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Metamath Proof Explorer

Theorem disjdifprg2

Description: A trivial partition of a set into its difference and intersection with another set. (Contributed by Thierry Arnoux, 25-Dec-2016)

Ref Expression
Assertion disjdifprg2 
|- ( A e. V -> Disj_ x e. { ( A \ B ) , ( A i^i B ) } x )

Proof

Step Hyp Ref Expression
1 inex1g 
 |-  ( A e. V -> ( A i^i B ) e. _V )
2 elex 
 |-  ( A e. V -> A e. _V )
3 disjdifprg 
 |-  ( ( ( A i^i B ) e. _V /\ A e. _V ) -> Disj_ x e. { ( A \ ( A i^i B ) ) , ( A i^i B ) } x )
4 1 2 3 syl2anc 
 |-  ( A e. V -> Disj_ x e. { ( A \ ( A i^i B ) ) , ( A i^i B ) } x )
5 difin 
 |-  ( A \ ( A i^i B ) ) = ( A \ B )
6 5 preq1i 
 |-  { ( A \ ( A i^i B ) ) , ( A i^i B ) } = { ( A \ B ) , ( A i^i B ) }
7 6 a1i 
 |-  ( A e. V -> { ( A \ ( A i^i B ) ) , ( A i^i B ) } = { ( A \ B ) , ( A i^i B ) } )
8 7 disjeq1d 
 |-  ( A e. V -> ( Disj_ x e. { ( A \ ( A i^i B ) ) , ( A i^i B ) } x <-> Disj_ x e. { ( A \ B ) , ( A i^i B ) } x ) )
9 4 8 mpbid 
 |-  ( A e. V -> Disj_ x e. { ( A \ B ) , ( A i^i B ) } x )