# Proof by Cases/Formulation 3

## Theorem

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

## Proof

By the tableau method of natural deduction:

$\paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r$
Line Pool Formula Rule Depends upon Notes
1 $\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$ Theorem Introduction (None) Constructive Dilemma: Formulation 3
2 $\paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies \paren {r \lor r}$ Substitution 1 $r \ / \ q$, $q \ / \ r$, $s \ / \ r$
3 3 $\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r}$ Assumption (None)
4 3 $r \lor r$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 3
5 3 $r$ Sequent Introduction 4 Rule of Idempotence: Disjunction
6 $\paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r$ Rule of Implication: $\implies \II$ 3 – 5 Assumption 3 has been discharged

$\blacksquare$