@VideCorSpoon,

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.

@SteveP phil,

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,???

@VideCorSpoon,

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