Key takeaways
- ¬∀x P(x) is equivalent to ∃x ¬P(x), and ¬∃x P(x) is equivalent to ∀x ¬P(x).
- For nested quantifiers, move the negation in one quantifier at a time, flipping each ∀ to ∃ and each ∃ to ∀, and keep their order unchanged.
- When the negation reaches the inside, apply the propositional rules: ¬(p → q) becomes p ∧ ¬q, and De Morgan's laws handle and and or.
- The negation of 'every student passed' is 'some student did not pass', not 'no student passed'.
- A counterexample is simply a proof of the negated statement, which is why one counterexample disproves a universal claim.
To negate a quantified statement, flip each quantifier (∀ becomes ∃, ∃ becomes ∀), keep them in the same order, and negate what follows. Every proof by contradiction and every counterexample starts with a negation like this, so the rule is short but worth practising, along with the handful of slips that break it.
The two basic rules
| Statement | Negation | In words |
|---|---|---|
| ∀x P(x) | ∃x ¬P(x) | Not everything has P: something lacks it |
| ∃x P(x) | ∀x ¬P(x) | Nothing has P: everything lacks it |
Applied Discrete Structures summarises it in one sentence: the negation of a universally quantified proposition is existentially quantified, and the negation of an existential one is universal. The pair is sometimes called De Morgan's laws for quantifiers, and the name fits. Over a finite domain {a, b, c}, ∀x P(x) is just P(a) ∧ P(b) ∧ P(c), and its negation by De Morgan's laws is ¬P(a) ∨ ¬P(b) ∨ ¬P(c), which is ∃x ¬P(x).
The everyday trap
The negation of "every student passed" is "some student did not pass". It is not "no student passed".
Suppose 29 of 30 students passed. Then "every student passed" is false, so its negation must be true. "Some student did not pass" is true; "no student passed" is false. Only the first can be the negation. The same mistake appears in the other direction: the negation of "some flights were delayed" is "no flights were delayed", which is ∀x ¬D(x), not "some flights were not delayed".
A quick test for any proposed negation: exactly one of the statement and its negation must be true in every situation. If you can imagine a situation where both are false, as with "all passed" and "none passed" when 29 of 30 pass, you have not found the negation.
Negating nested quantifiers: one layer at a time
For statements with several quantifiers, move the negation inward past one quantifier at a time. Each time it passes a quantifier, that quantifier flips. The order of the quantifiers never changes.
Example 1. Negate ∀x ∃y (x + y = 0).
- ¬∀x ∃y (x + y = 0)
- ≡ ∃x ¬∃y (x + y = 0)
- ≡ ∃x ∀y ¬(x + y = 0)
- ≡ ∃x ∀y (x + y ≠ 0)
In words: there is some x that no y can cancel. Over the integers the original is true (take y = −x), so this negation is false. Over the natural numbers {0, 1, 2, …} the original is false, because x = 1 has no partner, and the negation is true with x = 1 as the witness.
Example 2. Negate ∃y ∀x (y ≤ x), "there is a smallest element".
- ¬∃y ∀x (y ≤ x)
- ≡ ∀y ¬∀x (y ≤ x)
- ≡ ∀y ∃x ¬(y ≤ x)
- ≡ ∀y ∃x (x < y)
In words: for every y there is something smaller. Over the integers the negation is true (take x = y − 1), so the integers have no smallest element. Over the natural numbers the original is true with y = 0.
When the inside is a conditional or a compound
Once the negation reaches the predicate, use the propositional rules:
- ¬(p ∧ q) ≡ ¬p ∨ ¬q and ¬(p ∨ q) ≡ ¬p ∧ ¬q
- ¬(p → q) ≡ p ∧ ¬q
- ¬(a < b) is a ≥ b, and ¬(a = b) is a ≠ b
The conditional rule comes up constantly, because a statement of the form "every A is B" is translated with an arrow. Levin's open textbook uses this example: "every prime greater than 2 is odd" is ∀n ((Prime(n) ∧ n > 2) → Odd(n)). Its negation:
∃n (Prime(n) ∧ n > 2 ∧ ¬Odd(n))
that is, some prime greater than 2 is even. The arrow disappears; the negation of a conditional is never another conditional, a point made in the converse, inverse and contrapositive guide.
A harder example: the definition of a limit
The definition of a convergent sequence, from calculus and analysis, is a good test of the method. "The sequence aₙ converges to L" is
∀ε > 0 ∃N ∀n ≥ N (|aₙ − L| < ε).
Negate it one layer at a time:
- ∃ε > 0 ¬∃N ∀n ≥ N (|aₙ − L| < ε)
- ∃ε > 0 ∀N ¬∀n ≥ N (|aₙ − L| < ε)
- ∃ε > 0 ∀N ∃n ≥ N ¬(|aₙ − L| < ε)
- ∃ε > 0 ∀N ∃n ≥ N (|aₙ − L| ≥ ε)
In words: there is some distance ε such that, however far out you start, a later term is at least ε away from L. The restrictions "ε > 0" and "n ≥ N" stay attached to their quantifiers; they describe which values the quantifier ranges over, and they are not negated.
Why this is the basis of counterexamples
To disprove ∀x P(x), you prove its negation ∃x ¬P(x), and proving an existential statement means producing one example. That example is the counterexample. So "one counterexample is enough" is not a convention; it is the negation rule. By contrast, disproving ∃x P(x) means proving ∀x ¬P(x), which needs a general argument. A proof by contradiction starts by assuming a negation just like the ones above, so getting the negation right is the first line of the proof.
Mistakes to check for
- Changing the order of quantifiers. ¬∀x ∃y P becomes ∃x ∀y ¬P, never ∀y ∃x ¬P.
- Flipping only the first quantifier. Every quantifier the negation passes gets flipped.
- Leaving an arrow inside. ¬(P → Q) is P ∧ ¬Q.
- Negating the domain restriction. ¬∀x > 0 … becomes ∃x > 0 …, not ∃x ≤ 0 ….
- Using "no" for "not all".
If quantifiers themselves are unfamiliar, start with predicates and quantifiers. To check a negation you have written by hand, the tools section of Discrete Math: Study & Quiz includes a quantifier-negation tool that negates a quantified statement for you to compare against. Browse the discrete math hub for the full logic series, or our education category for more study apps.
Good to know
Frequently asked questions
What is the negation of 'for all x, P(x)'?
What is the negation of 'for every x there exists a y with P(x, y)'?
Is the negation of 'all students passed' 'no students passed'?
How do you negate a universal statement containing an implication?
Sources
- Quantifiers: The Negation of Quantified Propositions (Applied Discrete Structures) (opens in a new tab)discretemath.org — accessed
- Propositional Logic (Discrete Mathematics: An Open Introduction, 3rd edition, Oscar Levin) (opens in a new tab)Open Math Books — accessed
- 2.1 Statements and Quantifiers (Contemporary Mathematics) (opens in a new tab)OpenStax — accessed
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