Metamath Proof Explorer


Theorem sectfval

Description: Value of the section relation. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses issect.b B=BaseC
issect.h H=HomC
issect.o ·˙=compC
issect.i 1˙=IdC
issect.s S=SectC
issect.c φCCat
issect.x φXB
issect.y φYB
Assertion sectfval φXSY=fg|fXHYgYHXgXY·˙Xf=1˙X

Proof

Step Hyp Ref Expression
1 issect.b B=BaseC
2 issect.h H=HomC
3 issect.o ·˙=compC
4 issect.i 1˙=IdC
5 issect.s S=SectC
6 issect.c φCCat
7 issect.x φXB
8 issect.y φYB
9 1 2 3 4 5 6 7 7 sectffval φS=xB,yBfg|fxHygyHxgxy·˙xf=1˙x
10 simprl φx=Xy=Yx=X
11 simprr φx=Xy=Yy=Y
12 10 11 oveq12d φx=Xy=YxHy=XHY
13 12 eleq2d φx=Xy=YfxHyfXHY
14 11 10 oveq12d φx=Xy=YyHx=YHX
15 14 eleq2d φx=Xy=YgyHxgYHX
16 13 15 anbi12d φx=Xy=YfxHygyHxfXHYgYHX
17 10 11 opeq12d φx=Xy=Yxy=XY
18 17 10 oveq12d φx=Xy=Yxy·˙x=XY·˙X
19 18 oveqd φx=Xy=Ygxy·˙xf=gXY·˙Xf
20 10 fveq2d φx=Xy=Y1˙x=1˙X
21 19 20 eqeq12d φx=Xy=Ygxy·˙xf=1˙xgXY·˙Xf=1˙X
22 16 21 anbi12d φx=Xy=YfxHygyHxgxy·˙xf=1˙xfXHYgYHXgXY·˙Xf=1˙X
23 22 opabbidv φx=Xy=Yfg|fxHygyHxgxy·˙xf=1˙x=fg|fXHYgYHXgXY·˙Xf=1˙X
24 ovex XHYV
25 ovex YHXV
26 24 25 xpex XHY×YHXV
27 opabssxp fg|fXHYgYHXgXY·˙Xf=1˙XXHY×YHX
28 26 27 ssexi fg|fXHYgYHXgXY·˙Xf=1˙XV
29 28 a1i φfg|fXHYgYHXgXY·˙Xf=1˙XV
30 9 23 7 8 29 ovmpod φXSY=fg|fXHYgYHXgXY·˙Xf=1˙X