THEOREM

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P => Q | ~Q

PROOF

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Suppose to the contrary...

1 ~[P => Q | ~Q]

Premise

Apply Imply-And Rule (previously justified here)

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

Imply-And, 1

Apply De Morgan's Law

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

DeMorgan, 2

Remove ~~

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

Rem DNeg, 3

5 P & [~Q & ~~Q]

Rem DNeg, 4

6 P & [~Q & Q]

Rem DNeg, 5

7 P

Split, 6

Obtain contradiction...

8 ~Q & Q

Split, 6

Apply Conclusion Rule (Proof by contradiction)

9 ~~[P => Q | ~Q]

4 Conclusion, 1

Remove ~~

As Required:

10 P => Q | ~Q

Rem DNeg, 9