How to find my way in this proof (fitch natural deduction proof) P → ¬Q, ¬Q → P ∴ ¬(Q ↔︎ P)

Question

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.

Answer 1

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:

Answer 2

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

1. $P\rightarrow\sim Q$ (premise)

2. $\sim Q\rightarrow P$ (premise)

3. $P\leftrightarrow Q$ (assumption)

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

5. $P\rightarrow Q$ (simplification, 4)

6. $P$ (assumption)

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

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

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

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

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

12. $Q$ (Double Negation, 11)

13. $Q\rightarrow P$ (simplification, 4)

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

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

16. $\sim(P\leftrightarrow Q)$ (reductio ad absurdum, with lines 3--15 deleted)

Therefore, $P\rightarrow\sim Q,\,\sim Q\rightarrow P\models\sim(P\leftrightarrow Q)$

Answer 3

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*}