Help Needed: Proving Logical Equivalence of (C -> A) and (!C -> B)

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Discussion Overview

The discussion centers around proving the logical equivalence of the expressions (C -> A) and (!C -> B), exploring their transformation into alternative forms and the challenges faced in the proof process. The scope includes logical reasoning and mathematical manipulation of logical expressions.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant presents the expressions (C -> A) and (!C -> B) and claims they are logically equivalent to (A and C) or (!C and B), but struggles to prove this equivalence.
  • Another participant suggests a correction to the second expression, proposing it should be (A and C) or (!C or B), but later acknowledges that this is not valid under certain conditions.
  • A third participant provides a detailed expansion of the expressions into canonical form, showing the steps taken to manipulate the logical statements but does not assert that the equivalence is proven.
  • Participants express gratitude for assistance and clarification, indicating a collaborative effort in exploring the logical equivalence.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the logical equivalence of the expressions, with multiple competing views and corrections presented throughout the discussion.

Contextual Notes

Some expressions are noted to be invalid under specific conditions, such as when A, B, and C are all false, indicating limitations in the proposed equivalences.

flying2000
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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!
 
Last edited:
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flying2000 said:
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.
 
sorry, man, I made a mistake here,
the second one should be:
((A & C) v (~C & B)), it should be equivalent.


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

It fails if A=B=C=false.
 
(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
 
Last edited:
Thank you so much!

U r so helpful, man. thanx!

BicycleTree said:
(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
 

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