Metamath Proof Explorer


Theorem releq

Description: Equality theorem for the relation predicate. (Contributed by NM, 1-Aug-1994)

Ref Expression
Assertion releq ( 𝐴 = 𝐵 → ( Rel 𝐴 ↔ Rel 𝐵 ) )

Proof

Step Hyp Ref Expression
1 sseq1 ( 𝐴 = 𝐵 → ( 𝐴 ⊆ ( V × V ) ↔ 𝐵 ⊆ ( V × V ) ) )
2 df-rel ( Rel 𝐴𝐴 ⊆ ( V × V ) )
3 df-rel ( Rel 𝐵𝐵 ⊆ ( V × V ) )
4 1 2 3 3bitr4g ( 𝐴 = 𝐵 → ( Rel 𝐴 ↔ Rel 𝐵 ) )