Metamath Proof Explorer


Theorem seeq1

Description: Equality theorem for the set-like predicate. (Contributed by Mario Carneiro, 24-Jun-2015)

Ref Expression
Assertion seeq1 ( 𝑅 = 𝑆 → ( 𝑅 Se 𝐴 ↔ 𝑆 Se 𝐴 ) )

Proof

Step Hyp Ref Expression
1 eqimss2 ⊢ ( 𝑅 = 𝑆 → 𝑆 ⊆ 𝑅 )
2 sess1 ⊢ ( 𝑆 ⊆ 𝑅 → ( 𝑅 Se 𝐴 → 𝑆 Se 𝐴 ) )
3 1 2 syl ⊢ ( 𝑅 = 𝑆 → ( 𝑅 Se 𝐴 → 𝑆 Se 𝐴 ) )
4 eqimss ⊢ ( 𝑅 = 𝑆 → 𝑅 ⊆ 𝑆 )
5 sess1 ⊢ ( 𝑅 ⊆ 𝑆 → ( 𝑆 Se 𝐴 → 𝑅 Se 𝐴 ) )
6 4 5 syl ⊢ ( 𝑅 = 𝑆 → ( 𝑆 Se 𝐴 → 𝑅 Se 𝐴 ) )
7 3 6 impbid ⊢ ( 𝑅 = 𝑆 → ( 𝑅 Se 𝐴 ↔ 𝑆 Se 𝐴 ) )