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

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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 | $p \implies q$ | 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 |

$\blacksquare$

## Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle, by way of Double Negation Elimination.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this theorem from an intuitionistic perspective.

## Sources

- 1964: Donald Kalish and Richard Montague:
*Logic: Techniques of Formal Reasoning*... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T33}$