Question 332293: Please help me to derive the conclusion of the following using natural deduction. I'm not really sure what I'm doing on this one as far as double addition, etc.
1. (S > Q) ∙ (Q > ~S)
2. S v Q
3. ~Q /P ∙ R (hint: proof requires use of addition)
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
1. (S -> Q) * (Q -> ~S)
2. S v Q
3. ~Q /P * R
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4. S 2, 3 DS
5. S -> Q 1 Simp.
6. Q 5, 4 MP
7. Q * ~Q 6, 3 Conj.
8. F 7 Contradiction
9. F v P 8 Add.
10. P 10 See note below
11. F v R 8 Add.
12. R 11 See note below
13. P * R 10,12 Conj.
Note: p v q is true when either p or q is true. Put another way, p v q is only false when both are false. So for something like F v p (where F stands for false), the truth value of F v p depends entirely on p. So this means that F v p is logically equivalent to p (ie they have the same truth values). Therefore, F v P is equivalent to P, and F v R is equivalent to R.
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