Metamath Proof Explorer


Theorem seeq2

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

Ref Expression
Assertion seeq2 ( 𝐴 = 𝐵 → ( 𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqimss2 ⊢ ( 𝐴 = 𝐵 → 𝐵 ⊆ 𝐴 )
2 sess2 ⊢ ( 𝐵 ⊆ 𝐴 → ( 𝑅 Se 𝐴 → 𝑅 Se 𝐵 ) )
3 1 2 syl ⊢ ( 𝐴 = 𝐵 → ( 𝑅 Se 𝐴 → 𝑅 Se 𝐵 ) )
4 eqimss ⊢ ( 𝐴 = 𝐵 → 𝐴 ⊆ 𝐵 )
5 sess2 ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑅 Se 𝐵 → 𝑅 Se 𝐴 ) )
6 4 5 syl ⊢ ( 𝐴 = 𝐵 → ( 𝑅 Se 𝐵 → 𝑅 Se 𝐴 ) )
7 3 6 impbid ⊢ ( 𝐴 = 𝐵 → ( 𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵 ) )