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Metamath Proof Explorer

Theorem rabrab

Description: Abstract builder restricted to another restricted abstract builder. (Contributed by Thierry Arnoux, 30-Aug-2017)

Ref Expression
Assertion rabrab 
|- { x e. { x e. A | ph } | ps } = { x e. A | ( ph /\ ps ) }

Proof

Step Hyp Ref Expression
1 rabid 
 |-  ( x e. { x e. A | ph } <-> ( x e. A /\ ph ) )
2 1 anbi1i 
 |-  ( ( x e. { x e. A | ph } /\ ps ) <-> ( ( x e. A /\ ph ) /\ ps ) )
3 anass 
 |-  ( ( ( x e. A /\ ph ) /\ ps ) <-> ( x e. A /\ ( ph /\ ps ) ) )
4 2 3 bitri 
 |-  ( ( x e. { x e. A | ph } /\ ps ) <-> ( x e. A /\ ( ph /\ ps ) ) )
5 4 abbii 
 |-  { x | ( x e. { x e. A | ph } /\ ps ) } = { x | ( x e. A /\ ( ph /\ ps ) ) }
6 df-rab 
 |-  { x e. { x e. A | ph } | ps } = { x | ( x e. { x e. A | ph } /\ ps ) }
7 df-rab 
 |-  { x e. A | ( ph /\ ps ) } = { x | ( x e. A /\ ( ph /\ ps ) ) }
8 5 6 7 3eqtr4i 
 |-  { x e. { x e. A | ph } | ps } = { x e. A | ( ph /\ ps ) }