1
Each quantifier flips as the negation passes
∀ becomes ∃ and ∃ becomes ∀, in order, until the negation reaches the predicate. This is De Morgan generalised to arbitrary domains.
Lesson · Logic and Proof · Predicate logic
Negating Quantified Statements
¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Applied repeatedly, this converts any negated statement into one with the negation on the predicate — which is how you find out what a counterexample must look like.
¬∀x P(x) ≡ ∃x ¬P(x) "not all" = "some fails"
¬∃x P(x) ≡ ∀x ¬P(x) "none" = "all fail"
¬∀x ∃y P ≡ ∃x ∀y ¬P flip every quantifier in turn
¬(P → Q) ≡ P ∧ ¬Q inside the predicate
Generalised De Morgan
Core ideas
What this lesson establishes
Negation is mechanical here, and the mechanism is worth trusting: it tells you exactly what would have to be shown to refute a claim.
2
The result tells you what to look for
Negating a theorem statement produces the exact shape of a counterexample. That is the practical payoff of doing it mechanically.
3
Restrictions survive the flip
¬∀x (P(x) → Q(x)) ≡ ∃x (P(x) ∧ ¬Q(x)): the conditional becomes a conjunction, which is why restricted universals negate into restricted existentials.
The material
Pushing negation through quantifiers
The two rules, why they generalise De Morgan, and how to apply them in sequence.
Theorem · Quantifier negation
¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x), over any domain.
¬∀x P(x) says it is not the case that P holds throughout the domain — so some element fails it, which is ∃x ¬P(x). Conversely, if some c has ¬P(c) then ∀x P(x) cannot hold.
For a finite domain {a₁, …, aₙ} this is literally De Morgan: ¬(P(a₁) ∧ … ∧ P(aₙ)) ≡ ¬P(a₁) ∨ … ∨ ¬P(aₙ). The quantifier rules are the same statement freed from finiteness.
□
Applied to a nested statement, the rules are used one at a time from the outside in. Each application moves the negation one quantifier further right and flips that quantifier.
¬ ∀x ∃y ∀z P(x, y, z)
≡ ∃x ¬ ∃y ∀z P(x, y, z)
≡ ∃x ∀y ¬ ∀z P(x, y, z)
≡ ∃x ∀y ∃z ¬P(x, y, z)
Every quantifier has flipped and the negation now sits on the predicate, where the propositional rules of “Logical Equivalence” take over. If P is a conditional, ¬(A → B) ≡ A ∧ ¬B finishes the job.
Example · Negating a restricted universal
"Every prime greater than 2 is odd" is ∀x ((P(x) ∧ x > 2) → Odd(x)). Its negation is ∃x (P(x) ∧ x > 2 ∧ ¬Odd(x)) — "there is an even prime greater than 2". That is exactly the counterexample you would have to produce, and its shape came out of the mechanism rather than out of thinking about primes.
This is the lesson's real use. Faced with a claim you suspect is false, negate it mechanically and read what the negation demands. Very often the negation is easy to satisfy and the counterexample writes itself.
Theorem · Negating uniqueness
¬∃!x P(x) says either nothing satisfies P or at least two things do. Unfolding ∃! and negating gives exactly that disjunction — a good check on whether the mechanism has been applied correctly.
Interactive
Quantifier order
A finite universe you can edit
Method
Negating a statement
One quantifier at a time, outside in. Never guess the result.
1
Write the statement fully symbolically
Including the domain restriction as a conditional or conjunction. Negating English directly is where errors enter.
2
Flip the outermost quantifier
∀ to ∃ or ∃ to ∀, moving the negation inside it.
3
Repeat until the negation reaches the predicate
Each step flips exactly one quantifier. Do not skip ahead.
4
Simplify the predicate propositionally
¬(A → B) ≡ A ∧ ¬B, De Morgan for ∧/∨, double negation. Then read the result back into English as the description of a counterexample.
Negation
Score 0 / 0Choose the answer you can justify.
Negating the definition of continuity
f is continuous at a when ∀ε > 0 ∃δ > 0 ∀x (|x − a| < δ → |f(x) − f(a)| < ε).
¬ ∀ε ∃δ ∀x ( |x−a| < δ → |f(x)−f(a)| < ε )
≡ ∃ε ¬ ∃δ ∀x ( … )
≡ ∃ε ∀δ ¬ ∀x ( … )
≡ ∃ε ∀δ ∃x ¬( |x−a| < δ → |f(x)−f(a)| < ε )
≡ ∃ε ∀δ ∃x ( |x−a| < δ ∧ |f(x)−f(a)| ≥ ε )
Read it: there is a tolerance ε such that no matter how small δ is made, some point within δ of a is thrown further than ε away. That is precisely what discontinuity looks like, and the mechanism produced it without any geometric intuition being consulted.
Common mistakes
1
Negating "all are" as "none are""Not all" is ∃x ¬P(x); "none" is ∀x ¬P(x). The second is far stronger and is a different claim entirely.
2
Forgetting the conditional turns into a conjunction¬(P(x) → Q(x)) is P(x) ∧ ¬Q(x), not P(x) → ¬Q(x). Getting this wrong produces a negation that is satisfied by irrelevant elements.
3
Flipping only the outermost quantifierEvery quantifier the negation passes must flip. Stopping after the first leaves a statement that is neither the original nor its negation.
Check your understanding
✓
Finish when you can negate a three-quantifier definition without thinking about its meaning.
Negate ∀ε ∃N ∀n (n ≥ N → |aₙ − L| < ε), the definition of a convergent sequence, and read the answer back into English. If the mechanism carried you through, it will carry you through any definition you meet later.
Note: Doing this mechanically is a genuine technique, not a crutch. Mathematicians negate definitions symbolically precisely because intuition about what "not continuous" means is unreliable and the rules are not.