2

I need help with this question using the first 13 rules of inference. Here is what I have so far:

  1. ~(KvF)
  2. ~F=>(KvC)
  3. (GVC)=>~H / ~(KvH)
  4. ~Kv~F DM 1
  5. ~Fv~K Com 5
  • 3
    I think that the right startegy is to assume KvH and try to derive a contradiction... – Mauro ALLEGRANZA Oct 11 '16 at 19:30
  • 2
    4. is wrong : from 1., by De Morgan, we have ~K&~F. – Mauro ALLEGRANZA Oct 11 '16 at 19:36
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    Can't believe I missed that, I've solved it now – Nazmul Bhuiyan Oct 11 '16 at 19:48
1

Here is a proof.

It uses DeMorgan Rule (DeM), conjunction elimination (∧E), conditional elimination (→E), disjunctive syllogism (DS), disjunction introduction (∨I) and conjunction introduction (∧I). See *forall x: Calgary Remix for details about the rules.

enter image description here


References

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/

1
 1 |  ¬(K˅F)          premise
 2 |  ¬F→(K˅C)        premise
 3 |_ (G˅C)→¬H        premise
 4 |  |_ H˅K          assumption
 5 |  |  |_ K         assumption
 6 |  |  |  K˅F       ˅ introduction 5
 7 |  |  |  Ʇ         ¬ elimination 1,6
 8 |  |  K→Ʇ          → introduction 5-7
 9 |  |  |_ H         assumption
10 |  |  |  |_ F      assumption
11 |  |  |  |  K˅F    ˅ introduction 10
12 |  |  |  Ʇ         ¬ elimination 1,11
13 |  |  |  ¬F        ¬ introduction 10-12
14 |  |  |  K˅C       → elimination 2,13
15 |  |  |  |_ C      assumption
16 |  |  |  |  G˅C    ˅ introduction 15
17 |  |  |  |  ¬H     → elimination 3,16 
18 |  |  |  |  Ʇ      ¬ elimination 9,17
19 |  |  |  C→Ʇ       → introduction 15-18
20 |  |  |  Ʇ         ˅ elimination 14,8,19
21 |  |  H→Ʇ          → introduction 9-20
22 |  |  Ʇ            ˅ elimination 4,8,21
23 |  ¬(H˅K)          ¬ introduction 4-22

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