Metamath Proof Explorer


Theorem 2oppccomf

Description: The double opposite category has the same composition as the original category. Intended for use with property lemmas such as monpropd . (Contributed by Mario Carneiro, 3-Jan-2017)

Ref Expression
Hypothesis oppcbas.1 O = oppCat C
Assertion 2oppccomf comp 𝑓 C = comp 𝑓 oppCat O

Proof

Step Hyp Ref Expression
1 oppcbas.1 O = oppCat C
2 eqid Base C = Base C
3 1 2 oppcbas Base C = Base O
4 eqid comp O = comp O
5 eqid oppCat O = oppCat O
6 simpr1 x Base C y Base C z Base C x Base C
7 simpr2 x Base C y Base C z Base C y Base C
8 simpr3 x Base C y Base C z Base C z Base C
9 3 4 5 6 7 8 oppcco x Base C y Base C z Base C g x y comp oppCat O z f = f z y comp O x g
10 eqid comp C = comp C
11 2 10 1 8 7 6 oppcco x Base C y Base C z Base C f z y comp O x g = g x y comp C z f
12 9 11 eqtr2d x Base C y Base C z Base C g x y comp C z f = g x y comp oppCat O z f
13 12 ralrimivw x Base C y Base C z Base C g y Hom C z g x y comp C z f = g x y comp oppCat O z f
14 13 ralrimivw x Base C y Base C z Base C f x Hom C y g y Hom C z g x y comp C z f = g x y comp oppCat O z f
15 14 ralrimivvva x Base C y Base C z Base C f x Hom C y g y Hom C z g x y comp C z f = g x y comp oppCat O z f
16 eqid comp oppCat O = comp oppCat O
17 eqid Hom C = Hom C
18 eqidd Base C = Base C
19 1 2 2oppcbas Base C = Base oppCat O
20 19 a1i Base C = Base oppCat O
21 1 2oppchomf Hom 𝑓 C = Hom 𝑓 oppCat O
22 21 a1i Hom 𝑓 C = Hom 𝑓 oppCat O
23 10 16 17 18 20 22 comfeq comp 𝑓 C = comp 𝑓 oppCat O x Base C y Base C z Base C f x Hom C y g y Hom C z g x y comp C z f = g x y comp oppCat O z f
24 15 23 mpbird comp 𝑓 C = comp 𝑓 oppCat O
25 24 mptru comp 𝑓 C = comp 𝑓 oppCat O