Attempted refutation of Leibniz's Law using Cartwright's Puzzle

Leibniz's Law (LL): x = y ↔ [φ(x) ↔ ψ(y)]

Claim. LL does not hold generally.

Proof. Assume (to reach a contradiction) that LL is true. Let ‘A’ designate the sentence ‘The last word of A is obscene’ and let ‘B’ designate the sentence ‘The last word of A is obscene’. Let ‘G’ denote the property [λx](if x had quotes around its last word, then x would have been true). Since A = B, by LL it follows that G(A) ↔ G(B). A is not such that if it had quotes around its last word it would have been true (since if A had quotes around its last word its last word would have been “obscene” not ‘obscene’). Hence, ~G(B). But B is such that if it had quotes around its last word it would have been true (since if B had quotes around its last word it would have rightly said of A that its last word is ‘obscene’). Thus, G(B). Contradiction.


[There is a plausible premise to deny. Which is it?]