THEOREM

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~[P => Q] => P & ~Q

PROOF

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Suppose to the contrary...

1 ~[~[P => Q] => P & ~Q]

Premise

Apply Imply-And Rule (previously justified here)

2 ~[~~[P & ~Q] => P & ~Q]

Imply-And, 1

3 ~~[~~[P & ~Q] & ~[P & ~Q]]

Imply-And, 2

Remove ~~

4 ~~[P & ~Q] & ~[P & ~Q]

Rem DNeg, 3

Obtain contradiction...

5 P & ~Q & ~[P & ~Q]

Rem DNeg, 4

Apply Conclusion Rule (Proof by contradiction)

6 ~~[~[P => Q] => P & ~Q]

4 Conclusion, 1

Remove ~~

As Required:

7 ~[P => Q] => P & ~Q

Rem DNeg, 6