Consider the following true statement

(1) The set whose only members are the prime numbers between 6 and 12 is the same as the set whose only members are the solutions to the equation x^2-18x+77=0.

How do I translate statement (1) above to first-order naive set theory?

The book I am following seems to define a set as follows: ∃a ∀x (x ∈ a ↔ P(x)).

But doesn't this say rather that a set "a" exists such that its members have the property P?

If I write

∃a ∀x (x ∈ a ↔ Prime(x) & 6<x<12)

∃b ∀x (x ∈ b ↔ x^2-18x+77=0)

My attempt would be:

(∃a ∀x (x ∈ a ↔ Prime(x) & 6<x<12)) & (∃b ∀x (x ∈ b ↔ x^2-18x+77=0)) & a=b

Note that informally (ie outside of FOL) we write {x | P(x)} to mean the same thing as ∃a ∀x (x ∈ a ↔ P(x)).

Thus, informally, I believe statement (1) is {x | Prime(x) & 6<x<12} = {x | x^2-18x+77=0}.

Is this attempt at translating to first-order naive set theory correct?

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    Write each set using set builder notation, and put an equals/ identity sign between them. Jan 30 at 11:25
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    Your above set definition ∃a ∀x(x ∈ a ↔ P(x)) seems the axiom scheme of comprehension version of naive set theory, and together with another Axiom of Extensionality ∀a ∀b[∀x(x ∈ a ↔ x ∈ b) → a=b], proves there's a unique set conveniently notated by the usual set builder satisfying P(x). So you'd better revise your attempt to: ((∃a ∀x (x ∈ a ↔ Prime(x) & 6<x<12)) & (∃b ∀x (x ∈ b ↔ x^2-18x+77=0))) → (a=b) since the extensional equality between a and b is not given premise but can be easily proved via Axiom of Extensionality after exhaustively verified x ∈ a ↔ x ∈ b for all x in either a or b Jan 31 at 0:04
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    A set with brace notation is: A={x | P(x)}. Example: A={x | "x is Even"}. Jan 31 at 6:56
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    Thus, the first set of your example will be A={x | Prime(x) & (6<x<12)} while the second will be B= {x | x^2-18x+77=0}. That's all. Jan 31 at 7:16
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    If you are working into predicate logic (NOT set theory) and thus there is no "in" predicate and no "braces" what you can write is that two "predicates" have the same extension: ∀x[(Prime(x) & (6<x<12)) ↔ (x^2-18x+77=0)] Jan 31 at 10:09

2 Answers 2


No, this does not work. The problem is that a and b are bound to their respective statements. That is, when you say ∃a ∀x (x ∈ a ↔ Prime(x) & 6<x<12), you can't invoke 'a' outside of the scope of the existential statement. I don't quite know what the right answer is because I don't know what the notation is for naive set theory. I'm pretty sure that ∀a∀b((∀x (x ∈ a ↔ Prime(x) & 6<x<12) & ∀x (x ∈ b ↔ x^2-18x+77=0)) → a = b ) is true since conditionals are only ever false if the antecedent is true, and the consequent is false. This is impossible in this scenario, so the statement is true.

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    Your answer is as correct as could be expected at the level of the asker. Your criticism of the asker's attempt is also spot on.
    – user21820
    Feb 6 at 17:38

∃a∀x(x∈a ⇔ P(x)) says for every property there is a set which includes objects with that property. Basically saying you can have any sort of set you can think of. That's not really a formula for you to follow more of an "axiom".

Naive set−theory has the same logical and non−logical symbols as ZFC with the only difference between the two being that of axioms.

{x|P(x): 6 < x < 12} ={7,11}

You cannot state the above identity in a purely logical language. You need the extralogical bits because "<" is a binary non−logical relation symbol that says something about some set. If you want to do so without involving any particular semantic, then that makes it easy:

P(x): x is prime

G(x,y): x is greater than y

L(x,y): x is less than y





∀x[((P(x) ∧(G(x,a) ∧ L(x,b))) ⇔ [(x=c ∨ x=d) ∧ ¬(x=c ∧ x=d)]] you can also split it fi you want but you will be missing ALOT of information in either case. The non−logical symbols are to an extent essential.

I might be missing a clause in the formulation.

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