Metamath Proof Explorer


Theorem f1otrgds

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
Assertion f1otrgds φ X E Y = F X D 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 8 ralrimivva φ e B f B e E f = F e D F f
13 oveq1 e = X e E f = X E f
14 fveq2 e = X F e = F X
15 14 oveq1d e = X F e D F f = F X D F f
16 13 15 eqeq12d e = X e E f = F e D F f X E f = F X D F f
17 oveq2 f = Y X E f = X E Y
18 fveq2 f = Y F f = F Y
19 18 oveq2d f = Y F X D F f = F X D F Y
20 17 19 eqeq12d f = Y X E f = F X D F f X E Y = F X D F Y
21 16 20 rspc2v X B Y B e B f B e E f = F e D F f X E Y = F X D F Y
22 10 11 21 syl2anc φ e B f B e E f = F e D F f X E Y = F X D F Y
23 12 22 mpd φ X E Y = F X D F Y