Metamath Proof Explorer


Theorem scafeq

Description: If the scalar multiplication operation is already a function, the functionalization of it is equal to the original operation. (Contributed by Mario Carneiro, 5-Oct-2015)

Ref Expression
Hypotheses scaffval.b
|- B = ( Base ` W )
scaffval.f
|- F = ( Scalar ` W )
scaffval.k
|- K = ( Base ` F )
scaffval.a
|- .xb = ( .sf ` W )
scaffval.s
|- .x. = ( .s ` W )
Assertion scafeq
|- ( .x. Fn ( K X. B ) -> .xb = .x. )

Proof

Step Hyp Ref Expression
1 scaffval.b
 |-  B = ( Base ` W )
2 scaffval.f
 |-  F = ( Scalar ` W )
3 scaffval.k
 |-  K = ( Base ` F )
4 scaffval.a
 |-  .xb = ( .sf ` W )
5 scaffval.s
 |-  .x. = ( .s ` W )
6 1 2 3 4 5 scaffval
 |-  .xb = ( x e. K , y e. B |-> ( x .x. y ) )
7 fnov
 |-  ( .x. Fn ( K X. B ) <-> .x. = ( x e. K , y e. B |-> ( x .x. y ) ) )
8 7 biimpi
 |-  ( .x. Fn ( K X. B ) -> .x. = ( x e. K , y e. B |-> ( x .x. y ) ) )
9 6 8 eqtr4id
 |-  ( .x. Fn ( K X. B ) -> .xb = .x. )