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 𝐴𝑅 Fr 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqimss2 ( 𝐴 = 𝐵𝐵𝐴 )
2 frss ( 𝐵𝐴 → ( 𝑅 Fr 𝐴𝑅 Fr 𝐵 ) )
3 1 2 syl ( 𝐴 = 𝐵 → ( 𝑅 Fr 𝐴𝑅 Fr 𝐵 ) )
4 eqimss ( 𝐴 = 𝐵𝐴𝐵 )
5 frss ( 𝐴𝐵 → ( 𝑅 Fr 𝐵𝑅 Fr 𝐴 ) )
6 4 5 syl ( 𝐴 = 𝐵 → ( 𝑅 Fr 𝐵𝑅 Fr 𝐴 ) )
7 3 6 impbid ( 𝐴 = 𝐵 → ( 𝑅 Fr 𝐴𝑅 Fr 𝐵 ) )