Here are the following two sentences.

At least one person speaks English.


Exactly one person speaks English.

Instead of ∃𝑥E(x), what do I write?

1 Answer 1


You sometimes find the notation


as an abbreviation for "exactly one x".

With the standard symbol inventory, "exactly one" can be defined in terms of "at least one and not more than one" as follows:

∃x(E(x) ^ ¬∃y(E(y) ^ ¬(x = y)))
---------    --------------------------
existence       uniqueness

("There exists at least one person who speaks English, and there is noone who also speaks English but is different from that first person")

or more compactly

∃x∀y(E(y) ↔ y = x)

("There exists a person such that the people who speak English are exactly that person").

  • 5
    +1. Note to the OP that the notation "∃!" can be slightly misleading, since it hides the actual complexity of the quantifier - see e.g. here. Aug 9, 2020 at 22:23
  • @lemontree Take a set of objects a, b, c. Using your last statement, we have "∀y(E(y) ↔ y = a) v ∀y(E(y) ↔ y = b) v ∀y(E(y) ↔ y = c)". Then: "(E(a) ↔ a = a ^ E(b) ↔ b = a ^ E(c) ↔ c = a) v (E(a) ↔ a = b ^ E(b) ↔ b = b ^ E(c) ↔ c = b) v (E(a) ↔ a = c ^ E(b) ↔ b = c ^ E(c) ↔ c = c)". Here each of the three large compound statements that are ORed together could all possibly be true - inclusive OR doesn't forbid this. If so, then since a = a, b = b, c = c, we have E(a), E(b), and E(c), but the goal was to show exactly one person speaks English. So I think your statement doesn't forbid > 1 person. Aug 10, 2020 at 23:33
  • 1
    @Inertial Ignorance No, in this particular setting, the disjuncts can not be simultaneously true. Suppose the first disjunct (where x = a) is true. Then, in its second conjunct (where y = b), since b = a is false, with E(b) ↔ b = a, E(b) must be false. But then in the second disjunct (where x = b), with E(b) false and E(b) ↔ b = b, we have that b = b is false, which is a contradiction, so the second disjunct can not be true. Analogous for all other combinations. The equality constraints with the implication in both directions is precisely what forbids > 1. Aug 10, 2020 at 23:42
  • @lemontree Ah I see, you're right. Completely missed that. Aug 10, 2020 at 23:48
  • 1
    I.o.w, with the quantifiers multiplied-out like you did, the formula states that (Ea ^ ~Eb ^ ~Ec) v (~Ea ^ Eb ^ ~Ec) v (~Ea ^ ~Eb ^ Ec) -- that is, the ∀y part makes it such that each disjunct (each possible value for x) expresses that E holds of only one object . Aug 10, 2020 at 23:48

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