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 C_ B <-> ( B i^i A ) = A )

Proof

Step Hyp Ref Expression
1 df-ss 
 |-  ( A C_ B <-> ( A i^i B ) = A )
2 incom 
 |-  ( A i^i B ) = ( B i^i A )
3 2 eqeq1i 
 |-  ( ( A i^i B ) = A <-> ( B i^i A ) = A )
4 1 3 bitri 
 |-  ( A C_ B <-> ( B i^i A ) = A )