Predicate Logic · step 9 of 13

Equality & uniqueness (∃!)

First-order logic with equality treats = as a logical symbol interpreted as genuine identity. That single addition makes counting expressible: "at least two", "at most one" and "exactly one" all become formulas. ∃!x P(x) is shorthand, not a new quantifier — it abbreviates ∃x (P(x) ∧ ∀y (P(y) → y = x)), which is exactly existence plus uniqueness.

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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.

Expand the abbreviationWrite ∃!x P(x) as ∃x (P(x) ∧ ∀y (P(y) → y = x)) before doing anything technical with it.
Prove existence and uniqueness separatelyExhibit a witness, then take an arbitrary second object with the property and show it equals the first. Exams award these as two separate marks.
Express counting claims with equality"At least two" is ∃x ∃y (P(x) ∧ P(y) ∧ x ≠ y). "At most one" is ∀x ∀y ((P(x) ∧ P(y)) → x = y).
Use the equality axioms explicitlyReflexivity x = x, and substitution: from x = y and P(x) infer P(y). These are what make equality more than an ordinary relation.

Worked example

Over ℝ, show ∃!x (3x + 6 = 0) by the standard two-part method.

Existence: take x = −2. Then 3(−2) + 6 = 0, so the property holds.
Uniqueness: suppose y also satisfies 3y + 6 = 0.
Then 3y = −6, so y = −2.
Hence y = x, and the witness is unique.

Answer. Exactly one real satisfies 3x + 6 = 0, namely x = −2 — existence by witness, uniqueness by forcing any candidate to equal it.

Where marks get dropped

These are the specific errors that cost credit on equality & uniqueness (∃!) questions — QED's rubric penalises each of them separately.

Proving existence only and claiming ∃!. Uniqueness is a separate obligation and separately marked.
Proving uniqueness by assuming there are two and "seeing" they are the same, without deriving y = x from the property.
Writing "at least two" as ∃x ∃y (P(x) ∧ P(y)). Without x ≠ y this is satisfied by a single element used twice.

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Equality & uniqueness (∃!) — frequently asked questions

Is ∃! a real quantifier?

No — it is an abbreviation. Everything about it reduces to ∃, ∀ and =, which is why questions often ask you to expand it first.

Can I express "exactly two"?

Yes: ∃x ∃y (x ≠ y ∧ P(x) ∧ P(y) ∧ ∀z (P(z) → (z = x ∨ z = y))). Each extra number costs another clause, which is why counting is clumsy in first-order logic.

Why is equality a logical symbol?

Because its interpretation is fixed as true identity in every structure, unlike an ordinary predicate that can be interpreted freely. That fixing is what licenses the substitution rule.

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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