Metamath Proof Explorer


Theorem sseqin2

Description: A relationship between subclass and intersection. Similar to Exercise 9 of TakeutiZaring p. 18. (Contributed by NM, 17-May-1994)

Ref Expression
Assertion sseqin2 ⊢ A ⊆ B ↔ B ∩ A = A

Proof

Step Hyp Ref Expression
1 dfss2 ⊢ A ⊆ B ↔ A ∩ B = A
2 ineqcom ⊢ A ∩ B = A ↔ B ∩ A = A
3 1 2 bitri ⊢ A ⊆ B ↔ B ∩ A = A