# Factor Principles/Disjunction on Left/Formulation 1

## Theorem

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

## Proof 1

By the tableau method of natural deduction:

$p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 $r \implies r$ Theorem Introduction (None) Law of Identity: Formulation 2
3 1 $\paren {r \lor p} \implies \paren {r \lor q}$ Sequent Introduction 2, 1 Constructive Dilemma

$\blacksquare$

## Proof 2

By the tableau method of natural deduction:

$p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $r \lor p$ Assumption (None)
3 3 $r$ Assumption (None)
4 3 $r \lor q$ Rule of Addition: $\lor \II_1$ 3
5 5 $p$ Assumption (None)
6 1, 5 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 5
7 1, 5 $r \lor q$ Rule of Addition: $\lor \II_2$ 6
8 1, 2 $r \lor q$ Proof by Cases: $\text{PBC}$ 2, 3 – 4, 5 – 6 Assumptions 3 and 5 have been discharged
9 1 $\paren {r \lor p} \implies \paren {r \lor q}$ Rule of Implication: $\implies \II$ 2 – 7 Assumption 2 has been discharged

$\blacksquare$