Metamath Proof Explorer


Theorem fcoconst

Description: Composition with a constant function. (Contributed by Stefan O'Rear, 11-Mar-2015)

Ref Expression
Assertion fcoconst
|- ( ( F Fn X /\ Y e. X ) -> ( F o. ( I X. { Y } ) ) = ( I X. { ( F ` Y ) } ) )

Proof

Step Hyp Ref Expression
1 simplr
 |-  ( ( ( F Fn X /\ Y e. X ) /\ x e. I ) -> Y e. X )
2 fconstmpt
 |-  ( I X. { Y } ) = ( x e. I |-> Y )
3 2 a1i
 |-  ( ( F Fn X /\ Y e. X ) -> ( I X. { Y } ) = ( x e. I |-> Y ) )
4 dffn2
 |-  ( F Fn X <-> F : X --> _V )
5 4 birani
 |-  ( ( F Fn X /\ Y e. X ) -> F : X --> _V )
6 5 feqmptd
 |-  ( ( F Fn X /\ Y e. X ) -> F = ( y e. X |-> ( F ` y ) ) )
7 fveq2
 |-  ( y = Y -> ( F ` y ) = ( F ` Y ) )
8 1 3 6 7 fmptco
 |-  ( ( F Fn X /\ Y e. X ) -> ( F o. ( I X. { Y } ) ) = ( x e. I |-> ( F ` Y ) ) )
9 fconstmpt
 |-  ( I X. { ( F ` Y ) } ) = ( x e. I |-> ( F ` Y ) )
10 8 9 eqtr4di
 |-  ( ( F Fn X /\ Y e. X ) -> ( F o. ( I X. { Y } ) ) = ( I X. { ( F ` Y ) } ) )