In first-order logic, a primitive predicate "=" is introduced. As a result, "a = b" or "=(a,b)" is a primitive equal relation in first-order logic. In set theory, meanwhile, "=" can be defined as follows: $a = b$ if $$\forall x, x \in a \leftrightarrow x \in b.$$ I am wondering what is the meaning of the primitive equal sign, and what is the relation between the two equal signs respectively from first-order logic and set theory?
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See the post Why include equality in FOL for ZFC? We can use equality "inherited" from the underlying predicate logic (in which case we understood it as a general properties of every "object") or we can introduce it specifically for sets, in which case we define it in terms of the basic $\in$ relation and with Extensionality we prove that the defined $=$ relation has all the expected properties. – Mauro ALLEGRANZA Dec 22 '21 at 14:15
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See also this post as well as this one. – Mauro ALLEGRANZA Dec 22 '21 at 14:18
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The axiom of extensionality is not a definition, it's an axiom. It states that if two sets have the same elements, they satisfy the property of equality from first order logic.
David Lui
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The only criteria for an equality predicate are that it follows the "rules of equality".
The rules of equality are (the universal closure of) the following:
- For every $n$-ary predicate symbol $P$, $(\bigwedge\limits_{i = 1}^n x_i = y_i) \land P(x_1, \ldots, x_n) \to P(y_1, \ldots, y_n)$
- For every $n$-ary function symbol $f$, $(\bigwedge\limits_{i = 1}^n x_i = y_i) \to f(x_1, \ldots, x_n) = f(y_1, \ldots, y_n)$
- $x = x$
- $a = b \to b = a$
From here, we can prove the following metatheorem:
Substitution Principle of Equality: For every proposition $\phi$ where $y$ does not occur bound, the universal closure of $x = y \land \phi \to \phi[x \mapsto y]$ holds. Here, $\phi[x \mapsto y]$ is $\phi$ with every free occurence of $x$ replaced by $y$.
The proof is a straightforward structural induction on $\phi$.
Generally, first-order logic will come with equality (and the above rules) built in. But it's certainly possible to consider a formal theory without built-in equality.
Let's consider your "definition" of equality in set theory, which is
$$x = y :\equiv \forall z (z \in x \iff z \in y)$$
Note that without assuming any axioms of set theory, this does not suffice as a definition of equality because it does not meet criterion 1. We would have to formulate the axiom of extensionality as (the universal closure of)
$$(\forall z (z \in x \iff z \in y)) \land x \in k \to y \in k$$
to force rule 1 to hold. At that point, the logic of set theory becomes completely identical to the normal first-order theory with $=$ built-in.
Mark Saving
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