# Metamath Proof Explorer

## Theorem fcompt

Description: Express composition of two functions as a maps-to applying both in sequence. (Contributed by Stefan O'Rear, 5-Oct-2014) (Proof shortened by Mario Carneiro, 27-Dec-2014)

Ref Expression
Assertion fcompt
`|- ( ( A : D --> E /\ B : C --> D ) -> ( A o. B ) = ( x e. C |-> ( A ` ( B ` x ) ) ) )`

### Proof

Step Hyp Ref Expression
1 ffvelrn
` |-  ( ( B : C --> D /\ x e. C ) -> ( B ` x ) e. D )`
` |-  ( ( ( A : D --> E /\ B : C --> D ) /\ x e. C ) -> ( B ` x ) e. D )`
3 ffn
` |-  ( B : C --> D -> B Fn C )`
` |-  ( ( A : D --> E /\ B : C --> D ) -> B Fn C )`
5 dffn5
` |-  ( B Fn C <-> B = ( x e. C |-> ( B ` x ) ) )`
6 4 5 sylib
` |-  ( ( A : D --> E /\ B : C --> D ) -> B = ( x e. C |-> ( B ` x ) ) )`
7 ffn
` |-  ( A : D --> E -> A Fn D )`
` |-  ( ( A : D --> E /\ B : C --> D ) -> A Fn D )`
` |-  ( A Fn D <-> A = ( y e. D |-> ( A ` y ) ) )`
` |-  ( ( A : D --> E /\ B : C --> D ) -> A = ( y e. D |-> ( A ` y ) ) )`
` |-  ( y = ( B ` x ) -> ( A ` y ) = ( A ` ( B ` x ) ) )`
` |-  ( ( A : D --> E /\ B : C --> D ) -> ( A o. B ) = ( x e. C |-> ( A ` ( B ` x ) ) ) )`