Stuart’s proof by contradiction goes through, as far as I can see. (The speculation in the last paragraph doesn’t, as Paul notes, and I don’t know what “Hence P(‘G’)=1” is supposed to mean, but the proof by contradiction part does work.)
ETA: I now think that “Hence P(‘G’) = 1” is supposed to mean that this statement is implied by the first-order theory (T + the reflection axioms).
Stuart’s proof by contradiction goes through, as far as I can see. (The speculation in the last paragraph doesn’t, as Paul notes, and I don’t know what “Hence P(‘G’)=1” is supposed to mean, but the proof by contradiction part does work.)
ETA: I now think that “Hence P(‘G’) = 1” is supposed to mean that this statement is implied by the first-order theory (T + the reflection axioms).