well Thanks for the help, do you have a copy of a scratchpad you used? My test is tomorrow and I'd at least like to see how you went about TRYING those other two.

Sure. These are the two MOST LIKELY approaches to each of the problems I could come up with without nesting.

1.
1. A--> B / A --> ~ ( ~B & ~C)
2. A AP
3. B MP1
4. ~AvB IMP 1
5. ~B --> ~A Tans. 1
6. B v C ADD 2
7.???

2.
1. [A --> (B v C)] --> (D-->A)
2. ~A / ~D
3.~A v D ADD 2
4. A -->D IMP 3
5. ~D &~A Trans 4
6.~~D MT 2,5
7. D DNeg 6
8.~A MP 5,6
9.???
10.???
11. D-->A MP ???
12. ~D MP 2,???

Ah, thank you very much! these will help me tonight. I appreciate all your help!

Answer to #1

1. [ A ⊃ (B v C)] ⊃ (D⊃A)
2. ~A
3. ~(D ⊃ A) ⊃ ~ [A ⊃ (B v C)] 1 contra-position
4. ~(~D v A) ⊃ ~ [A ⊃ (B v C)] 3 implication
5. (~~D • ~A) ⊃ ~ [A ⊃ (B v C)] 4 DeMorgan's
6. (D • ~A) ⊃ ~ [A ⊃ (B v C)] 5 double negation
| -> 7. D AP assumed premise looking for a contradiction via Indirect Proof
| 8. D • ~A 2,7 conjunction
| 9. ~ [A ⊃ (B v C)] 6,8 Modus Ponens
| 10. ~ [~A v (B v C)] 9 implication
| 11. ~~A • ~(B v C) 10 DeMorgan's
| 12. ~~A 11 simplification
| 13. A 12 double negation
| 14. A • ~A 2,13 conjunction
| -------------------------------------------
15. ~D 7-14 Indirect Proof

This problem required the use of an indirect proof. Indirect proofs are used to find a contradiction by assuming the negation of the provided conclusion... in this case ~D is the provided conclusion therefore the negation would be D or ~~D which of course are equivalent. If you find a contradiction utilizing the negation of the conclusion that implies that the original conclusion must be true, since ~D is the logical opposite of D.

Cheers!!!

Thomas Belnap

1) A=>B/ A=>~(~B&~C)

2) A........................................................................Hypothesis for conditional proof

3)~~(~B&~C).........................................................Hypothesis for a contradiction

4) ~B&~C...............................................................3,double negation

5) ~B ......................................................................4, conjuction elimination

6) B..........................................................................1,2 M.Ponens

7) B&~B....................................................................5,6 conjuction introduction

8) ~(~B&~C)..........................................................By contradiction 3 to 7

9) A => ~(~B&~C)..................................................By conditionqal proof 2 to 8