Actually, I believe you’ve found a bug in the paper which everyone else seems to have missed so far, but fortunately it’s just a typo in the statement of the theorem! The quantification should be over L’, not over L, and the proof does prove this much stronger statement. The statement in the paper is indeed trivial for the reason you say.
Given the stronger statement, the reason you can’t just have P have value 0 or 1 is sentences like G ⇔ P(‘G’) < 0.5: if P(G) = 0, by reflection it would follow that P(G) = 1, and if P(G) = 1, then P(not G) = 0, and by reflection it would follow that P(not G) = 1.
Actually, I believe you’ve found a bug in the paper which everyone else seems to have missed so far, but fortunately it’s just a typo in the statement of the theorem! The quantification should be over L’, not over L, and the proof does prove this much stronger statement. The statement in the paper is indeed trivial for the reason you say.
Given the stronger statement, the reason you can’t just have P have value 0 or 1 is sentences like G ⇔ P(‘G’) < 0.5: if P(G) = 0, by reflection it would follow that P(G) = 1, and if P(G) = 1, then P(not G) = 0, and by reflection it would follow that P(not G) = 1.