Rule of Association

Definition
This rule is two-fold:


 * Conjunction is associative:
 * $p \land \left({q \land r}\right) \dashv \vdash \left({p \land q}\right) \land r$


 * Disjunction is associative:
 * $p \lor \left({q \lor r}\right) \dashv \vdash \left({p \lor q}\right) \lor r$

Its abbreviation in a tableau proof is $\textrm{Assoc}$.

Alternative rendition
These can alternatively be rendered as:


 * $\vdash \left({p \land \left({q \land r}\right)}\right) \iff \left({\left({p \land q}\right) \land r}\right)$
 * $\vdash \left({p \lor \left({q \lor r}\right)}\right) \iff \left({\left({p \lor q}\right) \lor r}\right)$

They can be seen to be logically equivalent to the forms above.

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $p$
 * $\land \mathcal E_1$
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $q \land r$
 * $\land \mathcal E_2$
 * 1
 * align="right" | 4 ||
 * align="right" | 1
 * $q$
 * $\land \mathcal E_1$
 * 3
 * align="right" | 5 ||
 * align="right" | 1
 * $r$
 * $\land \mathcal E_2$
 * 3
 * align="right" | 5 ||
 * align="right" | 1
 * $r$
 * $\land \mathcal E_2$
 * 3
 * $\land \mathcal E_2$
 * 3

$\left({p \land q}\right) \land r \vdash p \land \left({q \land r}\right)$

is proved similarly.


 * align="right" | 3 ||
 * align="right" | 2
 * $p \lor q$
 * $\lor \mathcal I_1$
 * 2
 * align="right" | 4 ||
 * align="right" | 2
 * $\left({p \lor q}\right) \lor r$
 * $\lor \mathcal I_1$
 * 2
 * ... we derive the conclusion.
 * $\lor \mathcal I_1$
 * 2
 * ... we derive the conclusion.


 * align="right" | 7 ||
 * align="right" | 6
 * $p \lor q$
 * $\lor \mathcal I_2$
 * 6
 * align="right" | 8 ||
 * align="right" | 6
 * $\left({p \lor q}\right) \lor r$
 * $\lor \mathcal I_1$
 * 7
 * ... we derive the conclusion.
 * $\lor \mathcal I_1$
 * 7
 * ... we derive the conclusion.


 * align="right" | 10 ||
 * align="right" | 9
 * $\left({p \lor q}\right) \lor r$
 * $\lor \mathcal I_2$
 * 9
 * ... and likewise we derive the same conclusion.
 * align="right" | 11 ||
 * align="right" | 5
 * $\left({p \lor q}\right) \lor r$
 * $\lor \mathcal E$
 * 5, 6-8, 9-10
 * align="right" | 12 ||
 * align="right" | 1
 * $\left({p \lor q}\right) \lor r$
 * $\lor \mathcal E$
 * 1, 2-4, 5-11
 * align="right" | 1
 * $\left({p \lor q}\right) \lor r$
 * $\lor \mathcal E$
 * 1, 2-4, 5-11

$\left({p \lor q}\right) \lor r \vdash p \lor \left({q \lor r}\right)$

is proved similarly.

Proof by Truth Table
We apply the Method of Truth Tables to the propositions in turn.

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

$\begin{array}{|ccccc||ccccc|} \hline p & \land & (q & \land & r) & (p & \land & q) & \land & r \\ \hline F & F & F & F & F & F & F & F & F & F \\ F & F & F & F & T & F & F & F & F & T \\ F & F & T & F & F & F & F & T & F & F \\ F & F & T & T & T & F & F & T & F & T \\ T & F & F & F & F & T & F & F & F & F \\ T & F & F & F & T & T & F & F & F & T \\ T & F & T & F & F & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

$\begin{array}{|ccccc||ccccc|} \hline p & \lor & (q & \lor & r) & (p & \lor & q) & \lor & r \\ \hline F & F & F & F & F & F & F & F & F & F \\ F & T & F & T & T & F & F & F & T & T \\ F & T & T & T & F & F & T & T & T & F \\ F & T & T & T & T & F & T & T & T & T \\ T & T & F & F & F & T & T & F & T & F \\ T & T & F & T & T & T & T & F & T & T \\ T & T & T & T & F & T & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$