Deductive Logic, Warren Goldfarb, Hackett Publishing, 2003, Part IV, Exercise 5(c), p. 285

Show that (x)(y)(x = y ⊃ Gxy) is equivalent to (x)Gxx. The example is a good illustration of how The Convention works (I use a slightly modified version of The Convention, Symbolic Logic, Copi, 5th edition, Chpt 4.5, Quantification Rules):

The instantiating variable must be free, or the instantiating constant must occur, in the instantiating statement in at least those places where the general variable is free in the decapitated general statement.

A 'decapitated' statement is a statement where the quantifier has been taken off.

1. * (x)(y)(x = y ⊃ Gxy) ......... ACP
2. * (y)(x = y ⊃ Gxy) ......... 1 UI x/x
3. * x = x ⊃ Gxx ......... 2 UI y/x
4. * x = x ......... Id
5. * Gxx ......... 4,3 MP
6. * (x)Gxx ......... 5 UG
7. (x)(y)(x = y ⊃ Gxy) ⊃ (x)Gxx ......... 1-6 CP
8. * (x)Gxx ......... 8 ACP
9. * Gxx ......... 8 UI x/x
10. * (y)Gxy ......... 9 UG
11. * Gxy ......... 10 UI y/y
12. * Gxy ∨¬ (x = y) ......... 11 Add.
13. * ¬ (x = y) ∨ Gxy ......... 12 Comm.
14. * x = y ⊃ Gxy ......... 13 MI
15. (y)(x = y ⊃ Gxy) ......... 14 UG
16. (x)(y)(x = y ⊃ Gxy) ......... 15 UG
17. (x)Gxx ⊃ (x)(y)(x = y ⊃ Gxy) .........8-16 CP
18. [(x)Gxx ⊃ (x)(y)(x = y ⊃ Gxy)] • [(x)(y)(x = y ⊃ Gxy) ⊃ (x)Gxx] ......... 7,17 Conj.
19. (x)(y)(x = y ⊃ Gxy) ≡ (x)Gxx ......... 18 AB