THEOREM

*******

~P => [P => Q]

PROOF

*****

Suppose to the contrary...

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

Premise

Apply Imply-And Rule (previously justified here)

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

Imply-And, 1

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

Imply-And, 2

Remove ~~

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

Rem DNeg, 3

Apply Split Rule

5 ~P & [P & ~Q]

Rem DNeg, 4

6 ~P

Split, 5

7 P & ~Q

Split, 5

8 P

Split, 7

9 ~Q

Split, 7

Obtain contradiction...

10 P & ~P

Join, 8, 6

Apply Conclusion Rule (Proof by contradiction)

11 ~~[~P => [P => Q]]

4 Conclusion, 1

Remove ~~

As Required:

12 ~P => [P => Q]

Rem DNeg, 11