Theorem zfregfr

Description: The membership relation is well-founded on any class. (Contributed by NM, 26-Nov-1995)

	Ref	Expression
Assertion	zfregfr	⊢ E Fr 𝐴

Proof

Step	Hyp	Ref	Expression
1		dfepfr	⊢ ( E Fr 𝐴 ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ( 𝑥 ∩ 𝑦 ) = ∅ ) )
2		vex	⊢ 𝑥 ∈ V
3		zfreg	⊢ ( ( 𝑥 ∈ V ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ( 𝑦 ∩ 𝑥 ) = ∅ )
4	2 3	mpan	⊢ ( 𝑥 ≠ ∅ → ∃ 𝑦 ∈ 𝑥 ( 𝑦 ∩ 𝑥 ) = ∅ )
5		incom	⊢ ( 𝑦 ∩ 𝑥 ) = ( 𝑥 ∩ 𝑦 )
6	5	eqeq1i	⊢ ( ( 𝑦 ∩ 𝑥 ) = ∅ ↔ ( 𝑥 ∩ 𝑦 ) = ∅ )
7	6	rexbii	⊢ ( ∃ 𝑦 ∈ 𝑥 ( 𝑦 ∩ 𝑥 ) = ∅ ↔ ∃ 𝑦 ∈ 𝑥 ( 𝑥 ∩ 𝑦 ) = ∅ )
8	4 7	sylib	⊢ ( 𝑥 ≠ ∅ → ∃ 𝑦 ∈ 𝑥 ( 𝑥 ∩ 𝑦 ) = ∅ )
9	8	adantl	⊢ ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ( 𝑥 ∩ 𝑦 ) = ∅ )
10	1 9	mpgbir	⊢ E Fr 𝐴