Metamath Proof Explorer


Theorem asclval

Description: Value of a mapped algebra scalar. (Contributed by Mario Carneiro, 8-Mar-2015)

Ref Expression
Hypotheses asclfval.a
|- A = ( algSc ` W )
asclfval.f
|- F = ( Scalar ` W )
asclfval.k
|- K = ( Base ` F )
asclfval.s
|- .x. = ( .s ` W )
asclfval.o
|- .1. = ( 1r ` W )
Assertion asclval
|- ( X e. K -> ( A ` X ) = ( X .x. .1. ) )

Proof

Step Hyp Ref Expression
1 asclfval.a
 |-  A = ( algSc ` W )
2 asclfval.f
 |-  F = ( Scalar ` W )
3 asclfval.k
 |-  K = ( Base ` F )
4 asclfval.s
 |-  .x. = ( .s ` W )
5 asclfval.o
 |-  .1. = ( 1r ` W )
6 oveq1
 |-  ( x = X -> ( x .x. .1. ) = ( X .x. .1. ) )
7 1 2 3 4 5 asclfval
 |-  A = ( x e. K |-> ( x .x. .1. ) )
8 ovex
 |-  ( X .x. .1. ) e. _V
9 6 7 8 fvmpt
 |-  ( X e. K -> ( A ` X ) = ( X .x. .1. ) )