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 )