Proof by Cases/Formulation 2/Reverse Implication

Theorem

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


Proof

By the tableau method of natural deduction:

$\vdash \paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} } $
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \lor q} \implies r$ Assumption (None)
2 1 $\paren {p \implies r} \land \paren {q \implies r}$ Sequent Introduction 1 Proof by Cases: Formulation 1: Reverse Implication
3 $\paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$

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