# Factor Principles/Conjunction on Right/Formulation 1

## Theorem

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

## Proof 1

By the tableau method of natural deduction:

$p \implies q \vdash \left({p \land r}\right) \implies \left ({q \land r}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 $r \implies r$ Law of Identity (None) This is a theorem so depends on nothing
3 1 $\left({p \implies q}\right) \land \left({r \implies r}\right)$ Rule of Conjunction: $\land \mathcal I$ 1, 2
4 1 $\left({p \land r}\right) \implies \left ({q \land r}\right)$ Sequent Introduction 3 Praeclarum Theorema

$\blacksquare$

## Proof 2

By the tableau method of natural deduction:

$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $p \land r$ Assumption (None)
3 2 $p$ Rule of Simplification: $\land \mathcal E_1$ 2
4 1, 2 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 3
5 2 $r$ Rule of Simplification: $\land \mathcal E_2$ 2
6 1, 2 $q \land r$ Rule of Conjunction: $\land \mathcal I$ 4, 5
7 1 $\paren {p \land r} \implies \paren {q \land r}$ Rule of Implication: $\implies \mathcal I$ 2 – 6 Assumption 2 has been discharged

$\blacksquare$

## Proof 3

Proof by Truth Table.

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

$\blacksquare$