How to find my way in this proof (fitch natural deduction proof) P → ¬Q, ¬Q → P ∴ ¬(Q ↔︎ P) 
Hello all, I am very stuck in this proof. I'm still pretty much new to logic but I'm trying to get better at proofs with doing a bunch of practice proofs and this is one of them. It seems like I just can't find my way. Can anyone show me how to continue or whether I am correct up to now to begin with? Thank you. I would really appreciate if someone could show me visually.
The rules I use: ∧Intro, Elim,
∨Intro, Elim,
Conditional and bi-conditional rules
Reductio, negation elim, X, DS.
 A: As correctly pointed out by the OP and in the comments here, this argument is provable. It was indeed possible to find a proof.
Here is one possibility using Fitch-style natural deduction system:

A: It may be proved by contradition. The following is a formal proof:

*

*$P\rightarrow\sim Q$ (premise)


*$\sim Q\rightarrow P$ (premise)


*$P\leftrightarrow Q$ (assumption)


*$(P\rightarrow Q)\wedge(Q\rightarrow P)$ (equivalent form of $\leftrightarrow$, 3)


*$P\rightarrow Q$ (simplification, 4)


*$P$ (assumption)


*$Q$ (M.P., 5 and 6)


*$\sim Q$ (M.P., 1 and 6)


*$\sim Q\wedge Q$ (adjunction, 7 and 8)


*$\sim P$ (reductio ad absurdum, with lines 6--9 deleted)


*$\sim\sim Q$ (M.T., 2 and 10)


*$Q$ (Double Negation, 11)


*$Q\rightarrow P$ (simplification, 4)


*$\sim Q$ (M.T. 10 and 13)


*$\sim Q\wedge Q$ (adjunction, 12 and 14)


*$\sim(P\leftrightarrow Q)$ (reductio ad absurdum, with lines
3--15 deleted)
Therefore, $P\rightarrow\sim Q,\,\sim Q\rightarrow P\models\sim(P\leftrightarrow Q)$
A: It suffices to prove that $$(P\rightarrow \neg Q) \wedge (\neg Q\rightarrow P) \equiv \neg(P\leftrightarrow Q).$$ In fact, $$(P\rightarrow \neg Q) \wedge (\neg Q\rightarrow P) \equiv  P\leftrightarrow \neg Q,$$ so it suffices to show that $$\neg(P\leftrightarrow Q)\equiv P\leftrightarrow \neg Q.$$ We will prove the equivalent fact that $$ \neg(P\leftrightarrow \neg Q)\equiv P\leftrightarrow Q.$$
Indeed, we can use well-known congruences to find that
\begin{align*}
\neg(P\leftrightarrow \neg Q) &\equiv \neg((P\rightarrow \neg Q)\wedge (\neg Q\rightarrow P))\\
&\equiv \neg((\neg P\vee \neg Q)\wedge (Q\vee P))\\
&\equiv (\neg P\wedge \neg Q)\vee (P\wedge Q)\\
&\equiv (\neg P\vee (P\wedge Q))\wedge (\neg Q\vee (P\wedge Q))\\
&\equiv (\neg P\vee P)\wedge (\neg P\vee Q)\wedge (\neg Q\vee P)\wedge(\neg Q\vee Q)\\
&\equiv \top\wedge (\neg P\vee Q)\wedge (\neg Q\vee P)\wedge\top\\
&\equiv (\neg P\vee Q)\wedge (\neg Q\vee P)\\
&\equiv (P\rightarrow Q)\wedge (Q\rightarrow P)\\
&\equiv P\leftrightarrow Q.
\end{align*}
