Here are the following two sentences.
At least one person speaks English.
∃𝑥E(x)
Exactly one person speaks English.
Instead of ∃𝑥E(x), what do I write?
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1 Answer
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)))
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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").
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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
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@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
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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
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@lemontree Ah I see, you're right. Completely missed that. Aug 10, 2020 at 23:48
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1I.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