Rule of Material Implication/Formulation 1/Reverse Implication/Proof 1
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Theorem
- $\neg p \lor q \vdash p \implies q$
Proof
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 |
$\blacksquare$
Sources
- 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction: Example $1.20$