# Proof by Cases/Formulation 3

## Theorem

$\vdash \left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r$

## Proof

By the tableau method of natural deduction:

$\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r$
Line Pool Formula Rule Depends upon Notes
1 $\left({\left({p \lor r}\right) \land \left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({q \lor s}\right)$ Theorem Introduction (None) Constructive Dilemma: Formulation 2
2 $\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({r \lor r}\right)$ Substitution 1 $r \ / \ q$, $q \ / \ r$, $s \ / \ r$
3 3 $\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)$ 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 $\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r$ Rule of Implication: $\implies \mathcal I$ 3 – 5 Assumption 3 has been discharged

$\blacksquare$