Proof by Cases

Context
The rule of or-elimination is one of the axioms of natural deduction.

The rule
If we can conclude $p \lor q$, and:


 * 1) By making the assumption $p$, we can conclude $r$;
 * 2) By making the assumption $q$, we can conclude $r$;

then we may infer $r$.


 * \(\displaystyle p \lor q, \left({p \vdash r}\right), \left({q \vdash r}\right) \vdash r\)

The conclusion does not depend upon either assumption $p$ or $q$.

This is also known as proof by cases, but this is also used for an extension of this concept.

It can be written:
 * $\displaystyle {p \lor q \quad \begin{array}{|c|} \hline p \\ \vdots \\ r \\ \hline \end{array} \quad \begin{array}{|c|} \hline q \\ \vdots \\ r \\ \hline \end{array} \over r} \lor_e$


 * Abbreviation: $\lor \mathcal E$
 * Deduced from: The pooled assumptions of:
 * 1) The instance of $p \lor q$;
 * 2) The instance of $r$ which was derived from the individual assumption $p$;
 * 3) The instance of $r$ which was derived from the individual assumption $q$.


 * Discharged Assumptions: The assumptions $p$ and $q$.
 * Depends on: The following three things:
 * 1) The line containing the instance of $p \lor q$;
 * 2) The series of lines from where the assumption $p$ was made to where $r$ was deduced;
 * 3) The series of lines from where the assumption $q$ was made to where $r$ was deduced.

Explanation
We know $p \lor q$, that is, either $p$ is true or $q$ is true, or both.

Suppose we assume that $p$ is true, and from that assumption we have managed to deduce that $r$ has to be true.

Then suppose we assume that $q$ is true, and from that assumption we have also managed to deduce that $r$ has to be true.

Therefore, it has to follow that the truth of $r$ follows from the fact of the truth of $p \lor q$.

Thus we can eliminate a disjunction from a sequent.

Demonstration by Truth Table
$\begin{array}{|ccc|ccc|ccc||c|} \hline p & \lor & q & p & \implies & r & q & \implies & r & r \\ \hline F & F & F & F & T & F & F & T & F & F \\ F & F & F & F & T & T & F & T & T & T \\ F & T & T & F & T & F & T & F & F & F \\ F & T & T & F & T & T & T & T & T & T \\ T & T & F & T & F & F & F & T & F & F \\ T & T & F & T & T & T & F & T & T & T \\ T & T & T & T & F & F & T & F & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

As can be seen, when $p \lor q$, $p \implies r$ and $q \implies r$ are all true, then so is $r$.