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Metamath Proof Explorer

Theorem ssequn1

Description: A relationship between subclass and union. Theorem 26 of Suppes p. 27. (Contributed by NM, 30-Aug-1993) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Assertion ssequn1 
|- ( A C_ B <-> ( A u. B ) = B )

Proof

Step Hyp Ref Expression
1 bicom 
 |-  ( ( x e. B <-> ( x e. A \/ x e. B ) ) <-> ( ( x e. A \/ x e. B ) <-> x e. B ) )
2 pm4.72 
 |-  ( ( x e. A -> x e. B ) <-> ( x e. B <-> ( x e. A \/ x e. B ) ) )
3 elun 
 |-  ( x e. ( A u. B ) <-> ( x e. A \/ x e. B ) )
4 3 bibi1i 
 |-  ( ( x e. ( A u. B ) <-> x e. B ) <-> ( ( x e. A \/ x e. B ) <-> x e. B ) )
5 1 2 4 3bitr4i 
 |-  ( ( x e. A -> x e. B ) <-> ( x e. ( A u. B ) <-> x e. B ) )
6 5 albii 
 |-  ( A. x ( x e. A -> x e. B ) <-> A. x ( x e. ( A u. B ) <-> x e. B ) )
7 df-ss 
 |-  ( A C_ B <-> A. x ( x e. A -> x e. B ) )
8 dfcleq 
 |-  ( ( A u. B ) = B <-> A. x ( x e. ( A u. B ) <-> x e. B ) )
9 6 7 8 3bitr4i 
 |-  ( A C_ B <-> ( A u. B ) = B )