Proof by Cases/Formulation 2
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Theorem
- $\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$
This can be expressed as two separate theorems:
Forward Implication
- $\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}$
Reverse Implication
- $\vdash \paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} }$
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 |
$\blacksquare$