Metamath Proof Explorer


Theorem sectpropd

Description: Two structures with the same base, hom-sets and composition operation have the same sections. (Contributed by Zhi Wang, 27-Oct-2025)

Ref Expression
Hypotheses sectpropd.1 φ Hom 𝑓 C = Hom 𝑓 D
sectpropd.2 φ comp 𝑓 C = comp 𝑓 D
Assertion sectpropd φ Sect C = Sect D

Proof

Step Hyp Ref Expression
1 sectpropd.1 φ Hom 𝑓 C = Hom 𝑓 D
2 sectpropd.2 φ comp 𝑓 C = comp 𝑓 D
3 1 2 sectpropdlem φ f Sect C f Sect D
4 1 eqcomd φ Hom 𝑓 D = Hom 𝑓 C
5 2 eqcomd φ comp 𝑓 D = comp 𝑓 C
6 4 5 sectpropdlem φ f Sect D f Sect C
7 3 6 impbida φ f Sect C f Sect D
8 7 eqrdv φ Sect C = Sect D