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