Proving (C → A) ∧ (¬C → B) is equivalent to (A ∧ C) ∨ (¬C ∧ B)

[flying2000]

1) (C -> A) and (!C -> B)
2) (A and C) or (!C and B)

I use the logic calculator and found these two formulas are logical equivalent, but I spend 2 hours there and still can't prove it because I can't eliminate one of the three items.

Can anyone help me?

Thanks in advance!

 

[Owen Holden]

1) (C -> A) and (!C -> B)
2) (A and C) or (!C or B)

I use the logic calculator and found these two formulas are logical equivalent, but I spend 2 hours there and still can't prove it because I can't eliminate one of the three items.

Can anyone help me?

Thanks in advance!

((C -> A) & (~C -> B)) <-> ((A & C) v (~C v B)), is not valid.

It fails if A=B=C=false.

 

[flying2000]

sorry, man, I made a mistake here,
the second one should be:
((A & C) v (~C & B)), it should be equivalent.

((C -> A) & (~C -> B)) <-> ((A & C) v (~C v B)), is not valid.

It fails if A=B=C=false.

 

[BicycleTree]

(C -> A) :: A v ~C
(~C -> B) :: C v B
So ((C -> A) & (~C -> B)) :: (A v ~C) & (C v B)
:: ((A v ~C) & C) v ((A v ~C) & B)
:: ((A & C) v (~C & C)) v ((A & B) v (~C & B))
:: (A & C) v (A & B) v (~C & B)
:: AC + AB + ~CB (change of notation)

I expanded it out into canonical form and combined terms:

:: AC(B + ~B) + AB(C + ~C) + ~CB(A + ~A)
:: ACB + AC~B + ~CBA + ~CB~A + ABC + AB~C
:: ABC + AC~B + ~CBA + ~CB~A
:: ~CB + AC

 

[flying2000]

Thank you so much!

U r so helpful, man. thanx!

(C -> A) :: A v ~C
(~C -> B) :: C v B
So ((C -> A) & (~C -> B)) :: (A v ~C) & (C v B)
:: ((A v ~C) & C) v ((A v ~C) & B)
:: ((A & C) v (~C & C)) v ((A & B) v (~C & B))
:: (A & C) v (A & B) v (~C & B)
:: AC + AB + ~CB (change of notation)

I expanded it out into canonical form and combined terms:

:: AC(B + ~B) + AB(C + ~C) + ~CB(A + ~A)
:: ACB + AC~B + ~CBA + ~CB~A + ABC + AB~C
:: ABC + AC~B + ~CBA + ~CB~A
:: ~CB + AC