Definition df-clel

Description: Define the membership connective between classes. Theorem 6.3 of Quine p. 41, or Proposition 4.6 of TakeutiZaring p. 13, which we adopt as a definition. See these references for its metalogical justification.

The hypotheses express that all instances of the conclusion where class variables are replaced with setvar variables hold. Therefore, this definition merely extends to class variables something that is true for setvar variables, hence is conservative. This is only a proof sketch of conservativity; for details see Appendix of Levy p. 357. This is the reason why we call this axiomatic statement a "definition", even though it does not have the usual form of a definition. If we required a definition to have the usual form, we would call df-clel an axiom.

See also comments under df-clab , df-cleq , and abeq2 .

Alternate characterizations of A e. B when either A or B is a set are given by clel2g , clel3g , and clel4g .

This is called the "axiom of membership" by Levy p. 338, who treats the theory of classes as an extralogical extension to our logic and set theory axioms.

While the three class definitions df-clab , df-cleq , and df-clel are eliminable and conservative and thus meet the requirements for sound definitions, they are technically axioms in that they do not satisfy the requirements for the current definition checker. The proofs of conservativity require external justification that is beyond the scope of the definition checker.

For a general discussion of the theory of classes, see mmset.html#class . (Contributed by NM, 26-May-1993) (Revised by BJ, 27-Jun-2019)

	Ref	Expression
Hypotheses	df-clel.1	$$⊢y\in z\leftrightarrow \exists u\left(u=y\wedge u\in z\right)$$
	df-clel.2	$$⊢t\in t\leftrightarrow \exists v\left(v=t\wedge v\in t\right)$$
Assertion	df-clel	$$⊢A\in B\leftrightarrow \exists x\left(x=A\wedge x\in B\right)$$

Detailed syntax breakdown

Step	Hyp	Ref	Expression
0		cA	$$classA$$
1		cB	$$classB$$
2	0 1	wcel	$$wffA\in B$$
3		vx	$$setvarx$$
4	3	cv	$$setvarx$$
5	4 0	wceq	$$wffx=A$$
6	4 1	wcel	$$wffx\in B$$
7	5 6	wa	$$wff\left(x=A\wedge x\in B\right)$$
8	7 3	wex	$$wff\exists x\left(x=A\wedge x\in B\right)$$
9	2 8	wb	$$wff\left(A\in B\leftrightarrow \exists x\left(x=A\wedge x\in B\right)\right)$$