Metamath Proof Explorer


Theorem freq2

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 3-Apr-1994)

Ref Expression
Assertion freq2 ⊢ A = B → R Fr A ↔ R Fr B

Proof

Step Hyp Ref Expression
1 eqimss2 ⊢ A = B → B ⊆ A
2 frss ⊢ B ⊆ A → R Fr A → R Fr B
3 1 2 syl ⊢ A = B → R Fr A → R Fr B
4 eqimss ⊢ A = B → A ⊆ B
5 frss ⊢ A ⊆ B → R Fr B → R Fr A
6 4 5 syl ⊢ A = B → R Fr B → R Fr A
7 3 6 impbid ⊢ A = B → R Fr A ↔ R Fr B