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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 dfss2 
 |-  ( A C_ B <-> ( A i^i B ) = A )
2 ineqcom 
 |-  ( ( A i^i B ) = A <-> ( B i^i A ) = A )
3 1 2 bitri 
 |-  ( A C_ B <-> ( B i^i A ) = A )