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Here are the following two sentences.

At least one person speaks English.

∃E(x)

Exactly one person speaks English.

Instead of ∃E(x), what do I write?

Abercrombie Dorfen
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1 Answers1

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You sometimes find the notation

∃!x

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").

Natalie Clarius
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    +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](https://math.stackexchange.com/q/3528948/28111). – Noah Schweber Aug 09 '20 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. – Inertial Ignorance Aug 10 '20 at 23:33
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    @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. – Natalie Clarius Aug 10 '20 at 23:42
  • @lemontree Ah I see, you're right. Completely missed that. – Inertial Ignorance Aug 10 '20 at 23:48
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    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 . – Natalie Clarius Aug 10 '20 at 23:48
  • Okay, glad I could resolve the issue. But nice try! – Natalie Clarius Aug 10 '20 at 23:49
  • @lemontree yes it's saying that just a, just b, or just c. Combining any of those can't happen, and therefore you have to only pick one. – Inertial Ignorance Aug 10 '20 at 23:51
  • Is there a way of expressing "There are exactly 6 people known to speak English." or at least 6 people or at most 6 people, etc? For example, would ∃!6x E(x) with either an absence of or the presence of a floor or ceiling symbol suffice? I apologize if I seem to be belaboring what could be interpreted as a pedantic point. I see it as significant. – Abercrombie Dorfen Sep 06 '20 at 21:59
  • Good question -- see here: https://math.stackexchange.com/a/3763423/344246 Or post it as a new question (a more detailed explanation on my side would get too long for a comment). – Natalie Clarius Sep 06 '20 at 22:17