F
T  F     T        T
F  T     T        T
F  F     F        F

Otherwise

We define $\Leftrightarrow$ to be $\neg(x\oplus y)$, where $\oplus$ is $(x\vee y)\wedge \neg(x\wedge y)$. $$\begin{align*}(\neg p\Leftrightarrow q)&\leftrightarrow \neg(p\Leftrightarrow q)&&\text{given}\\ &\leftrightarrow\neg(\neg(p\oplus q))&&\text{by definition}\\ &\leftrightarrow p\oplus q&&\text{double negation=positive}\ \end{align*}$$ and $$\begin{align*}(\neg p\Leftrightarrow q)&\leftrightarrow \neg(p\Leftrightarrow q)&&\text{given}\\ \neg(\neg p\oplus q)&\leftrightarrow&&\text{by definition}\\ p\oplus q&\leftrightarrow &&\text{double negation=positive}\ \end{align*}$$ From this, $p\oplus q\leftrightarrow p\oplus q$, therefore, $(\neg p\Leftrightarrow q)\leftrightarrow \neg(p\Leftrightarrow q) \>\>\blacksquare.$