Proof by Contradiction/Variant 3/Formulation 2/Proof 2

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Theorem

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


Proof

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


$\blacksquare$


Sources