Metamath Proof Explorer


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