# Factor Principles/Conjunction on Right/Formulation 1

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## Theorem

$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$

## Proof 1

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 $r \implies r$ Law of Identity (None) This is a theorem so depends on nothing
3 1 $\paren {p \implies q} \land \paren {r \implies r}$ Rule of Conjunction: $\land \II$ 1, 2
4 1 $\paren {p \land r} \implies \paren {q \land r}$ 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: $\land \EE_1$ 2
4 1, 2 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 3
5 2 $r$ Rule of Simplification: $\land \EE_2$ 2
6 1, 2 $q \land r$ Rule of Conjunction: $\land \II$ 4, 5
7 1 $\paren {p \land r} \implies \paren {q \land r}$ Rule of Implication: $\implies \II$ 2 – 6 Assumption 2 has been discharged

$\blacksquare$

## Proof by Truth Table

Proof by Truth Table:

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

$\blacksquare$