When does ℱx ≡ (x)ℱx ? What if x has the same domain in both?

Abbreviate Universal Generalisation to UG.

The longitude of this earlier question motivates me to pose further questions separately.
Source: A Concise Introduction to Logic (12 Ed, 2014) by Patrick Hurley

[p 465:] [...]   5. Py ⊃ Cy        3, 4, Hypothetical Syllogism
[ 5'. Px ⊃ Cx       x and y are Variables; so I can change y to x. ]
6. (x)(Px ⊃ Cx)       5, UG         [...]

Yet if we take certain liberties, we might characterize line 5 as saying “If it is a P, then it is a C, where “it” designates any item at random in the universe.
Line 6 can then be seen as reexpressing this sense of line 5.

[p 467:] [...]    5. Qy        3, 4, Modus Ponens       [...]

Line 5 states in effect that everything in the universe is a Q.

[p 484:] 1. (x)Rx ⊃ (x)Sx        [Conclusion:]  (x)(Rx ⊃ Sx)
| 2. Rx          Assumption for Conditional Proof
| 3. (x)Rx         2, UG (invalid)

1. I do not comprehend this answer: when does ℱx ≡ (x)ℱx? What if x represents the same set (or Domain/Universe of Discourse) in both?

2. For pp 465 and 467 above, does ℱx ≡ (x)ℱx? On p 465, the grey highlight state x as any item at random in the universe. On p 467, th grey states y as everything in the universe.

• It does not say that x is everything in the universe. x is a variable, it refers to one thing not everything. It's the quantifier (x) that allows to talk about everything. – Quentin Ruyant Jan 6 '16 at 22:10

Px is not equivalent to (x) Px.

If we know that "all (the men in the universe) are Philosopher" is true, we can certainly infer that "it is a Philosopher" is true, whoever the pronoun "it" denotes.

But if we know that "it is a Philosopher" is true, when e.g. we are "pointing at" Socrates, we are not at all entitled to infer that "all (the men in the universe) are Philosopher".

This fact is reflected into Hurley's proof system in the proviso of the UG rule
(see p 483, A Concise Introduction to Logic (12 Ed, 2014) by Patrick Hurley ):

Restriction: UG must not be used within the scope of an indented sequence if the instantial variable y is free in the first line of that sequence.

Note

When we "evaluate" the truth value of a formula, like Px ≡ (x) Px, of course the domain of the interpretation must be the same for all the variables (free and bound).

I'm not familiar with many of the abbreviations used in Hurley's text (I've only done a second year course "introduction to mathematical logic", but I hope to do a logic course in philosophy soon).

You appear to be asking why the UG of statement 6 from page 465 is valid, while the UG of statement 3 from page 484 is not valid.

In the example from page 465, the free variable (x or y) can indeed represent any x from the domain. Therefore, when statement 5 asserts that if Px holds, then so does Cx, we can legitimately conclude that this is true for any x.

In the example from page 484, the premise (statement 1) asserts that "if Rx holds for every x then Sx also holds for every x". It does not assert that Rx holds for some x or for all x. Note that statement 2 only says "assume that Rx holds for some x" and therefore we cannot generalise to conclude that Rx holds for every x. In other words, the application of UG in statement 3 is not valid.

This is an informal way of describing the distinction between the two examples given. The formal requirements for applying UG should be evident in the author's definition of UG.

Many FOL treatments (I was raised on Schoenfield's) find the opening (x) pointless, presuming open quantification whenever a variable is free. But then you have to be very careful about combining statements with free variables.

In marked quantification (x)(Ax) => (x)(Bx) does not mean the same thing as (x)(Ax => Bx), The latter is a much more powerful statement. But without explicit quantification, we are tempted to write them the same. What we really need to do with unmarked open quantification is to write the former as Ax => By, and the latter as Ax => Bx. (And we have to remember to do so every time.)

Explicit quantification which must be removed and then reintroduced avoids this potential confusion, at the cost of additional explicit steps. The rules for removing the quantification prompts the step of keeping your variables separated when they would otherwise get confused.