Practice Proofs -- Some Solutions and Hints (3.3) (P -> Q) -> Q |- P v Q 1. Assume (P -> Q) -> Q 2. Assume -(P v Q) for RAA. 3. Use SI(DeMorgan's) to derive -P & -Q. 4. Use SI(negative paradox) to derive P -> Q. 5. Use MPP on 1 and 4 to derive Q. (3.4) Chain Order: |- (P -> Q) v (Q -> P) Method A: 1. Show |- Q v -Q. 2. Assume Q. Use SI(positive paradox) to derive P -> Q. Use vI to derive (P -> Q) v (Q -> P). 3. Assume -Q. Use SI(negative paradox) to derive Q -> P. Use vI to derive (P -> Q) v (Q -> P). 4. Use vE to derive (P -> Q) v (Q -> P) Method B: 1. Use CP twice to show that |- Q -> (-P -> (P -> Q)). 2. Use SI(contrapositive) to derive Q -> (-(P -> Q) -> P). 3. Use SI(permutation) to derive -(P -> Q) -> (Q -> P). 4. Use SI(material implication) to finish the proof. Method C: Show first that -(P -> Q) -> (Q -> P), and then use SI(material implication) to finish the proof 1. Assume -(P -> Q) for CP. 2. Use SI(material implication) to derive -Q. 3. Use SI(negative paradox) to derive Q -> P. Method D: 1. Assume -[(P -> Q) v (Q -> P)] for RAA. 2. Use SI(DeMorgan's) to derive -(P -> Q). 3. Use SI(material implication) to derive -Q. 4. Use SI(negative paradox) to derive Q -> P. 5. Use vI to derive (P -> Q) v (Q -> P). (3.5) |- (P -> Q) v (Q -> R) Use Method A from Chain Order, with appropriate replacements of "R" for "P". (3.6) Material Implication P -> Q |- -P v Q 1 (1) P -> Q A 2 (2) -(-P v Q) A 3 (3) -Q A 1,3 (4) -P 1,3 MTT 1,3 (5) -P v Q 4 vI 1,2,3 (6) (-P v Q) & -(-P v Q) 5,2 &I 1,2 (7) --Q 3,6 RAA 1,2 (8) Q 7 DN 1,2 (9) -P v Q 8 vI 1,2 (10) (-P v Q) & -(-P v Q) 9,2 &I 1 (11) --(-P v Q) 2,10 RAA 1 (12) -P v Q 11 DN -P v Q |- P -> Q 1 (1) -P v Q A 2 (2) P A 3 (3) -P A 2,3 (4) Q 2,3 SI(Ex Falso) 5 (5) Q A 1,2 (6) Q 1,3,4,5,5 vE 1 (7) P -> Q 2,6 CP