Metamath Proof Explorer

Theorem riinint

Description: Express a relative indexed intersection as an intersection. (Contributed by Stefan O'Rear, 22-Feb-2015)

Ref Expression
Assertion riinint 
|- ( ( X e. V /\ A. k e. I S C_ X ) -> ( X i^i |^|_ k e. I S ) = |^| ( { X } u. ran ( k e. I |-> S ) ) )

Proof

Step Hyp Ref Expression
1 ssexg 
 |-  ( ( S C_ X /\ X e. V ) -> S e. _V )
2 1 expcom 
 |-  ( X e. V -> ( S C_ X -> S e. _V ) )
3 2 ralimdv 
 |-  ( X e. V -> ( A. k e. I S C_ X -> A. k e. I S e. _V ) )
4 3 imp 
 |-  ( ( X e. V /\ A. k e. I S C_ X ) -> A. k e. I S e. _V )
5 dfiin3g 
 |-  ( A. k e. I S e. _V -> |^|_ k e. I S = |^| ran ( k e. I |-> S ) )
6 4 5 syl 
 |-  ( ( X e. V /\ A. k e. I S C_ X ) -> |^|_ k e. I S = |^| ran ( k e. I |-> S ) )
7 6 ineq2d 
 |-  ( ( X e. V /\ A. k e. I S C_ X ) -> ( X i^i |^|_ k e. I S ) = ( X i^i |^| ran ( k e. I |-> S ) ) )
8 intun 
 |-  |^| ( { X } u. ran ( k e. I |-> S ) ) = ( |^| { X } i^i |^| ran ( k e. I |-> S ) )
9 intsng 
 |-  ( X e. V -> |^| { X } = X )
10 9 adantr 
 |-  ( ( X e. V /\ A. k e. I S C_ X ) -> |^| { X } = X )
11 10 ineq1d 
 |-  ( ( X e. V /\ A. k e. I S C_ X ) -> ( |^| { X } i^i |^| ran ( k e. I |-> S ) ) = ( X i^i |^| ran ( k e. I |-> S ) ) )
12 8 11 eqtrid 
 |-  ( ( X e. V /\ A. k e. I S C_ X ) -> |^| ( { X } u. ran ( k e. I |-> S ) ) = ( X i^i |^| ran ( k e. I |-> S ) ) )
13 7 12 eqtr4d 
 |-  ( ( X e. V /\ A. k e. I S C_ X ) -> ( X i^i |^|_ k e. I S ) = |^| ( { X } u. ran ( k e. I |-> S ) ) )