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 is Equivalent to Negation of Conditional
3 1 $p$ Rule of Simplification: $\land \EE_1$ 2
4 $\neg \paren {p \implies q} \implies p$ Rule of Implication: $\implies \II$ 1 – 3 Assumption 1 has been discharged

$\blacksquare$


Sources

  • 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T21}$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.