Q:

First-Order Logic: Formalisation of the phrase "Not all that glitters is gold"

I am trying to determine which of the following expressions are a valid formalisation of the phrase "Not all that glitters is gold". I have the following options:
(1) ∃x. ((x)→(x))
(2) ∀x. (¬((x))→(x))
(3) ¬∀x.((x)→(x))
(4) ∀x. (¬((x)→(x)))
I know that option 3 is valid since it can be rewritten as ∃x ¬((x)→(x)). However, I am not sure whether any of the other options are correct. I don't think option 1 is correct since it states that there exists some x such that if x glitters it is gold, but this doesn't show that "not all that glitters is gold". 
I don't believe option 2 is correct either, since it is saying that for all x if x does not glitter then it is gold, which doesn't show that that "not all that glitters is gold" either.
Option 4 is not correct either, since it is saying that for all x the statement "if x glitters then it is gold" is false. But for the phrase "not all that glitters is gold" there could be still be some x which glitters and is gold, it's just that not all x that glitters is gold.
However, I am not sure whether my reasoning about options (1), (2) and (4) being wrong are correct. Any insights are appreciated.

A:

As suggested by Dirk, for the formulas that you deem wrong, try to come up with countermodels where the sentence "Not all that glitters is gold" is true but that formula is false.  
Your paraphrases correctly capture what each of the formulas expresses, and you are also correct about your judgements that (3) is a correct formalization and that (1), (2), (4) are all incorrect -- but we need to show a bit more formally why this is the case. Actually proving that (3) is indeed a correct formalization of the sentence can be a bit cumbersome, so let's focus on showing why the other formulas are not, which is a relatively easy task since we all we need to do is provide one simple countermodel for each of the incorrect formalizations and we're done.  
As a first observation, remember that the correct formalization,

$¬∀x.((x)→(x))$

is, as you pointed out, logically equivalent to

$∃x ¬((x)→(x))$

which can be more easily rewritten as 

$∃x ((x)∧¬(x))$

so the sentence "Not all that glitters is gold" can be paraphrased as

"There is at least one thing which is glitter but not gold".

This is the scenario we need to keep true in all our structures while falsifying the incorrect formulas.
Let's have a look at the incorrect formalizations now. Reminder: What we're trying to do to show that (1), (2) and (4) do not capture the intended meaning of the sentence correctly is come up with structures where the situation "Not all that glitters is gold"/"There is at least one thing which glitters but is not gold" actually holds, but the putative formalization evaluates as false. So we want to find structures which validate (3) and at the same time falsify (1), (2), (4), respectively.  

(1) $∃x. ((x)→(x))$

To falsify that formula, we need that for all $x$ $(x)→(x)$ is false, which is the case when $(x)$ is true but $(x)$ is false, i.e. all our things must be glitter bot not gold. So we have a situation where not only not all, but in fact nothing that glitters is gold, which is consistent with the original sentence. Here is the simplest such structure:  
$\mathfrak{A}_1 = \langle A_1, \mathcal{I}_1 \rangle$ with
$A_1 = \{a\}$ and
$\mathcal{I}_1$ such that
     Glitter(x)  Gold(x)
x=a  true        false

You can easily verify that this structure is consistent with the sentence "Not all that glitters is gold" but not the formalization in (1), so we proved that (1) does not correctly capture the meaning of the sentence in question.  

(2) $∀x. (¬((x))→(x))$

To falsify this formula, we need to find at least one $x$ for which the implication $¬(x)→(x)$ is false, which is the case when both $(x)$ and $(x)$ are false. Put differently, by a few rewriting steps we can establish that (2) expresses "All things are either glitter or gold" so by finding a structure where for at least one element both predicates are false, we will render the formula false. Since, by the above equivalence, in order to keep our correct formalization true, we also need an element which is glitter (but not gold), our structure needs to have two elements and could look as follows:  
$\mathfrak{A}_2 = \langle A_2, \mathcal{I}_2 \rangle$ with
$A_2 = \{a, b\}$ and
$\mathcal{I}_2$ such that
     Glitter(x)  Gold(x)
x=a  true        false  
x=b  false       false

$a$ makes our intended scenario $∃x ((x)∧¬(x))$ true and $b$ provides a counterexample to $∀x. (¬((x))→(x))$, so here we have a structure which is consistent with the correct formalization (3) but not the formula (2).  

(4) $∀x. (¬((x)→(x)))$

This formula is equivalen to stating "All things are glitter but not gold" and is falsified when there is an element for which the negated implication is false, hence the implication $(x)→(x)$ true. We can now choose between three different scenarios: $(x)$ true and $(x)$ true, $(x)$ false and $(x)$ true, or $(x)$ false and $(x)$ false. To keep our original sentence true, we also need an element for which $(x)$ is true and $(x)$ false which is incompatible with the above three options, so we need a structure with two elements, one which validates the desired formalization and one which invalidates (4). Conveniently, the structure which we established for (2) fulfills all of our requirements, so we can just take $\mathfrak{A}_2$ as a countermodel for both the validity of (2) and (4).

