Theorem
 $\neg \left ({p \iff q}\right) \vdash \left({p \iff \neg q}\right)$
Proof
By the tableau method of natural deduction:
$\neg \left ({p \iff q}\right) \vdash \left({p \iff \neg q}\right)$
Line 

Pool

Formula

Rule

Depends upon

Notes

1 

1

$\neg \left ({p \iff q}\right)$

Premise

(None)


2 

1

$\neg \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)$

Sequent Introduction

1

Rule of Material Equivalence

3 

1

$\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$

Sequent Introduction

2

De Morgan's Laws: Disjunction of Negations

4 

4

$p$

Assumption

(None)


5 

4

$p$

Law of Identity

4


6 

6

$q$

Assumption

(None)


7 

6

$p \implies q$

Rule of Implication: $\implies \mathcal I$

4 – 6

Assumption 4 has been discharged

8 

6

$\neg \neg \left({p \implies q}\right)$

Double Negation Introduction: $\neg \neg \mathcal I$

7


9 

1, 6

$\neg \left({q \implies p}\right)$

Modus Tollendo Ponens $\mathrm{MTP}_1$

3, 8


10 

1, 6

$q \land \neg p$

Sequent Introduction

9

Conjunction with Negative Equivalent to Negation of Implication

11 

1, 6

$\neg p$

Rule of Simplification: $\land \mathcal E_2$

10


12 

1, 4, 6

$\bot$

Principle of NonContradiction: $\neg \mathcal E$

4, 11


13 

1, 4

$\neg q$

Proof by Contradiction: $\neg \mathcal I$

6 – 12

Assumption 6 has been discharged

14 

1

$p \implies \neg q$

Rule of Implication: $\implies \mathcal I$

4 – 13

Assumption 4 has been discharged

15 

15

$\neg q$

Assumption

(None)


16 

15

$\neg q$

Law of Identity

15


17 

17

$\neg p$

Assumption

(None)


18 

17

$\neg q \implies \neg p$

Rule of Implication: $\implies \mathcal I$

16 – 17

Assumption 16 has been discharged

19 

17

$p \implies q$

Sequent Introduction

18

Rule of Transposition

20 

1, 17

$\neg \left({q \implies p}\right)$

Modus Tollendo Ponens $\mathrm{MTP}_1$

3, 19


21 

1, 17

$q \land \neg p$

Sequent Introduction

20

Conjunction with Negative Equivalent to Negation of Implication

22 

1, 17

$q$

Rule of Simplification: $\land \mathcal E_1$

21


23 

1, 15, 17

$\bot$

Principle of NonContradiction: $\neg \mathcal E$

15, 22


24 

1, 15

$p$

Proof by Contradiction: $\neg \mathcal I$

17 – 23

Assumption 17 has been discharged

25 

1

$\neg q \implies p$

Rule of Implication: $\implies \mathcal I$

15 – 24

Assumption 15 has been discharged

26 

1

$\left({p \iff \neg q}\right)$

Biconditional Introduction: $\iff \mathcal I$

14, 25


$\blacksquare$
Law of the Excluded Middle
This proof depends on the Law of the Excluded Middle.
This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.
However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.