Metamath Proof Explorer


Theorem riin0

Description: Relative intersection of an empty family. (Contributed by Stefan O'Rear, 3-Apr-2015)

Ref Expression
Assertion riin0
|- ( X = (/) -> ( A i^i |^|_ x e. X S ) = A )

Proof

Step Hyp Ref Expression
1 iineq1
 |-  ( X = (/) -> |^|_ x e. X S = |^|_ x e. (/) S )
2 1 ineq2d
 |-  ( X = (/) -> ( A i^i |^|_ x e. X S ) = ( A i^i |^|_ x e. (/) S ) )
3 0iin
 |-  |^|_ x e. (/) S = _V
4 3 ineq2i
 |-  ( A i^i |^|_ x e. (/) S ) = ( A i^i _V )
5 inv1
 |-  ( A i^i _V ) = A
6 4 5 eqtri
 |-  ( A i^i |^|_ x e. (/) S ) = A
7 2 6 eqtrdi
 |-  ( X = (/) -> ( A i^i |^|_ x e. X S ) = A )