Metamath Proof Explorer


Theorem fcoss

Description: Composition of two mappings. Similar to fco , but with a weaker condition on the domain of F . (Contributed by Glauco Siliprandi, 3-Mar-2021)

Ref Expression
Hypotheses fcoss.f
|- ( ph -> F : A --> B )
fcoss.c
|- ( ph -> C C_ A )
fcoss.g
|- ( ph -> G : D --> C )
Assertion fcoss
|- ( ph -> ( F o. G ) : D --> B )

Proof

Step Hyp Ref Expression
1 fcoss.f
 |-  ( ph -> F : A --> B )
2 fcoss.c
 |-  ( ph -> C C_ A )
3 fcoss.g
 |-  ( ph -> G : D --> C )
4 3 2 fssd
 |-  ( ph -> G : D --> A )
5 fco
 |-  ( ( F : A --> B /\ G : D --> A ) -> ( F o. G ) : D --> B )
6 1 4 5 syl2anc
 |-  ( ph -> ( F o. G ) : D --> B )