# Principle of Dilemma/Formulation 2/Forward Implication/Proof 1

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## Theorem

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

## Proof

By the tableau method of natural deduction:

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 |

$\blacksquare$