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