THEOREM

*******

 

     A & ~B => ~[A => B]  (Truth Table, line 2)

 

PROOF

*****

 

    Suppose...

 

      1     A & ~B

            Premise

 

         Suppose...

 

            2     A => B

                  Premise

 

            3     A

                  Split, 1

 

            4     ~B

                  Split, 1

 

         Apply Detachment Rule

 

            5     B

                  Detach, 2, 3

 

         Obtain contradiction...

 

            6     B & ~B

                  Join, 5, 4

 

    Apply Conclusion Rule (By Contradiction)

 

      7     ~[A => B]

            4 Conclusion, 2

 

Apply Conclusion Rule (Direct Proof)

 

As Required:

 

8     A & ~B => ~[A => B]

      4 Conclusion, 1

 

 

 

 

Home