Rule of Association/Conjunction/Formulation 1

Theorem

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

Proof 1

By the tableau method of natural deduction:

$p \land \left({q \land r}\right) \vdash \left({p \land q}\right) \land r$
Line Pool Formula Rule Depends upon Notes
1 1 $p \land \left({q \land r}\right)$ Premise (None)
2 1 $p$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $q \land r$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $q$ Rule of Simplification: $\land \mathcal E_1$ 3
5 1 $r$ Rule of Simplification: $\land \mathcal E_2$ 3
6 1 $p \land q$ Rule of Conjunction: $\land \mathcal I$ 2, 4
7 1 $\left({p \land q}\right) \land r$ Rule of Conjunction: $\land \mathcal I$ 6, 5

By the tableau method of natural deduction:

$\left({p \land q}\right) \land r \vdash p \land \left({q \land r}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \land q}\right) \land r$ Premise (None)
2 1 $p \land q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $r$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $p$ Rule of Simplification: $\land \mathcal E_1$ 2
5 1 $q$ Rule of Simplification: $\land \mathcal E_2$ 2
6 1 $q \land r$ Rule of Conjunction: $\land \mathcal I$ 5, 3
7 1 $p \land \left({q \land r}\right)$ Rule of Conjunction: $\land \mathcal I$ 4, 6

$\blacksquare$

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 boolean interpretations.

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

$\blacksquare$