Proof by Contradiction/Variant 2/Formulation 2

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Theorem

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


Proof

By the tableau method of natural deduction:

$\vdash \left({\left({p \implies q}\right) \land \left({p \implies \neg q}\right)}\right) \implies \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies q}\right) \land \left({p \implies \neg q}\right)$ Assumption (None)
2 1 $p \implies q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $p \implies \neg q$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $\neg p$ Sequent Introduction 2, 3 Proof by Contradiction: Formulation 1
5 $\left({\left({p \implies q}\right) \land \left({p \implies \neg q}\right)}\right) \implies \neg p$ Rule of Implication: $\implies \mathcal I$ 1 – 4 Assumption 1 has been discharged


$\blacksquare$


Sources