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 ( 𝐴𝐵 ↔ ( 𝐵𝐴 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 df-ss ( 𝐴𝐵 ↔ ( 𝐴𝐵 ) = 𝐴 )
2 incom ( 𝐴𝐵 ) = ( 𝐵𝐴 )
3 2 eqeq1i ( ( 𝐴𝐵 ) = 𝐴 ↔ ( 𝐵𝐴 ) = 𝐴 )
4 1 3 bitri ( 𝐴𝐵 ↔ ( 𝐵𝐴 ) = 𝐴 )