SOLUTION: Premise: ~(A->(B&C)) Conclusion: ~(A<->(B&C)) I've tried to Provisionally assume A<->(B&C), but I don't understand double arrow rules that well.

Algebra ->  Proofs -> SOLUTION: Premise: ~(A->(B&C)) Conclusion: ~(A<->(B&C)) I've tried to Provisionally assume A<->(B&C), but I don't understand double arrow rules that well.      Log On


   

Question 952302: Premise: ~(A->(B&C))
Conclusion: ~(A<->(B&C))
I've tried to Provisionally assume A<->(B&C), but I don't understand double arrow rules that well.
Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Hint:

A<->(B&C) breaks down into [A -> (B&C)] & [(B&C) -> A] using the material equivalence rule.


material equivalence rule: P <--> Q turns into [P -> Q] & [Q -> P]

------------------------------------------------------------------------------------------------------------------------

If you need more one-on-one help, email me at jim_thompson5910@hotmail.com. You can ask me a few more questions for free, but afterwards, I would charge you ($2 a problem to have steps shown or $1 a problem for answer only).

Alternatively, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Any amount is greatly appreciated as it helps me a lot. This donation is to support free tutoring. Thank you.

Jim
------------------------------------------------------------------------------------------------------------------------

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
This argument is NOT VALID!

The double arrow is the biconditional

P<->Q by definition is (P-Q)&(Q->P)

It is sometimes called "equivalence",

which means they have the same truth value.

But your conclusion is not valid.

We can show that it is not valid by a truth table:

            ?
~[A->(B&C)] => ~[A<->(B&C)]

Since (B&C) appears in both, we can let D = (B&C)

           ?
   ~(A->D) => ~(A<->~D)
-----------------------
     T  T       T    T     
     T  F       T    F
     F  T       F    T
     F  F       F    F

becomes:

           ?
   ~(A->D) => ~(A<->~D)
-----------------------
       T          F      
       F          T 
       T          T 
       T          T  

which becomes:
           
           ?
   ~(A->D) => ~(A<->~D)
-----------------------
   F          T          
   T          F     
   F          T     
   F          T    

which becomes:

           ?
   ~(A->D) => ~(A<->~D)
-----------------------
           T             
           F        
           T        
           T    

It is not valid because it fails in case 2.

Edwin