Meaning of FOL sentences with multiple quantifiers one variable letter

Question

Consider the following FOL sentence:

∀x(∃yLeftOf(x,y) ∧ ∃xRed(x))

I am aware that this technically is a sentence since every variable is bound but I'm struggling to parse what it means, especially the second reference to x. Does this just mean the the same as if it wasn't there, or does this refer to a new variable which is also called x but could be different to the one with the larger scope. Clarification would be much appreciated.

On top of this, mention of the meanings of ∀x(...∀x...), ∃x(...∃x...) and ∃x(...∀x...) would be helpful.

Answer 1

A multiple quantifier on the same variable has no effect.

$\forall x \ \text {Red}(x)$ means : "everything is Red". This is a sentence with no free occurrence of the variable $x$.

Thus, also if syntactically admissible, adding a new quantifier in front, like e.g. : $\exists x \ \forall x \ \text {Red}(x)$, does not change the meaning of the sentence.

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Regading your example, we have to note that $\forall$ distribute over "and".

Thus, $∀x \ (∃y \ \text {LeftOf}(x,y) ∧ ∃x \ \text {Red}(x))$ is equivalent to :

$∀x \ ∃y \ \text {LeftOf}(x,y) ∧ ∀x \ ∃x \ \text {Red}(x)$,

which in turn is simply :

$∀x \ ∃y \ \text {LeftOf}(x,y) ∧ ∃x \ \text {Red}(x)$.

Answer 2

In the sentence

$$\forall x(\exists y \ LeftOf(x,y) \land \exists x \ Red(x))$$

the $x$ inside $LeftOf(x,y)$ is quantified by the $\forall x$, while the $x$ in $Red(x)$ is quantified by the $\exists x$.

Now, that the $x$ inside $|eftOf(x,y)$ is quantified by the $\forall x$ is obvious: the $x$ occurs within the scope of that quantifier (as indicated by the parentheses), and it does not occur within the scope of thre $\exists x$.

However, the $x$ in $Red(x)$ occurs within the scope of both quantifiers. So, what to do here? Well, if you think about how the sentence is constructed, it should be clear that we started with $Red(x)$, then quantified with $\exists x$, then conjuncted that with $LeftOf(x,y)$, and finally quantified that whole forumla with the $\forall x$. So ... it should be clear that it was the $\exists x$ that quantified the $x$ in $Red(x)$ ... and the fact that that whole formula ended up within the scope of the $\forall x$ did not take away from that.

Indeed, while we ended up using another $x$ in $LeftOf(x,y)$, it should be clear that that $x$ had nothing to do with thew $x$ in $Red(x)$. As such, it would have been more clear to just use a different variable, so you would have gotten something like:

$$\forall z(\exists y \ LeftOf(z,y) \land \exists x \ Red(x))$$

Not to say that it is wrong to reuse an $x$ either. Indeed:

$$\forall x(\exists y \ LeftOf(x,y) \land \exists x \ Red(x))$$

and

$$\forall z(\exists y \ LeftOf(z,y) \land \exists x \ Red(x))$$

are logically equivalent.

Answer 3

$\forall x~(\exists y\operatorname{LeftOf}(x,y) \land \exists x\operatorname{Red}(x))$

"For everything: there is something that it is left of, and there is something that is red."

Or more elegantly: "Everything is left of something. Also, something is red."

The $x$ in the $\operatorname{Red}$ predicate is bound to scope of the existential quantifier.   Every occurance of $x$ inside that scope is quanteened from use of $x$ within the scope of the universal quantifier.

By the alpha-replacement rule of substitution, it is equivalent to $$\forall x~(\exists y\operatorname{LeftOf}(x,y) \land \exists z\operatorname{Red}(z))$$