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Universal & existential quantifiers

∀x P(x) asserts that P holds for every element of the domain; ∃x P(x) asserts it holds for at least one. Neither claim means anything until the domain is fixed — ∀x (x² ≥ 0) is true over ℝ but ∃x (x² = −1) is false over ℝ and true over ℂ. The pairing with connectives is also fixed by convention: ∀ almost always pairs with →, and ∃ with ∧.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. State the domain firstWrite "domain: the integers" or restrict inside the formula. Marks are routinely lost for quantified claims with no domain.
  2. Choose the right partner connective"All P are Q" is ∀x (P(x) → Q(x)). "Some P is Q" is ∃x (P(x) ∧ Q(x)). Swapping these is the standard error.
  3. Evaluate ∀ by attempting a counterexampleA universal claim over a finite domain is checked element by element; over an infinite one you argue generally or produce one failing element.
  4. Evaluate ∃ by producing a witnessOne concrete element satisfying the predicate settles an existential claim completely.

Worked example

Over the domain ℤ, decide the truth of ∀x ∃y (x + y = 0) and of ∃y ∀x (x + y = 0).

  1. First statement: given any integer x, choose y = −x. Then x + y = 0.
  2. The choice of y is allowed to depend on x, so the claim holds.
  3. Second statement: one fixed y must work for every x. Take y = c; then x = 1 gives 1 + c = 0 forcing c = −1, but x = 2 gives 2 + c = 0 forcing c = −2.
  4. No single c satisfies both.

Answer. ∀x ∃y (x + y = 0) is true; ∃y ∀x (x + y = 0) is false. Order of quantifiers changes the meaning.

Where marks get dropped

These are the specific errors that cost credit on universal & existential quantifiers questions — QED's rubric penalises each of them separately.

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Universal & existential quantifiers — frequently asked questions

Why does ∀ pair with → and ∃ with ∧?

Because ∀ restricts a claim to the P-things without asserting any exist, while ∃ must assert both that something exists and that it is Q. The mismatched pairings produce statements nobody intends.

Is ∀x P(x) true when the domain is empty?

Yes — vacuously. There is no counterexample available. ∃x P(x) is false over an empty domain, which is why the two are not symmetric.

Can I quantify over predicates?

Not in first-order logic, where quantifiers range only over domain elements. Quantifying over predicates puts you in second-order logic, which loses completeness.

The rest of Predicate Logic

Quantifiers, predicates, binding, and validity. Each subtopic below has its own method, worked example and mark-losing traps.

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