Nesting Conditional Proofs

Here we’re looking at the idea of nesting conditional proofs. This means to nest one assumption inside of another, closing off the scope of the inner assumption before we close off the scope of the outer assumption. We may nest as many conditional proofs as we like, though there are restrictions on what we can do with them. Something may be derived from a previous line in a proof only if that previous line is still part of an open scope. If it belongs to a conditional proof that has since been closed, it is no longer available. This is what keeps us from using conditional proof to prove anything at all. The whole point of conditional proof is prove a conditional by showing that a particular assumption would lead to a particular conclusion. What we can conclude from this is that if our assumption were true, then the conclusion following from it would be true. Nested conditional proofs will be illustrated with a proof of the rule of exportation. You can see this covered in the video, try to do the proof on your own, or just look at the answer below.

Prove the rule of Exportation without using it:

((P & Q) ⊃ R) ≡ (P ⊃ (Q ⊃ R))

 1. | (P & Q) ⊃ R // Assumption
 2. || P // Assumption
 3. ||| Q // Assumption
 4. ||| P & Q // 2,3 Conjunction
 5. ||| R // 1,4 Modus Ponens
 6. || Q ⊃ R // 3-5 Conditional Proof
 7. | P ⊃ (Q ⊃ R) // 2-6 Conditional Proof
 8. ((P & Q) ⊃ R) ⊃ (P ⊃ (Q ⊃ R)) // 1-7 Conditional Proof
 9. | P ⊃ (Q ⊃ R) // Assumption
10. || P & Q // Assumption
11. || P // 10 Simplification
12. || Q ⊃ R // 9,11 Modus Ponens
13. || Q // 10 Simplification
14. || R // 12,13 Modus Ponens
15. | (P & Q) ⊃ R // 10-14 Conditional Proof
16. (P ⊃ (Q ⊃ R)) ⊃ ((P & Q) ⊃ R) // 9-15 Conditional Proof
17. ((P & Q) ⊃ R) ≡ (P ⊃ (Q ⊃ R)) // 8,16 Biconditional Intro.

For Next Time:

Prove the disjunctive forms of

Commutation
(P ∨ Q) ≡ (Q ∨ P)

Association
(P ∨ (Q ∨ R)) ≡ ((P ∨ Q) ∨ R)