Metamath Proof Explorer


Theorem f1otrgitv

Description: Convenient lemma for f1otrg . (Contributed by Thierry Arnoux, 19-Mar-2019)

Ref Expression
Hypotheses f1otrkg.p P = Base G
f1otrkg.d D = dist G
f1otrkg.i I = Itv G
f1otrkg.b B = Base H
f1otrkg.e E = dist H
f1otrkg.j J = Itv H
f1otrkg.f φ F : B 1-1 onto P
f1otrkg.1 φ e B f B e E f = F e D F f
f1otrkg.2 φ e B f B g B g e J f F g F e I F f
f1otrgitv.x φ X B
f1otrgitv.y φ Y B
f1otrgitv.z φ Z B
Assertion f1otrgitv φ Z X J Y F Z F X I F Y

Proof

Step Hyp Ref Expression
1 f1otrkg.p P = Base G
2 f1otrkg.d D = dist G
3 f1otrkg.i I = Itv G
4 f1otrkg.b B = Base H
5 f1otrkg.e E = dist H
6 f1otrkg.j J = Itv H
7 f1otrkg.f φ F : B 1-1 onto P
8 f1otrkg.1 φ e B f B e E f = F e D F f
9 f1otrkg.2 φ e B f B g B g e J f F g F e I F f
10 f1otrgitv.x φ X B
11 f1otrgitv.y φ Y B
12 f1otrgitv.z φ Z B
13 9 ralrimivvva φ e B f B g B g e J f F g F e I F f
14 oveq1 e = X e J f = X J f
15 14 eleq2d e = X g e J f g X J f
16 fveq2 e = X F e = F X
17 16 oveq1d e = X F e I F f = F X I F f
18 17 eleq2d e = X F g F e I F f F g F X I F f
19 15 18 bibi12d e = X g e J f F g F e I F f g X J f F g F X I F f
20 oveq2 f = Y X J f = X J Y
21 20 eleq2d f = Y g X J f g X J Y
22 fveq2 f = Y F f = F Y
23 22 oveq2d f = Y F X I F f = F X I F Y
24 23 eleq2d f = Y F g F X I F f F g F X I F Y
25 21 24 bibi12d f = Y g X J f F g F X I F f g X J Y F g F X I F Y
26 eleq1 g = Z g X J Y Z X J Y
27 fveq2 g = Z F g = F Z
28 27 eleq1d g = Z F g F X I F Y F Z F X I F Y
29 26 28 bibi12d g = Z g X J Y F g F X I F Y Z X J Y F Z F X I F Y
30 19 25 29 rspc3v X B Y B Z B e B f B g B g e J f F g F e I F f Z X J Y F Z F X I F Y
31 10 11 12 30 syl3anc φ e B f B g B g e J f F g F e I F f Z X J Y F Z F X I F Y
32 13 31 mpd φ Z X J Y F Z F X I F Y