Principle of Dilemma

Formulation 1

 * $p \implies q, \neg p \implies q \dashv \vdash q$

Formulation 2

 * $\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \iff q$

Proof
By the tableau method of natural deduction:

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:
 * $p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$

From the Law of Excluded Middle, we have:
 * $\vdash p \lor \neg p$

From the Rule of Idempotence we have:
 * $q \lor q \vdash q$

and the result follows by Hypothetical Syllogism.

Proof 2
We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all models.

$\begin{array}{|cccccccc||c|} \hline (p & \implies & q) & \land & (\neg & p & \implies & q) & q \\ \hline F & T & F & F & T & F & F & F & F \\ F & T & T & T & T & F & T & T & T \\ T & F & F & F & F & T & T & F & F \\ T & T & T & T & F & T & T & T & T \\ \hline \end{array}$

Also known as
This proof structure is also referred to as proof by cases but it is usual to reserve that name for a more general concept.

Also see

 * Proof by Cases

Note
This depends on the Law of the Excluded Middle, which is disallowed by the intuitionist school.