Metamath Proof Explorer


Theorem xpord3lem

Description: Lemma for triple ordering. Calculate the value of the relationship. (Contributed by Scott Fenton, 21-Aug-2024)

Ref Expression
Hypothesis xpord3.1 U = x y | x A × B × C y A × B × C 1 st 1 st x R 1 st 1 st y 1 st 1 st x = 1 st 1 st y 2 nd 1 st x S 2 nd 1 st y 2 nd 1 st x = 2 nd 1 st y 2 nd x T 2 nd y 2 nd x = 2 nd y x y
Assertion xpord3lem a b c U d e f a A b B c C d A e B f C a R d a = d b S e b = e c T f c = f a d b e c f

Proof

Step Hyp Ref Expression
1 xpord3.1 U = x y | x A × B × C y A × B × C 1 st 1 st x R 1 st 1 st y 1 st 1 st x = 1 st 1 st y 2 nd 1 st x S 2 nd 1 st y 2 nd 1 st x = 2 nd 1 st y 2 nd x T 2 nd y 2 nd x = 2 nd y x y
2 opex a b c V
3 opex d e f V
4 eleq1 x = a b c x A × B × C a b c A × B × C
5 vex a V
6 vex b V
7 vex c V
8 5 6 7 ot21std x = a b c 1 st 1 st x = a
9 8 breq1d x = a b c 1 st 1 st x R 1 st 1 st y a R 1 st 1 st y
10 8 eqeq1d x = a b c 1 st 1 st x = 1 st 1 st y a = 1 st 1 st y
11 9 10 orbi12d x = a b c 1 st 1 st x R 1 st 1 st y 1 st 1 st x = 1 st 1 st y a R 1 st 1 st y a = 1 st 1 st y
12 5 6 7 ot22ndd x = a b c 2 nd 1 st x = b
13 12 breq1d x = a b c 2 nd 1 st x S 2 nd 1 st y b S 2 nd 1 st y
14 12 eqeq1d x = a b c 2 nd 1 st x = 2 nd 1 st y b = 2 nd 1 st y
15 13 14 orbi12d x = a b c 2 nd 1 st x S 2 nd 1 st y 2 nd 1 st x = 2 nd 1 st y b S 2 nd 1 st y b = 2 nd 1 st y
16 opex a b V
17 16 7 op2ndd x = a b c 2 nd x = c
18 17 breq1d x = a b c 2 nd x T 2 nd y c T 2 nd y
19 17 eqeq1d x = a b c 2 nd x = 2 nd y c = 2 nd y
20 18 19 orbi12d x = a b c 2 nd x T 2 nd y 2 nd x = 2 nd y c T 2 nd y c = 2 nd y
21 11 15 20 3anbi123d x = a b c 1 st 1 st x R 1 st 1 st y 1 st 1 st x = 1 st 1 st y 2 nd 1 st x S 2 nd 1 st y 2 nd 1 st x = 2 nd 1 st y 2 nd x T 2 nd y 2 nd x = 2 nd y a R 1 st 1 st y a = 1 st 1 st y b S 2 nd 1 st y b = 2 nd 1 st y c T 2 nd y c = 2 nd y
22 neeq1 x = a b c x y a b c y
23 21 22 anbi12d x = a b c 1 st 1 st x R 1 st 1 st y 1 st 1 st x = 1 st 1 st y 2 nd 1 st x S 2 nd 1 st y 2 nd 1 st x = 2 nd 1 st y 2 nd x T 2 nd y 2 nd x = 2 nd y x y a R 1 st 1 st y a = 1 st 1 st y b S 2 nd 1 st y b = 2 nd 1 st y c T 2 nd y c = 2 nd y a b c y
24 4 23 3anbi13d x = a b c x A × B × C y A × B × C 1 st 1 st x R 1 st 1 st y 1 st 1 st x = 1 st 1 st y 2 nd 1 st x S 2 nd 1 st y 2 nd 1 st x = 2 nd 1 st y 2 nd x T 2 nd y 2 nd x = 2 nd y x y a b c A × B × C y A × B × C a R 1 st 1 st y a = 1 st 1 st y b S 2 nd 1 st y b = 2 nd 1 st y c T 2 nd y c = 2 nd y a b c y
25 eleq1 y = d e f y A × B × C d e f A × B × C
26 vex d V
27 vex e V
28 vex f V
29 26 27 28 ot21std y = d e f 1 st 1 st y = d
30 29 breq2d y = d e f a R 1 st 1 st y a R d
31 29 eqeq2d y = d e f a = 1 st 1 st y a = d
32 30 31 orbi12d y = d e f a R 1 st 1 st y a = 1 st 1 st y a R d a = d
33 26 27 28 ot22ndd y = d e f 2 nd 1 st y = e
34 33 breq2d y = d e f b S 2 nd 1 st y b S e
35 33 eqeq2d y = d e f b = 2 nd 1 st y b = e
36 34 35 orbi12d y = d e f b S 2 nd 1 st y b = 2 nd 1 st y b S e b = e
37 opex d e V
38 37 28 op2ndd y = d e f 2 nd y = f
39 38 breq2d y = d e f c T 2 nd y c T f
40 38 eqeq2d y = d e f c = 2 nd y c = f
41 39 40 orbi12d y = d e f c T 2 nd y c = 2 nd y c T f c = f
42 32 36 41 3anbi123d y = d e f a R 1 st 1 st y a = 1 st 1 st y b S 2 nd 1 st y b = 2 nd 1 st y c T 2 nd y c = 2 nd y a R d a = d b S e b = e c T f c = f
43 neeq2 y = d e f a b c y a b c d e f
44 42 43 anbi12d y = d e f a R 1 st 1 st y a = 1 st 1 st y b S 2 nd 1 st y b = 2 nd 1 st y c T 2 nd y c = 2 nd y a b c y a R d a = d b S e b = e c T f c = f a b c d e f
45 25 44 3anbi23d y = d e f a b c A × B × C y A × B × C a R 1 st 1 st y a = 1 st 1 st y b S 2 nd 1 st y b = 2 nd 1 st y c T 2 nd y c = 2 nd y a b c y a b c A × B × C d e f A × B × C a R d a = d b S e b = e c T f c = f a b c d e f
46 2 3 24 45 1 brab a b c U d e f a b c A × B × C d e f A × B × C a R d a = d b S e b = e c T f c = f a b c d e f
47 ot2elxp a b c A × B × C a A b B c C
48 ot2elxp d e f A × B × C d A e B f C
49 5 6 7 otthne a b c d e f a d b e c f
50 49 anbi2i a R d a = d b S e b = e c T f c = f a b c d e f a R d a = d b S e b = e c T f c = f a d b e c f
51 47 48 50 3anbi123i a b c A × B × C d e f A × B × C a R d a = d b S e b = e c T f c = f a b c d e f a A b B c C d A e B f C a R d a = d b S e b = e c T f c = f a d b e c f
52 46 51 bitri a b c U d e f a A b B c C d A e B f C a R d a = d b S e b = e c T f c = f a d b e c f