Proof by Cases/Formulation 2/Proof

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:

$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r} $
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 $\text{T50}$
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