Here we look at how to prove disjunctions through conditional proof. This lesson is available as a video or as text. Scroll past the video for the text version.

First, let's take a look at the rule of material implication.

Material Implication:

(P ⊃ Q) ≡ (~P ∨ Q)

This rule shows that a conditional is materially equivalent to a disjunction where the first disjunct is the negation of the conditional's antecedent, and the second disjunct is the consequent of the conditional. Remember that the conditional here is just a material conditional. It expresses relations between truth values, not cause and effect. Even though it doesn't richly capture what we always mean by "If ... then ...", it normally captures the logically relevant characteristics of if-then statements. A conditional is true for three combinations of truth values, and so is a disjunction. When we negate the first disjunct of a disjunction, its truth values line up with those of a conditional, as seen in this truth table for material implication.

( | P | ⊃ | Q | ) | ≡ | ( | ~ | P | ∨ | Q | ) |
---|---|---|---|---|---|---|---|---|---|---|---|

T | T | T | T | F | T | T | T | ||||

T | F | F | T | F | T | F | F | ||||

F | T | T | T | T | F | T | T | ||||

F | T | F | T | T | F | T | F |

Using the rule of material implication, we can prove a disjunction like so:

To Prove ~P ∨ Q:

- Assume P
- Derive Q
- Infer P ⊃ Q with Conditional Proof
- Infer ~P ∨ Q with Material Implication

This works well for a disjunction that is already in the form that corresponds to a conditional. But what if we want to prove one of the form P ∨ Q instead?

To Prove P ∨ Q

- Assume ~P
- Derive Q
- Infer ~P ⊃ Q with Conditional Proof
- Infer ~~P ∨ Q with Material Implication
- Infer P ∨ Q with Double Negation

This is essentially the same with the additional step of double negation.

Without using Commutation for disjunctions, prove it:

(P ∨ Q) ≡ (Q ∨ P)

Without using Association,
Prove the rule of Association for disjunctions:

Prove: (P ∨ (Q ∨ R)) ≡ ((P ∨ Q) ∨ R)

Distribution: (P & (Q ∨ R)) ≡ ((P & Q) ∨ (P & R)) (P ∨ (Q & R)) ≡ ((P ∨ Q) & (P ∨ R)) Material Implication (~P ∨ Q) ≡ (P ⊃ Q)