Question:
Elementary logic proving?
-XPgeometry-
2010-09-25 20:31:46 UTC
If A then B
If C then D
therefore: If (If ~A then C) then (If ~D then B)
.
.
How do i prove this using natural deduction? Help me please!!
Three answers:
Ex Falso Quodlibet
2010-09-26 20:23:25 UTC
A > B, C > D /- (~A > C) > (~D > B)



(1) 1. A > B Premise

(2) 2. C > D Premise

(3) 3. ~A > C Assumption

(4) 4. ~D Assumption

(2,4) 5. ~C 2,4 MT

(2,3,4) 6. ~~A 3,5 MT

(2,3,4) 7. A 6 DNE

(1,2,3,4) 8. B 1,7 MP

(1,2,3) 9. ~D > B 4,8 CP

(1,2) 10. (~A > C) > (~D > B) 3,9 CP



1. A > B Premise

2. C > D Premise

3. ~A v B 1 Material Implication

4. (~A v B) v D 3 Addition

5. ~A v (B v D) 4 Association

6. ~A v (D v B) 5 Commutation

7. ~C v D 2 Material Implication

8. (~C v D) v B 7 Addition

9. ~C v (D v B) 8 Association

10. (D v B) v ~A 6 Commutation

11. (D v B) v ~C 9 Commutation

12. ((D v B) v ~A) & ((D v B) v ~C) 10,11 Conjunction

13. (D v B) v (~A & ~C) 12 Distribution

14. (~A & ~C) v (D v B) 13 Commutation

15. (~A & ~C) v (~~D v B) 14 Double Negation

16. (~A & ~C) v (~D > B) 15 Material Implication

17. ~(A v C) v (~D > B) 16 De Morgan's Law

18. (A v C) > (~D > B) 17 Material Implication

19. (~~A v C) > (~D > B) 18 Double Negation

20. (~A > C) > (~D > B) 19 Material Implication
?
2010-09-26 03:40:27 UTC
Cannot be done, as A+C and B+D (respectively, pairwise) have no relationship to each other, and therefore cannot generate any logical deduction.
?
2010-09-26 03:36:57 UTC
basically you're just saying that if a is b and c is d and at some point a is equal to c, then what is equal to a is equal to equal to c, therefore b is equal to d, it must be. Transitive Property.


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