Predicate Logic · step 4 of 13

Negating quantified statements

Negation flips quantifiers: ¬∀x P(x) ≡ ∃x ¬P(x), and ¬∃x P(x) ≡ ∀x ¬P(x). Applying these repeatedly pushes a negation all the way to the atomic predicates, which is what "negate and simplify" questions demand. In words: the denial of "everything is P" is "something is not P", not "nothing is P".

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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.

Flip the outermost quantifierMove ¬ inside and change ∀ to ∃ or ∃ to ∀. Repeat for each quantifier in turn, left to right.
Apply propositional De Morgan to the matrixOnce past the quantifiers, negate the body: ¬(A ∧ B) becomes ¬A ∨ ¬B, and ¬(A → B) becomes A ∧ ¬B.
Watch the restricted forms¬∀x (P(x) → Q(x)) becomes ∃x (P(x) ∧ ¬Q(x)) — the implication turns into a conjunction. ¬∃x (P(x) ∧ Q(x)) becomes ∀x (P(x) → ¬Q(x)).
Stop only at literalsThe simplified form should have ¬ applied to atomic predicates and nowhere else.

Worked example

Negate and fully simplify ∀x (Student(x) → ∃y (Course(y) ∧ Takes(x, y))).

Flip the ∀: ∃x ¬(Student(x) → ∃y (Course(y) ∧ Takes(x, y))).
Negate the implication: ∃x (Student(x) ∧ ¬∃y (Course(y) ∧ Takes(x, y))).
Flip the ∃: ∃x (Student(x) ∧ ∀y ¬(Course(y) ∧ Takes(x, y))).
De Morgan on the body: ∃x (Student(x) ∧ ∀y (Course(y) → ¬Takes(x, y))).

Answer. ∃x (Student(x) ∧ ∀y (Course(y) → ¬Takes(x, y))) — "some student takes no course".

Where marks get dropped

These are the specific errors that cost credit on negating quantified statements questions — QED's rubric penalises each of them separately.

Negating ∀x P(x) to ∀x ¬P(x). That says everything fails P, which is much stronger than the denial.
Forgetting that the restricting → becomes ∧ under negation. ∃x (P(x) → ¬Q(x)) is not the negation of ∀x (P(x) → Q(x)).
Stopping halfway with a ¬ still in front of a quantifier. "Fully simplified" means negations sit only on predicates.

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Negating quantified statements — frequently asked questions

What is the negation of "every student passed"?

"Some student did not pass" — ∃x (Student(x) ∧ ¬Passed(x)). It is not "no student passed", which would be a much stronger claim.

How do I negate a statement with three quantifiers?

Flip them one at a time from the outside in, keeping their order, then De Morgan the matrix. ¬∀x∃y∀z P becomes ∃x∀y∃z ¬P.

Does this work for ∃!?

Uniqueness must be expanded first. ∃!x P(x) unfolds to ∃x (P(x) ∧ ∀y (P(y) → y = x)), and you negate that expansion.

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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