1

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?

0implies0
  • 542
  • While it may be technically correct (not sure), to avoid confusion, you should write $\forall x P(x) \rightarrow Q(y)$ where $y$ is some variable other than $x$. – Dan Christensen Nov 15 '20 at 16:53

3 Answers3

3

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.

Mauro ALLEGRANZA
  • 94,169
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
ryang
  • 38,879
  • 14
  • 81
  • 179
0

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.