Negation of Conditional implies Antecedent

Theorem

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

Proof

By the tableau method of natural deduction:

$\vdash \neg \paren {p \implies q} \implies p$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \implies q}$ Assumption (None)
2 1 $p \land \neg q$ Sequent Introduction 1 Conjunction with Negative Equivalent to Negation of Implication
3 1 $p$ Rule of Simplification: $\land \mathcal E_1$ 2
4 $\neg \paren {p \implies q} \implies p$ Rule of Implication: $\implies \mathcal I$ 1 – 3 Assumption 1 has been discharged

$\blacksquare$