# Proof by Contradiction/Variant 3/Formulation 2

## Theorem

$\vdash \paren {p \implies \neg p} \implies \neg p$

## Proof 1

By the tableau method of natural deduction:

$\vdash \left({p \implies \neg p}\right) \implies \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \neg p$ Assumption (None)
2 1 $\neg p$ Sequent Introduction 1 Proof by Contradiction: Variant 3: Formulation 1
3 $\left({p \implies \neg p}\right) \implies \neg p$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$

## Proof 2

This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.

By the tableau method:

$\vdash \left({p \implies \neg p}\right) \implies \neg p$
Line Pool Formula Rule Depends upon Notes
1 $\left({p \lor p}\right) \implies p$ Axiom $A1$
2 $\left({\neg p \lor \neg p}\right) \implies \neg p$ Rule $RST \, 1$ 1 $\neg p \, / \, p$
3 $\left({p \implies \neg p}\right) \implies \neg p$ Rule $RST \, 2 \, (2)$ 2

$\blacksquare$