Opps, I made an error when I said: “It is clear that (X->Y)->X does not imply (not X)->X, so your little trick doesn’t work.” This is false. This implication does exist, so we get:
((not ◻C)->(◻C))
This sentence is true just in case C is in fact provable, so we don’t have a problem with your 2=1 cases.
Opps, I made an error when I said: “It is clear that (X->Y)->X does not imply (not X)->X, so your little trick doesn’t work.” This is false. This implication does exist, so we get:
((not ◻C)->(◻C))
This sentence is true just in case C is in fact provable, so we don’t have a problem with your 2=1 cases.