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:
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: $\neg \EE$ | 3, 2 | ... and demonstrate a contradiction | |
5 | 2, 3 | $q$ | Rule of Explosion: $\bot \EE$ | 4 | ... from a falsehood, any statement can be derived - pick $q$ | |
6 | 2 | $p \implies q$ | Rule of Implication: $\implies \II$ | 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: $\implies \II$ | 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:
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 | Conditional is Equivalent to Negation of Conjunction with Negative: $\neg \paren {p \land \neg q} \vdash p \implies q$ | |
4 | 4 | $p$ | Assumption | (None) | ||
5 | 4 | $\neg \neg p$ | Double Negation Introduction: $\neg \neg \II$ | 4 | ||
6 | 1, 4 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 3, 5 | ||
7 | 1 | $p \implies q$ | Rule of Implication: $\implies \II$ | 4 – 6 | Assumption 4 has been discharged |
$\blacksquare$