(F&~H)→G

~A→~H

~HvC

~(B→E)

C→D

----------

{~G→[F→(D&A)]}&B

Can someone solve this, I believe the answer is that it is valid. But I don't know how to negate a ~A on a separate branch. I have to work through it some more but would love some help thanks in advance =)

## Challenging Truth Tree problem

### Re: Challenging Truth Tree problem

I am late on this, and I hope that the poster found a solution to his/her problem. Here is what I came up with.

Code: Select all

```
1. (1) (F & ~H)-> G Premise
2. (2) ~A -> ~H Premise
3. (3) ~H v C Premise
4. (4) ~(B ->E) Premise
5. (5) C-> D Premise
6 (6) ~G Assumption
7 (7) F Assumption
1,6 (8) ~(F&~H) 1, 6 Paradox of material implication
1,6 (9) ~F v ~~H 8 DeMorgan
1,6 (10) ~F v H 9 Subformula double negation
1,6,7 (11) H 7,10 disjunctive syllogism (DS)
1,3,6,7(12) C 3,11 DS
1,3,5,6,7 (13) D 5,12 MP
1,2,6,7(14) A 2,11 PMI
1,2,3,5,6,7 (15) D&A 13,14 &I
1,2,3,5,6(16) F -> (D&A) 7,15 ->I
1,2,3,5 (17) ~G-> F -> (D&A) 6,13 ->I
4 (18)B & ~E 4, Negation implication
4(19) B 18, &E
1,2,3,4,5 (20){~G->[F->(D &A)]} & B 17,19 &I
```

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