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 C_ A )
2 frss 
 |-  ( B C_ A -> ( R Fr A -> R Fr B ) )
3 1 2 syl 
 |-  ( A = B -> ( R Fr A -> R Fr B ) )
4 eqimss 
 |-  ( A = B -> A C_ B )
5 frss 
 |-  ( A C_ 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 ) )