# Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication

## Theorem

$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$

## Proof

By the tableau method of natural deduction:

$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \iff q}$ Premise (None)
2 1 $\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }$ Sequent Introduction 1 Rule of Material Equivalence
3 1 $\neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ 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: $\implies \II$ 4 – 6 Assumption 4 has been discharged
8 6 $\neg \neg \paren {p \implies q}$ Double Negation Introduction: $\neg \neg \II$ 7
9 1, 6 $\neg \paren {q \implies p}$ 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: $\land \EE_2$ 10
12 1, 4, 6 $\bot$ Principle of Non-Contradiction: $\neg \EE$ 4, 11
13 1, 4 $\neg q$ Proof by Contradiction: $\neg \II$ 6 – 12 Assumption 6 has been discharged
14 1 $p \implies \neg q$ Rule of Implication: $\implies \II$ 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: $\implies \II$ 16 – 17 Assumption 16 has been discharged
19 17 $p \implies q$ Sequent Introduction 18 Rule of Transposition
20 1, 17 $\neg \paren {q \implies p}$ 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: $\land \EE_1$ 21
23 1, 15, 17 $\bot$ Principle of Non-Contradiction: $\neg \EE$ 15, 22
24 1, 15 $p$ Proof by Contradiction: $\neg \II$ 17 – 23 Assumption 17 has been discharged
25 1 $\neg q \implies p$ Rule of Implication: $\implies \II$ 15 – 24 Assumption 15 has been discharged
26 1 $\paren {p \iff \neg q}$ Biconditional Introduction: $\iff \II$ 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.