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 $$r$$.