A-1. (x)–(Hx⊃Hx)∨(∃y)–(Hy⊃Hy)
1. (x)–(Hx⊃Hx)∨(∃y)–(Hy⊃Hy) | Premise |
2. (∃y)–(Hy⊃Hy)∨(x)–(Hx⊃Hx) | 1, Commutation |
3. (∃y)[–(Hy⊃Hy)∨(x)–(Hx⊃Hx)] | 2, Q.E. |
4. –(Ha⊃Ha)∨(x)–(Hx⊃Hx) | 3, E.I. |
5. (x)–(Hx⊃Hx)∨–(Ha⊃Ha) | 4, Commutation |
6. (x)[–(Hx⊃Hx)∨–(Ha⊃Ha)] | 5, Q.E. |
7. –(Ha⊃Ha)∨–(Ha⊃Ha) | 6, U.I. |
8. –(Ha⊃Ha) | 7, Idempotent law |
9. Ha&–Ha | 8, Equivalence for negative conditional and conjunction |
以''(x)–(Hx⊃Hx)∨(∃y)–(Hy⊃Hy)''作為前提可明顯地推出矛盾''Ha&–Ha'',可知''(x)–(Hx⊃Hx)∨(∃y)–(Hy⊃Hy)''為矛盾句。
C-3. Everything is physical and destructible. Therefore, everything is physical, and everything is destructible.
前提:Everything is physical and destructible.每樣東西都是物質的且可壞朽的。
結論:everything is physical, and everything is destructible.
每樣東西是物質的,且每樣東西是可壞朽的。
述詞設置:''Px'': ''x是物質的'';''Dx'': ''x是可壞朽的''
1. 每樣東西都是物質的且可壞朽的 | 2. 每樣東西是物質的 | 3. 每樣東西是可壞朽的 |
1.1 每樣東西,它是物質的且它是可壞朽的 | 2.1每樣東西,它是物質的 | 3.1每樣東西,它是可壞朽的 |
1.2 每樣東西x,x是物質的且x是可壞朽的 | 2.2每樣東西x,x是物質的 | 3.2每樣東西x,x是可壞朽的 |
1.3 (x)(x是物質的且x是可壞朽的) | 2.3(x)(x是物質的) | 3.3(x)(x是可壞朽的) |
1.4(x)(Px&Dx) | 2.4(x)Px | 3.4(x)Dx |
| 將2.4與3.4以連言符號連結:(x)Px&(x)Dx |
改寫後,
前提:(x)(Px&Dx);結論:(x)Px&(x)Dx,以間接證法證明此論證有效:
1. (x)(Px&Dx) | Premise |
2. –[(x)Px&(x)Dx] | Premise |
3. –(x)Px∨–(x)Dx | 2, De Morgan's laws |
4. (∃x)–Px∨–(x)Dx | 3, Q.E. |
5. (∃x)–Px∨(∃x)–Dx | 4, Q.E. |
6. (∃x)[–Px∨(∃x)–Dx] | 5, Q.E. |
7. –Pa∨(∃x)–Dx | 6, E.I. |
8. (∃x)–Dx∨–Pa | 7, Commutation |
9. (∃x)[–Dx∨–Pa] | 8, Q.E. |
10. –Db∨–Pa | 9, E.I. |
11. Pa&Da | 1, U.I. |
12. Pb&Db | 11, U.I. |
13. Db | 12, Simplification |
14. –Pa | 10, 13, Disjunctive arguments |
15. Pa | 11, Simplification |
16. Pa&–Pa | 15, 14, adjunction |
從推導出矛盾可知前提不可能全部為真,若原始論證的前提皆為真,則否定結論而設的前提必定為假,所以原結論為真。故原始論證為有效論證。 |
D-1. (∃y)(Gy&–Gy)∨(z)(Hz≡–Hz)
1. (∃y)(Gy&–Gy)∨(z)(Hz≡–Hz) | Premiese |
2. (∃y)[(Gy&–Gy)∨(z)(Hz≡–Hz)] | 1, Q.E. |
3. (Ga&–Ga)∨(z)(Hz≡–Hz) | 2, E.I. |
4. –(Ga&–Ga) | 3, Tautology |
5. (z)(Hz≡–Hz) | 3,4, Disjunctive arguments |
6. Hb≡–Hb | 5, U.I. |
7. (Hb⊃–Hb)&(–Hb⊃Hb) | 6, Equivalence for biconditional and conditional |
8. Hb⊃–Hb | 7, Simplification |
9. –Hb | 8, Reductio ad absurdum |
10. –Hb⊃Hb | 7, Simplification |
11. Hb | 10. Reductio ad absurdum |
12. Hb&–Hb | 9, 11, adjunction |
以''(∃y)(Gy&–Gy)∨(z)(Hz≡–Hz)''作為前提可明顯地推出矛盾''Hb&–Hb'',可知''(∃y)(Gy&–Gy)∨(z)(Hz≡–Hz)''為矛盾句。 |