作者MathTurtle (恩典)
看板logic
标题Re: [请益] 逻辑证明
时间Fri Jun 17 18:42:01 2011
※ 引述《a53530 (Susu)》之铭言:
: 1.(x)(Px≡Qx)
: 2.(x)[Px→(Qx→Rx)]
: 3.(Ex)Qx v (Ex)Px
: /(Ex)Rx v (Ex)Sx
: 用→替代horseshoe
: 用v 替代或
: 用^ 替代且
: 用(Ex)替代有些x
: 我算了好久算不出来
: 请强者帮忙解答
: 将以1000P币作为酬谢
我来试试。(有些推论规则的名称我忘了)
1.(x)(Px≡Qx) Ass
2.(x)[Px→(Qx→Rx)] Ass
3.(Ex)Qx v (Ex)Px Ass
4. ~(Ex)Px Ass for RAA
5. (Ex)Qx 4, 3. DS.
6. Qa 5. EI
7. Pa≡Qa 1. UI
8. (Pa→Qa) ^ (Qa→Pa) from 7
9. (Qa→Pa) 8 Simp
10. Pa 6, 9. MP
11. (x)~Px from 4
12. ~Pa 11. UI
13. Pa^~Pa 10, 12
14. (Ex)Px 4-13. RAA
15. Pb 14, EI
16. (Pb≡Qb) 1. UI
17. (Pb→Qb)^(Qb→Pb) from 16
18. (Pb→Qb) 17 Simp
19. Qb 15, 18 MP
20. [Pb→(Qb→Rb)] 2. UI
21. (Qb→Rb) 15, 20 MP
22. Rb 19, 21 MP
23. (Ex)Rx 22. EU (note that b was introduced
in 15 by EI, so that's fine)
24. (Ex)Rx v (Ex)Sx 23, Add
--
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◆ From: 86.30.200.58
1F:推 dementia:"Ass"改成"Premise"会不会比较好?? 06/18 03:50
2F:→ dementia:"EI"通常会被当成一种assumption 是不是有步骤省略了?? 06/18 03:57
3F:推 dementia:就我所知 应该用Constructive Dilemma比较轻松 推出 06/18 04:01
4F:→ dementia:"(Ex)Rx v (Ex)Rx"再用Tautology得到"(Ex)Rx"就可以了 06/18 04:02
5F:→ MathTurtle:楼上的解法很漂亮...我都忘了有constructive dilemma 06/18 04:15
6F:→ MathTurtle:可以用了... 06/18 04:15