# Factor Principles/Conjunction on Left/Formulation 1/Proof 1

## Theorem

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

## Proof

By the tableau method of natural deduction:

$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$
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 $\paren {r \implies r} \land \paren {p \implies q}$ Rule of Conjunction: $\land \mathcal I$ 2, 1
4 1 $\paren {r \land p} \implies \paren {r \land q}$ Sequent Introduction 3 Praeclarum Theorema

$\blacksquare$