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 ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 )
fcoss.c ⊢ ( 𝜑 → 𝐶 ⊆ 𝐴 )
fcoss.g ⊢ ( 𝜑 → 𝐺 : 𝐷 ⟶ 𝐶 )
Assertion fcoss ( 𝜑 → ( 𝐹 ∘ 𝐺 ) : 𝐷 ⟶ 𝐵 )

Proof

Step Hyp Ref Expression
1 fcoss.f ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 )
2 fcoss.c ⊢ ( 𝜑 → 𝐶 ⊆ 𝐴 )
3 fcoss.g ⊢ ( 𝜑 → 𝐺 : 𝐷 ⟶ 𝐶 )
4 3 2 fssd ⊢ ( 𝜑 → 𝐺 : 𝐷 ⟶ 𝐴 )
5 fco ⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐺 : 𝐷 ⟶ 𝐴 ) → ( 𝐹 ∘ 𝐺 ) : 𝐷 ⟶ 𝐵 )
6 1 4 5 syl2anc ⊢ ( 𝜑 → ( 𝐹 ∘ 𝐺 ) : 𝐷 ⟶ 𝐵 )