Question 952302
<pre>
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</pre>