Rule of Material Implication/Formulation 1/Reverse Implication/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

$\neg p \lor q \vdash p \implies q$


Proof

By the tableau method of natural deduction:

$\neg p \lor q \vdash p \implies q$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg p \lor q$ Premise (None)
2 1 $\neg \paren {\neg \neg p \land \neg q}$ Sequent Introduction 1 De Morgan's Laws: Disjunction
3 1 $\neg \neg p \implies q$ Sequent Introduction 2 Implication Equivalent to Negation of Conjunction with Negative: $\neg \left({p \land \neg q}\right) \vdash p \implies q$
4 4 $p$ Assumption (None)
5 4 $\neg \neg p$ Double Negation Introduction: $\neg \neg \mathcal I$ 4
6 1, 4 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 3, 5
7 1 $p \implies q$ Rule of Implication: $\implies \mathcal I$ 4 – 6 Assumption 4 has been discharged

$\blacksquare$


Sources