Metamath Proof Explorer


Theorem diophrex

Description: Projecting a Diophantine set by removing a coordinate results in a Diophantine set. (Contributed by Stefan O'Rear, 10-Oct-2014)

Ref Expression
Assertion diophrex N 0 M N S Dioph M t | u S t = u 1 N Dioph N

Proof

Step Hyp Ref Expression
1 eqeq1 a = t a = b 1 N t = b 1 N
2 1 rexbidv a = t b S a = b 1 N b S t = b 1 N
3 reseq1 b = u b 1 N = u 1 N
4 3 eqeq2d b = u t = b 1 N t = u 1 N
5 4 cbvrexvw b S t = b 1 N u S t = u 1 N
6 2 5 bitrdi a = t b S a = b 1 N u S t = u 1 N
7 6 cbvabv a | b S a = b 1 N = t | u S t = u 1 N
8 rexeq S = d | e 0 d = e 1 M c e = 0 b S a = b 1 N b d | e 0 d = e 1 M c e = 0 a = b 1 N
9 8 abbidv S = d | e 0 d = e 1 M c e = 0 a | b S a = b 1 N = a | b d | e 0 d = e 1 M c e = 0 a = b 1 N
10 9 adantl N 0 M N S Dioph M c mzPoly S = d | e 0 d = e 1 M c e = 0 a | b S a = b 1 N = a | b d | e 0 d = e 1 M c e = 0 a = b 1 N
11 eqeq1 d = b d = e 1 M b = e 1 M
12 11 anbi1d d = b d = e 1 M c e = 0 b = e 1 M c e = 0
13 12 rexbidv d = b e 0 d = e 1 M c e = 0 e 0 b = e 1 M c e = 0
14 13 rexab b d | e 0 d = e 1 M c e = 0 a = b 1 N b e 0 b = e 1 M c e = 0 a = b 1 N
15 r19.41v e 0 b = e 1 M c e = 0 a = b 1 N e 0 b = e 1 M c e = 0 a = b 1 N
16 15 exbii b e 0 b = e 1 M c e = 0 a = b 1 N b e 0 b = e 1 M c e = 0 a = b 1 N
17 rexcom4 e 0 b b = e 1 M c e = 0 a = b 1 N b e 0 b = e 1 M c e = 0 a = b 1 N
18 anass b = e 1 M c e = 0 a = b 1 N b = e 1 M c e = 0 a = b 1 N
19 18 exbii b b = e 1 M c e = 0 a = b 1 N b b = e 1 M c e = 0 a = b 1 N
20 vex e V
21 20 resex e 1 M V
22 reseq1 b = e 1 M b 1 N = e 1 M 1 N
23 22 eqeq2d b = e 1 M a = b 1 N a = e 1 M 1 N
24 23 anbi2d b = e 1 M c e = 0 a = b 1 N c e = 0 a = e 1 M 1 N
25 21 24 ceqsexv b b = e 1 M c e = 0 a = b 1 N c e = 0 a = e 1 M 1 N
26 19 25 bitri b b = e 1 M c e = 0 a = b 1 N c e = 0 a = e 1 M 1 N
27 ancom c e = 0 a = e 1 M 1 N a = e 1 M 1 N c e = 0
28 simpl2 N 0 M N S Dioph M c mzPoly M N
29 fzss2 M N 1 N 1 M
30 resabs1 1 N 1 M e 1 M 1 N = e 1 N
31 28 29 30 3syl N 0 M N S Dioph M c mzPoly e 1 M 1 N = e 1 N
32 31 eqeq2d N 0 M N S Dioph M c mzPoly a = e 1 M 1 N a = e 1 N
33 32 anbi1d N 0 M N S Dioph M c mzPoly a = e 1 M 1 N c e = 0 a = e 1 N c e = 0
34 27 33 syl5bb N 0 M N S Dioph M c mzPoly c e = 0 a = e 1 M 1 N a = e 1 N c e = 0
35 26 34 syl5bb N 0 M N S Dioph M c mzPoly b b = e 1 M c e = 0 a = b 1 N a = e 1 N c e = 0
36 35 rexbidv N 0 M N S Dioph M c mzPoly e 0 b b = e 1 M c e = 0 a = b 1 N e 0 a = e 1 N c e = 0
37 17 36 bitr3id N 0 M N S Dioph M c mzPoly b e 0 b = e 1 M c e = 0 a = b 1 N e 0 a = e 1 N c e = 0
38 16 37 bitr3id N 0 M N S Dioph M c mzPoly b e 0 b = e 1 M c e = 0 a = b 1 N e 0 a = e 1 N c e = 0
39 14 38 syl5bb N 0 M N S Dioph M c mzPoly b d | e 0 d = e 1 M c e = 0 a = b 1 N e 0 a = e 1 N c e = 0
40 39 abbidv N 0 M N S Dioph M c mzPoly a | b d | e 0 d = e 1 M c e = 0 a = b 1 N = a | e 0 a = e 1 N c e = 0
41 eldioph3 N 0 c mzPoly a | e 0 a = e 1 N c e = 0 Dioph N
42 41 3ad2antl1 N 0 M N S Dioph M c mzPoly a | e 0 a = e 1 N c e = 0 Dioph N
43 40 42 eqeltrd N 0 M N S Dioph M c mzPoly a | b d | e 0 d = e 1 M c e = 0 a = b 1 N Dioph N
44 43 adantr N 0 M N S Dioph M c mzPoly S = d | e 0 d = e 1 M c e = 0 a | b d | e 0 d = e 1 M c e = 0 a = b 1 N Dioph N
45 10 44 eqeltrd N 0 M N S Dioph M c mzPoly S = d | e 0 d = e 1 M c e = 0 a | b S a = b 1 N Dioph N
46 eldioph3b S Dioph M M 0 c mzPoly S = d | e 0 d = e 1 M c e = 0
47 46 simprbi S Dioph M c mzPoly S = d | e 0 d = e 1 M c e = 0
48 47 3ad2ant3 N 0 M N S Dioph M c mzPoly S = d | e 0 d = e 1 M c e = 0
49 45 48 r19.29a N 0 M N S Dioph M a | b S a = b 1 N Dioph N
50 7 49 eqeltrrid N 0 M N S Dioph M t | u S t = u 1 N Dioph N