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Metamath Proof Explorer

Theorem reldisj

Description: Two ways of saying that two classes are disjoint, using the complement of B relative to a universe C . (Contributed by NM, 15-Feb-2007) (Proof shortened by Andrew Salmon, 26-Jun-2011) Avoid ax-12 . (Revised by GG, 28-Jun-2024)

Ref Expression
Assertion reldisj 
|- ( A C_ C -> ( ( A i^i B ) = (/) <-> A C_ ( C \ B ) ) )

Proof

Step Hyp Ref Expression
1 df-ss 
 |-  ( A C_ C <-> A. x ( x e. A -> x e. C ) )
2 eleq1w 
 |-  ( x = y -> ( x e. A <-> y e. A ) )
3 eleq1w 
 |-  ( x = y -> ( x e. C <-> y e. C ) )
4 2 3 imbi12d 
 |-  ( x = y -> ( ( x e. A -> x e. C ) <-> ( y e. A -> y e. C ) ) )
5 4 spw 
 |-  ( A. x ( x e. A -> x e. C ) -> ( x e. A -> x e. C ) )
6 pm5.44 
 |-  ( ( x e. A -> x e. C ) -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> ( x e. C /\ -. x e. B ) ) ) )
7 eldif 
 |-  ( x e. ( C \ B ) <-> ( x e. C /\ -. x e. B ) )
8 7 imbi2i 
 |-  ( ( x e. A -> x e. ( C \ B ) ) <-> ( x e. A -> ( x e. C /\ -. x e. B ) ) )
9 6 8 bitr4di 
 |-  ( ( x e. A -> x e. C ) -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> x e. ( C \ B ) ) ) )
10 5 9 syl 
 |-  ( A. x ( x e. A -> x e. C ) -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> x e. ( C \ B ) ) ) )
11 1 10 sylbi 
 |-  ( A C_ C -> ( ( x e. A -> -. x e. B ) <-> ( x e. A -> x e. ( C \ B ) ) ) )
12 11 albidv 
 |-  ( A C_ C -> ( A. x ( x e. A -> -. x e. B ) <-> A. x ( x e. A -> x e. ( C \ B ) ) ) )
13 disj1 
 |-  ( ( A i^i B ) = (/) <-> A. x ( x e. A -> -. x e. B ) )
14 df-ss 
 |-  ( A C_ ( C \ B ) <-> A. x ( x e. A -> x e. ( C \ B ) ) )
15 12 13 14 3bitr4g 
 |-  ( A C_ C -> ( ( A i^i B ) = (/) <-> A C_ ( C \ B ) ) )