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Metamath Proof Explorer

Theorem unab

Description: Union of two class abstractions. (Contributed by NM, 29-Sep-2002) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Assertion unab 
|- ( { x | ph } u. { x | ps } ) = { x | ( ph \/ ps ) }

Proof

Step Hyp Ref Expression
1 sbor 
 |-  ( [ y / x ] ( ph \/ ps ) <-> ( [ y / x ] ph \/ [ y / x ] ps ) )
2 df-clab 
 |-  ( y e. { x | ( ph \/ ps ) } <-> [ y / x ] ( ph \/ ps ) )
3 df-clab 
 |-  ( y e. { x | ph } <-> [ y / x ] ph )
4 df-clab 
 |-  ( y e. { x | ps } <-> [ y / x ] ps )
5 3 4 orbi12i 
 |-  ( ( y e. { x | ph } \/ y e. { x | ps } ) <-> ( [ y / x ] ph \/ [ y / x ] ps ) )
6 1 2 5 3bitr4ri 
 |-  ( ( y e. { x | ph } \/ y e. { x | ps } ) <-> y e. { x | ( ph \/ ps ) } )
7 6 uneqri 
 |-  ( { x | ph } u. { x | ps } ) = { x | ( ph \/ ps ) }