In the sentence, "
∀
𝑥
𝑃
𝑥
⟶
∃
𝑦
𝑅
𝑥
𝑦
" every instance of x is bound.

Question

In the sentence, "
∀
𝑥
𝑃
𝑥
⟶
∃
𝑦
𝑅
𝑥
𝑦
" every instance of x is bound.

BlackTom AI · Accepted Answer

A quantifier binds only occurrences of its variable that fall within its scope; here ∀x binds x in the subformula that immediately follows it, and ∃y binds y in the subformula that immediately follows it.

1. Relevant knowledge points and concepts
- Bound occurrence: an occurrence of a variable is bound if it lies within the scope of a corresponding quantifier (e.g., ∀x or ∃x).
- Free occurrence: an occurrence not within any quantifier for that variable is free.
- Scope convention: without parentheses, quantifiers bind only the formula directly to their right; an expression like ∀x P x ⟶ ∃y R x y is typically parsed as (∀x P x) ⟶ (∃y R x y).

2. Step-by-step reasoning
- Parse the formula according to the usual convention: (∀x P x) ⟶ (∃y R x y).
- Identify scopes:
  - Scope(∀x) = P x, so the x in P x is bound by ∀x.
  - Scope(∃y) = R x y, so ∃y binds the y in R x y.
- Examine every occurrence of x:
  - x in P x: bound (inside Scope(∀x)).
  - x in R x y: not inside Scope(∀x) (because ∀x’s scope ended before the implication), and there is no quantifier ∀x or ∃x covering it on the right; therefore this occurrence of x is free.

3. Intermediate “calculation” steps (scoping):
- Original: ∀x P x ⟶ ∃y R x y
- Parsed: (∀x P x) ⟶ (∃y R x y)
- Bound x’s: {x in P x}
- Free x’s: {x in R x y}

4. Why the correct answer is correct
Because at least one occurrence of x (the one inside R x y) is not within the scope of ∀x, not every instance of x is bound. That makes the assertion "every instance of x is bound" false.

5. Why the other option ("True") is incorrect
"True" would require that ∀x’s scope include the whole implication so that x in R x y is bound; that only holds if the formula were written ∀x (P x ⟶ ∃y R x y). As written, the convention gives a free x, so "True" is incorrect.

6. Conclusion
Therefore the correct evaluation is False, matching the provided answer.

In the sentence, " ∀ 𝑥 𝑃 𝑥 ⟶ ∃ 𝑦 𝑅 𝑥 𝑦 " every instance of x is bound.判断题

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In the sentence, "" every instance of x is bound.

The expression, "" expresses “there is something with property P.”

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