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)

For the next two lessons, try to prove these:

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