Polyadic Universal Generalization: Generalizing $Axx$ to $(z)(y)Ayz$

Question

In a logic course lecture (https://legacy.earlham.edu/~peters/courses/log/deriv2.htm) based on the book: Copi, Irving M. "Symbolic logic." (1954)., Peter Suber says:

If the convention gives you little flexibility in instantiating (x)Axx, it gives you unexpected flexibility in generalizing Axx. You may generalize this to (y)Ayx, and then generalize it again to (z)(y)Ayz.

I do not understand this. Is this saying that if Axx was $$x=x,$$ then we could generalize to $$ \forall y: y=x,$$ and then to $$ \forall z \forall y: y=z \quad ???$$

Surely I am wrong, but what am I missing?

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For reference, the larger context of the quote is this paragraph:

In working with polyadic predicates, generally avoid generalizing different variables to the same variable in the same expression.

For example, it is legal to generalize Axy to (x)Axx. But unless the latter is just what you need to finish the proof, it is usually unwise to generalize in this way. The reason is that the convention will prevent you from instantiating (x)Axx to Axy again; you will only be able to instantiate it to Axx, Ayy, Azz and so on.

If the convention gives you little flexibility in instantiating (x)Axx, it gives you unexpected flexibility in generalizing Axx. You may generalize this to (y)Ayx, and then generalize it again to (z)(y)Ayz.

Apart from these obstacles and opportunities, derivations with polyadic predicates are not very different from derivations with monadic predicates.

Answer 1

You are absolutely correct, and it should indeed not be possible to infer $(z)(y)Ayz$ from $Axx$. In fact, inferring $(y) Ayx$ from $Axx$ is already a mistake.

I wonder if Peter Suber meant that you can infer $(\exists z)(\exists y)Ayz$ from $Axx$.