Rule of Addition

Axiom
The rule of addition is one of the axioms of natural deduction.

This is two axioms in one.
 * $(1): \quad$ If we can conclude $p$, then we may infer $p \lor q$.
 * $(2): \quad$ If we can conclude $p$, then we may infer $q \lor p$.

Sequent Form
The rule of addition is symbolised by the sequents:


 * $(1): \quad p \vdash p \lor q$
 * $(2): \quad p \vdash q \lor p$

It can be written:
 * $\displaystyle {p \over p \lor q} \lor_{i_1} \qquad \qquad {q \over p \lor q} \lor_{i_2}$

Tableau Form
In a tableau proof, the rule of addition can be invoked in the following manner:


 * Abbreviation: $\lor \mathcal I_1$ or $\lor \mathcal I_2$
 * Deduced from: The pooled assumptions of $p$.
 * Depends on: The line containing $p$.

Explanation
Note that there are two axioms here in one. The first of the two tells us that, given a statement, we may infer a disjunction where the given statement is the first of the disjuncts, while the second says that, given a statement, we may infer a disjunction where the given statement is the second of the disjuncts.

At this stage, such attention to detail is important.

The statement $q$ being added may be any statement at all. It does not matter what its truth value is. If $p$ is true, then $p \vdash p \lor q$ is true, whatever $q$ may be.

This may seem a bewildering and perhaps paradoxical axiom to admit. How can you deduce a valid argument from a statement form that can deliberately be used to include a statement whose truth value can be completely arbitrary? Or even blatantly false?

But consider the common (although admittedly rhetorical) figure of speech which goes:


 * "Reading Town are going up this season or I'm a Dutchman."

Also known as
This is sometimes known as the rule of or-introduction.

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

As can be seen, whenever either $p$ or $q$ (or both) are true, then so is $p \lor q$.