Is it really problematic for Universal Instantiation to precede Existential Instantiation?

Question

Abbreviate: Existential Statement to ES, Existential Instantiation to EI, Universal Instantiation to UI, and Universal Statement to US.

_{Source: A Concise Introduction to Logic (12 Ed, 2014) by Patrick Hurley}

[p 468:] [...] Notice that the line involving existential instantiation is listed before the line involving universal instantiation. There is a reason for this. If the order were reversed, the existential instantiation step would rest on the illicit assumption that the something that is both an A and a G has thesamename as the name used in the earlier universal instantiation step. In other words, it would involve the assumption that the something that is both an A and a G is the very same something named in the line Ad ⊃ Cd. Of course, no such assumption is legitimate. To keep this mistake from happening, we introduce
[...]
[p 469:] These two restrictions can easily be combined into a single restriction that requires that the name introduced by existential instantiation be a new name that has not occurred in any previous line, including the line adjacent to the last premise that indicates the conclusion to be derived.

I recognise the possible problem that I coloured grey in the above quote; but how is it a genuine problem? And why does it necessitate EI always to precede UI?
I agree that writing EI before UI is visually convenient, but does the grey above exaggerate?

To me, the illicit assumption can be evaded simply by knowledge of EI vs UI. Only US (Line 1) are true for ANY value of x; so only US empower you to choose whatever Constant (ie: letter) you need, such as the specific Constant needed to match the Constants in any subsequent PS. One must start always by applying EI to the ES (Line 2) which generates the first Constant, before one can even know which Constant needs to be matched from applying UI to the US.

So knowing the previous sentences, how can someone be confused by whether the 'something' in EI is identical to what is named in the UI? If they are not identical, then EI and UI would use different Instantial Letters to name the 'something'!

Answer 1

Is it really problematic for Universal Instantiation to precede Existential Instantiation?

Of course not.

The issue is that if we start "instantiating" ∀x first, then when instantiating ∃x we cannot use the same name used into UI, in order to comply with the proviso of EI that:

"the name introduced by existential instantiation [must] be a new name".

This means that we will have e.g.

i) Ad ⊃ Cd --- from 1. by UI

and

ii) Ae ∧ Be --- from 2. by EI [e new, by the proviso]

and we cannot conclude; in order to "move on", we have to instantiate again 1. with the newly introduced name e.

Answer 2

A problem arises when reasoning with doubly quantified sentences and binary predicates.

For example:

1. AxEy R(x,y) (Premise)
2. Ey R(a,y) (UI 1)
3. R(a,a) (EI without respecting the prohibition on reusing names 2)
4. Therefore, Ex R(x,x) (EG 3)

Now take the domain to be the natural numbers and R to be the 'is less than' relation, <. Obviously, the premise will be true (for all natural numbers there's a greater number). However, the conclusion will not be true: there does not exist a natural number which is less than itself.

I was not able to find a counterexample with sentences with only one quantifier and only unary predicates, and I conjecture that such counterexamples do not exist. There's too little dependence between the universally quantified and existentially quantified sentences in that case.

Answer 3

The answer provided seems only to recapitulate the rule the rationale for which is being questioned. It makes sense to prohibit using the same instantial letter in two uses of EI within the same proof (so that we don't end up assuming that (say) the thing that is round and the thing that is square are the same thing). But why extend the restriction further than that? Why prohibit using the same instantial letter for EI that was previously used in UI? What's wrong with the proof below?

1. (x)(Ax > Bx)
2. (E!x)Ax / (E!x)Bx
3. Ac > Bc 1, UI
4. Ac 2, EI
5. Bc 3, 4 MP
6. (E!x)Bx 5, EG

If there's an invalid step here, then prove it with a counterexample.