# Factor Principles/Disjunction on Right/Formulation 1

## Theorem

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

## Proof 1

By the tableau method of natural deduction:

$p \implies q \vdash \left({p \lor r}\right) \implies \left ({q \lor r}\right)$
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 $\left({p \lor r}\right) \implies \left ({q \lor r}\right)$ Sequent Introduction 1, 2 Constructive Dilemma

$\blacksquare$

## Proof 2

By the tableau method of natural deduction:

$p \implies q \vdash \left({p \lor r}\right) \implies \left ({q \lor r}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $p \lor r$ Assumption (None)
3 3 $r$ Assumption (None)
4 3 $q \lor r$ Rule of Addition: $\lor \mathcal I_2$ 3
5 5 $p$ Assumption (None)
6 1, 5 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 5
7 1, 5 $q \lor r$ Rule of Addition: $\lor \mathcal I_1$ 6
8 1, 2 $q \lor r$ Proof by Cases: $\text{PBC}$ 2, 5 – 7, 3 – 4 Assumptions 5 and 3 have been discharged
9 1 $\left({p \lor r}\right) \implies \left ({q \lor r}\right)$ Rule of Implication: $\implies \mathcal I$ 2 – 8 Assumption 2 has been discharged

$\blacksquare$

## Proof 3

By the tableau method of natural deduction:

$p \implies q \vdash \left({p \lor r}\right) \implies \left ({q \lor r}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $p \lor r$ Assumption (None)
3 2 $r \lor p$ Sequent Introduction 2 Disjunction is Commutative
4 1 $\left({r \lor p}\right) \implies \left ({r \lor q}\right)$ Sequent Introduction 1 Factor Principles/Disjunction on Left/Formulation 1
5 1,2 $r \lor q$ Modus Ponendo Ponens: $\implies \mathcal E$ 4, 3
6 1,2 $q \lor r$ Sequent Introduction 5 Disjunction is Commutative
7 1 $\left({p \lor r}\right) \implies \left ({q \lor r}\right)$ Rule of Implication: $\implies \mathcal I$ 2 – 6 Assumption 2 has been discharged

$\blacksquare$

## Proof 4

As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the left hand side is $T$, that under the one on the right hand side is also $T$:

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

$\blacksquare$