Scope in $\forall x P(x) \rightarrow Q(x)$

Question

For $ \exists x P(x) \wedge Q(\color{blue}x) $

The scope of the quantifier, is just the predicate $P(x)$, thus the occurrence of x in blue is free.

I am unsure exactly the scope of the quantifier in this formula:

$ \forall x P(x) \rightarrow Q(\color{red}x) $

Based off the previous formula, I think the occurrence of the x in red is free but I'm not sure.

Is the occurrence of $\color{red}{x}$ free or bound? Would this vary based off convention?

Answer 1

You are right.

The "usual" convention is that quantifiers, like negation, apply as little as possible.

Thus, $∃xα∧β$ is $(∃xα∧β)$, and not $∃x(α∧β)$.

The same for $∀xP(x) → Q(x)$: without parentheses, the scope of $∀x$ is only $P(x)$. Thus, the occurrence of $x$ into $Q(x)$ is free.

Answer 2

In first-order logic, quantifiers bind more strongly than logical connectives; the precedence convention is:

1. quantification
2. negation
3. conjunction, disjunction
4. conditional, biconditional

Answer 3

For the first one you can think of it as the same as $\exists \color{red}x(P(\color{red}x) \land Q(y))$ because they're under different domains.

The second one I am guessing that they are both under different domains. What makes me say this is that quantifiers are usually before a set of parentheses. $$(1) \space \forall \color{red}x(P(\color{red}x) \rightarrow Q(\color{red}x)) \text{ vs. } (2)\space \forall \color{red}x(P(\color{red}x) \rightarrow Q(\color{blue}y))$$

I think $(2)$ is your answer.

Answer 4

As I wrote here, the convention varies by field, so the blanket statements in the other answers are a bit strong. In most of mathematics, the quantifier binds strongly, so the red $x$ would be free. But in fields like type theory, quantifiers capture as much as possible while still respecting parentheses; in those areas of mathematics, we would read the red $x$ as bound by the quantifier.