# Factor Principles/Conjunction on Left/Formulation 1

## Theorem

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

## Proof 1

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

$\blacksquare$

## Proof 2

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

$\blacksquare$