Metamath Proof Explorer


Theorem brco3f1o

Description: Conditions allowing the decomposition of a binary relation. (Contributed by RP, 8-Jun-2021)

Ref Expression
Hypotheses brco3f1o.c φ C : Y 1-1 onto Z
brco3f1o.d φ D : X 1-1 onto Y
brco3f1o.e φ E : W 1-1 onto X
brco3f1o.r φ A C D E B
Assertion brco3f1o φ C -1 B C B D -1 C -1 B D C -1 B A E D -1 C -1 B

Proof

Step Hyp Ref Expression
1 brco3f1o.c φ C : Y 1-1 onto Z
2 brco3f1o.d φ D : X 1-1 onto Y
3 brco3f1o.e φ E : W 1-1 onto X
4 brco3f1o.r φ A C D E B
5 f1ocnv E : W 1-1 onto X E -1 : X 1-1 onto W
6 f1ofn E -1 : X 1-1 onto W E -1 Fn X
7 3 5 6 3syl φ E -1 Fn X
8 f1ocnv D : X 1-1 onto Y D -1 : Y 1-1 onto X
9 f1of D -1 : Y 1-1 onto X D -1 : Y X
10 2 8 9 3syl φ D -1 : Y X
11 f1ocnv C : Y 1-1 onto Z C -1 : Z 1-1 onto Y
12 f1of C -1 : Z 1-1 onto Y C -1 : Z Y
13 1 11 12 3syl φ C -1 : Z Y
14 relco Rel C D E
15 14 relbrcnv B C D E -1 A A C D E B
16 cnvco C D E -1 = E -1 C D -1
17 cnvco C D -1 = D -1 C -1
18 17 coeq2i E -1 C D -1 = E -1 D -1 C -1
19 16 18 eqtri C D E -1 = E -1 D -1 C -1
20 19 breqi B C D E -1 A B E -1 D -1 C -1 A
21 coass C D E = C D E
22 21 breqi A C D E B A C D E B
23 15 20 22 3bitr3ri A C D E B B E -1 D -1 C -1 A
24 4 23 sylib φ B E -1 D -1 C -1 A
25 7 10 13 24 brcofffn φ B C -1 C -1 B C -1 B D -1 D -1 C -1 B D -1 C -1 B E -1 A
26 f1orel C : Y 1-1 onto Z Rel C
27 relbrcnvg Rel C B C -1 C -1 B C -1 B C B
28 1 26 27 3syl φ B C -1 C -1 B C -1 B C B
29 f1orel D : X 1-1 onto Y Rel D
30 relbrcnvg Rel D C -1 B D -1 D -1 C -1 B D -1 C -1 B D C -1 B
31 2 29 30 3syl φ C -1 B D -1 D -1 C -1 B D -1 C -1 B D C -1 B
32 f1orel E : W 1-1 onto X Rel E
33 relbrcnvg Rel E D -1 C -1 B E -1 A A E D -1 C -1 B
34 3 32 33 3syl φ D -1 C -1 B E -1 A A E D -1 C -1 B
35 28 31 34 3anbi123d φ B C -1 C -1 B C -1 B D -1 D -1 C -1 B D -1 C -1 B E -1 A C -1 B C B D -1 C -1 B D C -1 B A E D -1 C -1 B
36 25 35 mpbid φ C -1 B C B D -1 C -1 B D C -1 B A E D -1 C -1 B