Predicate Logic · step 6 of 13

Nested quantifiers & quantifier order

When quantifiers of different kinds are nested, the order encodes a dependence. ∀x ∃y R(x, y) lets y depend on x — "everyone has a mother". ∃y ∀x R(x, y) demands one y that works for all x — "someone is everyone’s mother". The second always implies the first, never the reverse, and that asymmetry is the single most examined idea in predicate logic.

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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.

Read left to right as a dialogueEach ∀ is a challenge from an opponent, each ∃ your reply. Later choices may depend on earlier ones, never on later ones.
Identify what may depend on whatWrite y = f(x) if y comes after x. If a proposed witness needs information from a variable quantified later, the statement is false.
Compare adjacent same-kind quantifiers∀x ∀y and ∀y ∀x are interchangeable, as are ∃x ∃y and ∃y ∃x. Only mixed pairs are order-sensitive.
Test both orders on a concrete structureOver ℤ or a two-element domain, evaluate both readings to confirm which one the English actually means.

Worked example

Over the positive reals with R(x, y) meaning x < y, evaluate ∀x ∃y R(x, y) and ∃y ∀x R(x, y).

First: given x > 0, take y = x + 1 > x. The witness depends on x, which is permitted.
So ∀x ∃y (x < y) is true — there is no largest positive real.
Second: a single y must exceed every positive x. But x = y + 1 is a positive real not less than y.
So no such y exists.

Answer. ∀x ∃y (x < y) is true; ∃y ∀x (x < y) is false. Swapping the quantifiers turns a true statement into a false one.

Where marks get dropped

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

Assuming ∀x ∃y and ∃y ∀x are interchangeable. The ∃∀ form is strictly stronger, and this is where most marks are lost.
Choosing a witness that secretly depends on a later-quantified variable — the definition of an invalid proof at this level.
Misreading English scope. "Every lock is opened by some key" is ∀ lock ∃ key; "some key opens every lock" is ∃ key ∀ lock.

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Nested quantifiers & quantifier order — frequently asked questions

Does ∃y ∀x R(x, y) imply ∀x ∃y R(x, y)?

Yes, always. A single universal witness serves as an individual witness for each x. The converse fails, as the example above shows.

Can I swap two quantifiers of the same type?

Yes. ∀x ∀y P is equivalent to ∀y ∀x P, and likewise for two ∃. Only alternating pairs carry dependence information.

How does this relate to ε–δ definitions?

Directly. Continuity is ∀ε ∃δ ∀x (…), and uniform continuity moves δ outside the ∀x. The entire difference between the two concepts is quantifier order.

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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