THEOREM

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~P & P => Q

PROOF

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Suppose to the contrary...

1 ~[~P & P => Q]

Premise

Apply Imply-And Rule (previously justified here)

2 ~~[~P & P & ~Q]

Imply-And, 1

Remove ~~

3 ~P & P & ~Q

Rem DNeg, 2

4 ~P

Split, 3

5 P

Split, 3

6 ~Q

Split, 3

Obtain contradiction

7 P & ~P

Join, 5, 4

Apply Conclusion Rule (Proof by contradiction)

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

4 Conclusion, 1

Remove ~~

As Required:

9 ~P & P => Q

Rem DNeg, 8