Proof by Cases/Formulation 2/Proof
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Theorem
- $\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}$ | Theorem Introduction | (None) | Proof by Cases: Forward Implication | ||
2 | $\paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} }$ | Theorem Introduction | (None) | Proof by Cases: Reverse Implication | ||
3 | $\paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$ | Biconditional Introduction: $\iff \II$ | 1, 2 |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T50}$