Logic Proof Help

Get Email Updates Email this Topic Print this Page

SteveP phil
 
Reply Sun 8 Jun, 2008 10:02 pm
@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. Very Happy
 
VideCorSpoon
 
Reply Sun 8 Jun, 2008 10:43 pm
@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,???
 
SteveP phil
 
Reply Mon 9 Jun, 2008 12:35 am
@VideCorSpoon,
Ah, thank you very much! these will help me tonight. I appreciate all your help!
 
tlbelnap
 
Reply Fri 4 Feb, 2011 06:22 pm
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
 
triclino
 
Reply Mon 15 Aug, 2011 07:25 pm
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
 
 

 
Copyright © 2019 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.02 seconds on 12/11/2019 at 01:58:24