# Proof by Cases/Formulation 1/Forward Implication

## Theorem

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

## Proof 1

By the tableau method of natural deduction:

$\left({p \implies r}\right) \land \left({q \implies r}\right) \vdash \left({p \lor q}\right) \implies r$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies r}\right) \land \left({q \implies r}\right)$ Premise (None)
2 1 $p \implies r$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $q \implies r$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $p \lor q \implies r \lor r$ Sequent Introduction 2, 3 Constructive Dilemma
5 5 $p \lor q$ Assumption (None)
6 1, 5 $r \lor r$ Modus Ponendo Ponens: $\implies \mathcal E$ 4, 5
7 1, 5 $r$ Sequent Introduction 6 Rule of Idempotence: Disjunction
8 1 $p \lor q \implies r$ Rule of Implication: $\implies \mathcal I$ 5 – 7 Assumption 5 has been discharged

$\blacksquare$

## Proof 2

From the Constructive Dilemma we have:

$p \implies q, r \implies s \vdash p \lor r \implies q \lor s$

from which, changing the names of letters strategically:

$p \implies r, q \implies r \vdash p \lor q \implies r \lor r$

From the Rule of Idempotence we have:

$r \lor r \vdash r$

and the result follows by Hypothetical Syllogism.

$\blacksquare$