# Principle of Dilemma/Formulation 2/Forward Implication

## Theorem

$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$

## Proof 1

By the tableau method of natural deduction:

$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \land \paren {\neg p \implies q}$ Assumption (None)
2 1 $q$ Sequent Introduction 1 Principle of Dilemma: Formulation 1
3 $\paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$

## Proof 2

By the tableau method of natural deduction:

$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \land \paren {\neg p \implies q}$ Assumption (None)
2 1 $p \implies q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $\neg p \implies q$ Rule of Simplification: $\land \mathcal E_2$ 1
4 4 $\neg q$ Assumption (None)
5 1, 4 $\neg p$ Modus Tollendo Tollens (MTT) 2, 4
6 1, 4 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 3, 5
7 1, 4 $\bot$ Principle of Non-Contradiction: $\neg \mathcal E$ 4, 6
8 1 $\neg \neg q$ Proof by Contradiction: $\neg \mathcal I$ 4 – 7 Assumption 4 has been discharged
9 1 $q$ Double Negation Elimination: $\neg \neg \mathcal E$ 8
10 $\paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ Rule of Implication: $\implies \mathcal I$ 1 – 9 Assumption 1 has been discharged

$\blacksquare$