SOLUTION: I have bee having trouble solving these problems and been at it for hours I need help. Anyone's help would be greatly appreciated. I have to prove them.
Problem #1
1) (B^A)->D
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-> SOLUTION: I have bee having trouble solving these problems and been at it for hours I need help. Anyone's help would be greatly appreciated. I have to prove them.
Problem #1
1) (B^A)->D
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Question 1038664: I have bee having trouble solving these problems and been at it for hours I need help. Anyone's help would be greatly appreciated. I have to prove them.
Problem #1
1) (B^A)->D
2)A
3) C->~D/~(C^B)
4) :. (C^B)
Problem #2
1) (D^S)^T
2) [(Sv~F)vZ]->C/C^T
3) :. C^T
Problem #3
1) A->B
2) D->E
3) ~(B^E)/ ~Dv~A
4) :. ~Dv~A
Problem #4
1)~[(A^B) v (Cv~R)]
2) (T^~S)-> (Cv~R)
3) ~S/~T
4) ~T
Problem #5
1) (~Z^W)->Q
2) ~Z
3) R<-> (W^~Q)/~R
4) ~R Answer by solver91311(24713) (Show Source):
For the first one you need to do a conditional proof. Note that by DeMorgan, your conclusion is equivalent to ~C or ~B.
First assume B, then, having been given A, you have B and A, and then you have D by Modus Ponens. But D is the same as ~~D. ~~D makes ~C out of statement 3 by Modus Tollens. And ~C gives you ~C or ~B.
Second assume ~B which gives you ~C or ~B directly.
So either way, you get the desired conclusion.
John
My calculator said it, I believe it, that settles it
