Factor Principles/Disjunction on Right/Formulation 1
Theorem
- $p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$
Proof 1
By the tableau method of natural deduction:
| 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 | $\paren {p \lor r} \implies \paren {q \lor r}$ | Sequent Introduction | 1, 2 | Constructive Dilemma |
$\blacksquare$
Proof 2
By the tableau method of natural deduction:
| 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: $\lor \II_2$ | 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: $\lor \II_1$ | 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 | $\paren {p \lor r} \implies \paren {q \lor r}$ | Rule of Implication: $\implies \II$ | 2 – 8 | Assumption 2 has been discharged |
$\blacksquare$
Proof 3
By the tableau method of natural deduction:
| 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 | $\paren {r \lor p} \implies \paren {r \lor q}$ | 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 | $\paren {p \lor r} \implies \paren {q \lor r}$ | Rule of Implication: $\implies \II$ | 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 $T$:
- $\begin{array}{|ccc||ccc||ccccccc|} \hline
p & q & r & (p & \implies & q) & (p & \lor & r) & \implies & (q & \lor & r) \\ \hline F & F & F & F & T & F & F & F & F & T & F & F & F \\ F & F & T & F & T & F & F & T & T & T & F & T & T \\ F & T & F & F & T & T & F & F & F & T & T & T & F \\ F & T & T & F & T & T & F & T & T & T & T & T & T \\ T & F & F & T & F & F & T & T & F & F & F & F & F \\ T & F & T & T & F & F & T & T & T & T & F & T & T \\ T & T & F & T & T & T & T & T & F & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$
$\blacksquare$