Metamath Proof Explorer


Theorem ofcof

Description: Relate function operation with operation with a constant. (Contributed by Thierry Arnoux, 3-Oct-2018)

Ref Expression
Hypotheses ofcof.1
|- ( ph -> F : A --> B )
ofcof.2
|- ( ph -> A e. V )
ofcof.3
|- ( ph -> C e. W )
Assertion ofcof
|- ( ph -> ( F oFC R C ) = ( F oF R ( A X. { C } ) ) )

Proof

Step Hyp Ref Expression
1 ofcof.1
 |-  ( ph -> F : A --> B )
2 ofcof.2
 |-  ( ph -> A e. V )
3 ofcof.3
 |-  ( ph -> C e. W )
4 1 ffnd
 |-  ( ph -> F Fn A )
5 eqidd
 |-  ( ( ph /\ x e. A ) -> ( F ` x ) = ( F ` x ) )
6 4 2 3 5 ofcfval
 |-  ( ph -> ( F oFC R C ) = ( x e. A |-> ( ( F ` x ) R C ) ) )
7 fnconstg
 |-  ( C e. W -> ( A X. { C } ) Fn A )
8 3 7 syl
 |-  ( ph -> ( A X. { C } ) Fn A )
9 inidm
 |-  ( A i^i A ) = A
10 fvconst2g
 |-  ( ( C e. W /\ x e. A ) -> ( ( A X. { C } ) ` x ) = C )
11 3 10 sylan
 |-  ( ( ph /\ x e. A ) -> ( ( A X. { C } ) ` x ) = C )
12 4 8 2 2 9 5 11 offval
 |-  ( ph -> ( F oF R ( A X. { C } ) ) = ( x e. A |-> ( ( F ` x ) R C ) ) )
13 6 12 eqtr4d
 |-  ( ph -> ( F oFC R C ) = ( F oF R ( A X. { C } ) ) )