Principle of Dilemma/Formulation 1/Reverse Implication
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Theorem
- $q \vdash \left({p \implies q}\right) \land \left({\neg p \implies q}\right)$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $q$ | Premise | (None) | ||
2 | 1 | $p \implies q$ | Sequent Introduction | 1 | True Statement is implied by Every Statement | |
3 | 1 | $\neg p \implies q$ | Sequent Introduction | 1 | True Statement is implied by Every Statement | |
4 | 1 | $\left({p \implies q}\right) \land \left({\neg p \implies q}\right)$ | Rule of Conjunction: $\land \II$ | 2, 3 |
$\blacksquare$