Metamath Proof Explorer


Theorem sectcan

Description: If G is a section of F and F is a section of H , then G = H . Proposition 3.10 of Adamek p. 28. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses sectcan.b B=BaseC
sectcan.s S=SectC
sectcan.c φCCat
sectcan.x φXB
sectcan.y φYB
sectcan.1 φGXSYF
sectcan.2 φFYSXH
Assertion sectcan φG=H

Proof

Step Hyp Ref Expression
1 sectcan.b B=BaseC
2 sectcan.s S=SectC
3 sectcan.c φCCat
4 sectcan.x φXB
5 sectcan.y φYB
6 sectcan.1 φGXSYF
7 sectcan.2 φFYSXH
8 eqid HomC=HomC
9 eqid compC=compC
10 eqid IdC=IdC
11 1 8 9 10 2 3 4 5 issect φGXSYFGXHomCYFYHomCXFXYcompCXG=IdCX
12 6 11 mpbid φGXHomCYFYHomCXFXYcompCXG=IdCX
13 12 simp1d φGXHomCY
14 1 8 9 10 2 3 5 4 issect φFYSXHFYHomCXHXHomCYHYXcompCYF=IdCY
15 7 14 mpbid φFYHomCXHXHomCYHYXcompCYF=IdCY
16 15 simp1d φFYHomCX
17 15 simp2d φHXHomCY
18 1 8 9 3 4 5 4 13 16 5 17 catass φHYXcompCYFXYcompCYG=HXXcompCYFXYcompCXG
19 15 simp3d φHYXcompCYF=IdCY
20 19 oveq1d φHYXcompCYFXYcompCYG=IdCYXYcompCYG
21 12 simp3d φFXYcompCXG=IdCX
22 21 oveq2d φHXXcompCYFXYcompCXG=HXXcompCYIdCX
23 18 20 22 3eqtr3d φIdCYXYcompCYG=HXXcompCYIdCX
24 1 8 10 3 4 9 5 13 catlid φIdCYXYcompCYG=G
25 1 8 10 3 4 9 5 17 catrid φHXXcompCYIdCX=H
26 23 24 25 3eqtr3d φG=H