Metamath Proof Explorer


Theorem dmdbr3

Description: Binary relation expressing the dual modular pair property. This version quantifies an equality instead of an inference. (Contributed by NM, 6-Jul-2004) (New usage is discouraged.)

Ref Expression
Assertion dmdbr3 A C B C A 𝑀 * B x C x B A B = x B A B

Proof

Step Hyp Ref Expression
1 dmdbr A C B C A 𝑀 * B y C B y y A B = y A B
2 chub2 B C x C B x B
3 2 ancoms x C B C B x B
4 chjcl x C B C x B C
5 sseq2 y = x B B y B x B
6 ineq1 y = x B y A = x B A
7 6 oveq1d y = x B y A B = x B A B
8 ineq1 y = x B y A B = x B A B
9 7 8 eqeq12d y = x B y A B = y A B x B A B = x B A B
10 5 9 imbi12d y = x B B y y A B = y A B B x B x B A B = x B A B
11 10 rspcv x B C y C B y y A B = y A B B x B x B A B = x B A B
12 4 11 syl x C B C y C B y y A B = y A B B x B x B A B = x B A B
13 3 12 mpid x C B C y C B y y A B = y A B x B A B = x B A B
14 13 ex x C B C y C B y y A B = y A B x B A B = x B A B
15 14 com3l B C y C B y y A B = y A B x C x B A B = x B A B
16 15 ralrimdv B C y C B y y A B = y A B x C x B A B = x B A B
17 chlejb2 B C x C B x x B = x
18 17 biimpa B C x C B x x B = x
19 18 ineq1d B C x C B x x B A = x A
20 19 oveq1d B C x C B x x B A B = x A B
21 18 ineq1d B C x C B x x B A B = x A B
22 20 21 eqeq12d B C x C B x x B A B = x B A B x A B = x A B
23 22 biimpd B C x C B x x B A B = x B A B x A B = x A B
24 23 ex B C x C B x x B A B = x B A B x A B = x A B
25 24 com23 B C x C x B A B = x B A B B x x A B = x A B
26 25 ralimdva B C x C x B A B = x B A B x C B x x A B = x A B
27 sseq2 x = y B x B y
28 ineq1 x = y x A = y A
29 28 oveq1d x = y x A B = y A B
30 ineq1 x = y x A B = y A B
31 29 30 eqeq12d x = y x A B = x A B y A B = y A B
32 27 31 imbi12d x = y B x x A B = x A B B y y A B = y A B
33 32 cbvralvw x C B x x A B = x A B y C B y y A B = y A B
34 26 33 syl6ib B C x C x B A B = x B A B y C B y y A B = y A B
35 16 34 impbid B C y C B y y A B = y A B x C x B A B = x B A B
36 35 adantl A C B C y C B y y A B = y A B x C x B A B = x B A B
37 1 36 bitrd A C B C A 𝑀 * B x C x B A B = x B A B