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

## Theorem

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

## Proof

By the tableau method of natural deduction:

$\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies q}\right) \land \left({\neg p \implies q}\right)$ Premise (None)
2 1 $p \implies q$ Rule of Simplification: $\land \EE_1$ 1
3 1 $\neg p \implies q$ Rule of Simplification: $\land \EE_2$ 1
4 $p \lor \neg p$ Law of Excluded Middle (None)
5 5 $p$ Assumption (None)
6 1, 5 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 5
7 7 $\neg p$ Assumption (None)
8 1, 7 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 3, 7
9 1 $q$ Proof by Cases: $\text{PBC}$ 4, 5 – 6, 7 – 8 Assumptions 5 and 7 have been discharged

$\blacksquare$

## Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle.

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.