Metamath Proof Explorer


Theorem xpord3lem

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

Ref Expression
Hypothesis xpord3.1 U=xy|xA×B×CyA×B×C1st1stxR1st1sty1st1stx=1st1sty2nd1stxS2nd1sty2nd1stx=2nd1sty2ndxT2ndy2ndx=2ndyxy
Assertion xpord3lem abcUdefaAbBcCdAeBfCaRda=dbSeb=ecTfc=fadbecf

Proof

Step Hyp Ref Expression
1 xpord3.1 U=xy|xA×B×CyA×B×C1st1stxR1st1sty1st1stx=1st1sty2nd1stxS2nd1sty2nd1stx=2nd1sty2ndxT2ndy2ndx=2ndyxy
2 otex abcV
3 otex defV
4 eleq1 x=abcxA×B×CabcA×B×C
5 2fveq3 x=abc1st1stx=1st1stabc
6 vex aV
7 vex bV
8 vex cV
9 ot1stg aVbVcV1st1stabc=a
10 6 7 8 9 mp3an 1st1stabc=a
11 5 10 eqtrdi x=abc1st1stx=a
12 11 breq1d x=abc1st1stxR1st1styaR1st1sty
13 11 eqeq1d x=abc1st1stx=1st1stya=1st1sty
14 12 13 orbi12d x=abc1st1stxR1st1sty1st1stx=1st1styaR1st1stya=1st1sty
15 2fveq3 x=abc2nd1stx=2nd1stabc
16 ot2ndg aVbVcV2nd1stabc=b
17 6 7 8 16 mp3an 2nd1stabc=b
18 15 17 eqtrdi x=abc2nd1stx=b
19 18 breq1d x=abc2nd1stxS2nd1stybS2nd1sty
20 18 eqeq1d x=abc2nd1stx=2nd1styb=2nd1sty
21 19 20 orbi12d x=abc2nd1stxS2nd1sty2nd1stx=2nd1stybS2nd1styb=2nd1sty
22 fveq2 x=abc2ndx=2ndabc
23 ot3rdg cV2ndabc=c
24 23 elv 2ndabc=c
25 22 24 eqtrdi x=abc2ndx=c
26 25 breq1d x=abc2ndxT2ndycT2ndy
27 25 eqeq1d x=abc2ndx=2ndyc=2ndy
28 26 27 orbi12d x=abc2ndxT2ndy2ndx=2ndycT2ndyc=2ndy
29 14 21 28 3anbi123d x=abc1st1stxR1st1sty1st1stx=1st1sty2nd1stxS2nd1sty2nd1stx=2nd1sty2ndxT2ndy2ndx=2ndyaR1st1stya=1st1stybS2nd1styb=2nd1stycT2ndyc=2ndy
30 neeq1 x=abcxyabcy
31 29 30 anbi12d x=abc1st1stxR1st1sty1st1stx=1st1sty2nd1stxS2nd1sty2nd1stx=2nd1sty2ndxT2ndy2ndx=2ndyxyaR1st1stya=1st1stybS2nd1styb=2nd1stycT2ndyc=2ndyabcy
32 4 31 3anbi13d x=abcxA×B×CyA×B×C1st1stxR1st1sty1st1stx=1st1sty2nd1stxS2nd1sty2nd1stx=2nd1sty2ndxT2ndy2ndx=2ndyxyabcA×B×CyA×B×CaR1st1stya=1st1stybS2nd1styb=2nd1stycT2ndyc=2ndyabcy
33 eleq1 y=defyA×B×CdefA×B×C
34 2fveq3 y=def1st1sty=1st1stdef
35 vex dV
36 vex eV
37 vex fV
38 ot1stg dVeVfV1st1stdef=d
39 35 36 37 38 mp3an 1st1stdef=d
40 34 39 eqtrdi y=def1st1sty=d
41 40 breq2d y=defaR1st1styaRd
42 40 eqeq2d y=defa=1st1stya=d
43 41 42 orbi12d y=defaR1st1stya=1st1styaRda=d
44 2fveq3 y=def2nd1sty=2nd1stdef
45 ot2ndg dVeVfV2nd1stdef=e
46 35 36 37 45 mp3an 2nd1stdef=e
47 44 46 eqtrdi y=def2nd1sty=e
48 47 breq2d y=defbS2nd1stybSe
49 47 eqeq2d y=defb=2nd1styb=e
50 48 49 orbi12d y=defbS2nd1styb=2nd1stybSeb=e
51 fveq2 y=def2ndy=2nddef
52 ot3rdg fV2nddef=f
53 52 elv 2nddef=f
54 51 53 eqtrdi y=def2ndy=f
55 54 breq2d y=defcT2ndycTf
56 54 eqeq2d y=defc=2ndyc=f
57 55 56 orbi12d y=defcT2ndyc=2ndycTfc=f
58 43 50 57 3anbi123d y=defaR1st1stya=1st1stybS2nd1styb=2nd1stycT2ndyc=2ndyaRda=dbSeb=ecTfc=f
59 neeq2 y=defabcyabcdef
60 58 59 anbi12d y=defaR1st1stya=1st1stybS2nd1styb=2nd1stycT2ndyc=2ndyabcyaRda=dbSeb=ecTfc=fabcdef
61 33 60 3anbi23d y=defabcA×B×CyA×B×CaR1st1stya=1st1stybS2nd1styb=2nd1stycT2ndyc=2ndyabcyabcA×B×CdefA×B×CaRda=dbSeb=ecTfc=fabcdef
62 2 3 32 61 1 brab abcUdefabcA×B×CdefA×B×CaRda=dbSeb=ecTfc=fabcdef
63 otelxp abcA×B×CaAbBcC
64 otelxp defA×B×CdAeBfC
65 6 7 8 otthne abcdefadbecf
66 65 anbi2i aRda=dbSeb=ecTfc=fabcdefaRda=dbSeb=ecTfc=fadbecf
67 63 64 66 3anbi123i abcA×B×CdefA×B×CaRda=dbSeb=ecTfc=fabcdefaAbBcCdAeBfCaRda=dbSeb=ecTfc=fadbecf
68 62 67 bitri abcUdefaAbBcCdAeBfCaRda=dbSeb=ecTfc=fadbecf