# Rule of Material Implication/Formulation 1/Reverse Implication

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## Theorem

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

## Proof 1

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 2 $\neg p$ Assumption (None) Pick the first of the disjuncts ...
3 3 $p$ Assumption (None) Assume its negation ...
4 2, 3 $\bot$ Principle of Non-Contradiction: $\neg \mathcal E$ 3, 2 ... and demonstrate a contradiction
5 2, 3 $q$ Rule of Explosion: $\bot \mathcal E$ 4 ... from a falsehood, any statement can be derived - pick $q$
6 2 $p \implies q$ Rule of Implication: $\implies \mathcal I$ 3 – 5 Assumption 3 has been discharged
7 7 $q$ Assumption (None) Pick the second of the disjuncts ...
8 8 $p$ Assumption (None) ... again assume $p$ ...
9 7 $q$ Law of Identity 7 The truth of $q$ still holds
10 7 $p \implies q$ Rule of Implication: $\implies \mathcal I$ 8 – 9 Assumption 8 has been discharged
11 1 $p \implies q$ Proof by Cases: $\text{PBC}$ 1, 2 – 6, 7 – 10 Assumptions 2 and 7 have been discharged

$\blacksquare$

## Proof 2

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

$\blacksquare$