Proof by Cases/Formulation 1/Reverse Implication

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Theorem

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


Proof

By the tableau method of natural deduction:

$\paren {p \lor q} \implies r \vdash \paren {p \implies r} \land \paren {q \implies r} $
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \lor q} \implies r$ Premise (None)
2 2 $p$ Assumption (None)
3 2 $p \lor q$ Rule of Addition: $\lor \II_1$ 2
4 1, 2 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 3
5 1 $p \implies r$ Rule of Implication: $\implies \II$ 2 – 4 Assumption 2 has been discharged
6 6 $q$ Assumption (None)
7 6 $p \lor q$ Rule of Addition: $\lor \II_2$ 6
8 1, 6 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 7
9 1 $q \implies r$ Rule of Implication: $\implies \II$ 6 – 8 Assumption 6 has been discharged
10 1 $\paren {p \implies r} \land \paren {q \implies r}$ Rule of Conjunction: $\land \II$ 5, 9

$\blacksquare$