Exactly One in Predicate Logic: Existence Plus Uniqueness

Require a witness and rule out a second P object

Exactly one combines existence with at-most-one. A convenient formula is ∃x(Px∧∀y(Py→y=x)). The chosen x is P, and any y that is P must be the very same object.

Exactly one P:
∃x(Px ∧ ∀y(Py → y=x))

At most one P:
∀x∀y((Px ∧ Py) → x=y)

Identity x=y means sameness of object, not merely sharing properties. We use a nonempty domain and ordinary classical first-order logic with identity. Variables can range over the same object; having two variable letters does not force distinctness.

Definitions and rules: [1] [2].

Check zero, one and two P objects

All three original models have the complete domain {A,B}. Their only difference is P’s extension: empty, {A}, or {A,B}. A and B identify distinct domain objects in these model descriptions.

Download this original labelled diagram

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Download every P membership in the three models

With zero P, the at-most-one conditional never encounters two P objects, so it is true. Existence fails. With one P, choose that object as x; every P y is identical to it. With two P objects, whichever x you choose, the other P y violates y=x.

Read each part of the compact formula

The outer ∃x supplies a proposed unique P. The first conjunct Px ensures the proposed object really has the property. The inner ∀y then examines all domain objects, but its conditional restricts the identity demand to objects satisfying Py.

Removing Px loses existence of a P. The formula ∃x∀y(Py→y=x) can be true when there are no P objects: choose any domain object for x and all the inner conditionals have false antecedents. Removing the inner universal loses uniqueness: ∃xPx only says one or more P objects exist.

Replacing Py→y=x with Py∧y=x would require every domain object to be P and identical to x. That restricts the whole domain to a singleton, which is stronger than exactly one P in an otherwise larger domain.

Definitions and rules: [1].

Do not count names as distinct objects

Suppose constants a and b both refer to A in a domain containing A and B, and P holds only of A. Then Pa and Pb are both true, but there is still just one P object. The interpretation permits more than one name for the same object.

At least two P objects:
∃x∃y(Px ∧ Py ∧ x≠y)

The inequality is essential. ∃x∃y(Px∧Py) can use the same object for both witnesses. For three distinct witnesses, every pair must be unequal; merely writing x≠y and y≠z does not rule out x=z.

Definitions and rules: [1].

In ordinary reading, a story’s two character names or a spreadsheet’s two labels might refer to the same entity. Verify the identity question before treating label count as object count. The formal example is not an entity-resolution method for real data.

Keep property uniqueness separate from domain size

Exactly one P does not say exactly one object exists. In the middle model, two objects exist and only A is P. A large or infinite domain can likewise contain exactly one P.

Conversely, a one-object domain does not by itself ensure exactly one P: its sole object might not be P. You need the property claim too. Keep the domain and P extension as separate lines in the interpretation key.

Use the full formula when a later statement depends on the unique object, such as Russell’s analysis of a definite description. Do not assume a phrase like “the P” has already secured existence and uniqueness under every theory of language.

Try three identity checks

Assess objects and predicate membership, rather than the number of written labels.