Metamath Proof Explorer


Theorem gzrngunitlem

Description: Lemma for gzrngunit . (Contributed by Mario Carneiro, 4-Dec-2014)

Ref Expression
Hypothesis gzrng.1 Z = fld 𝑠 i
Assertion gzrngunitlem A Unit Z 1 A

Proof

Step Hyp Ref Expression
1 gzrng.1 Z = fld 𝑠 i
2 sq1 1 2 = 1
3 ax-1ne0 1 0
4 gzsubrg i SubRing fld
5 1 subrgring i SubRing fld Z Ring
6 eqid Unit Z = Unit Z
7 subrgsubg i SubRing fld i SubGrp fld
8 cnfld0 0 = 0 fld
9 1 8 subg0 i SubGrp fld 0 = 0 Z
10 4 7 9 mp2b 0 = 0 Z
11 cnfld1 1 = 1 fld
12 1 11 subrg1 i SubRing fld 1 = 1 Z
13 4 12 ax-mp 1 = 1 Z
14 6 10 13 0unit Z Ring 0 Unit Z 1 = 0
15 4 5 14 mp2b 0 Unit Z 1 = 0
16 3 15 nemtbir ¬ 0 Unit Z
17 1 subrgbas i SubRing fld i = Base Z
18 4 17 ax-mp i = Base Z
19 18 6 unitcl A Unit Z A i
20 gzabssqcl A i A 2 0
21 19 20 syl A Unit Z A 2 0
22 elnn0 A 2 0 A 2 A 2 = 0
23 21 22 sylib A Unit Z A 2 A 2 = 0
24 23 ord A Unit Z ¬ A 2 A 2 = 0
25 gzcn A i A
26 19 25 syl A Unit Z A
27 26 abscld A Unit Z A
28 27 recnd A Unit Z A
29 sqeq0 A A 2 = 0 A = 0
30 28 29 syl A Unit Z A 2 = 0 A = 0
31 26 abs00ad A Unit Z A = 0 A = 0
32 eleq1 A = 0 A Unit Z 0 Unit Z
33 32 biimpcd A Unit Z A = 0 0 Unit Z
34 31 33 sylbid A Unit Z A = 0 0 Unit Z
35 30 34 sylbid A Unit Z A 2 = 0 0 Unit Z
36 24 35 syld A Unit Z ¬ A 2 0 Unit Z
37 16 36 mt3i A Unit Z A 2
38 37 nnge1d A Unit Z 1 A 2
39 2 38 eqbrtrid A Unit Z 1 2 A 2
40 26 absge0d A Unit Z 0 A
41 1re 1
42 0le1 0 1
43 le2sq 1 0 1 A 0 A 1 A 1 2 A 2
44 41 42 43 mpanl12 A 0 A 1 A 1 2 A 2
45 27 40 44 syl2anc A Unit Z 1 A 1 2 A 2
46 39 45 mpbird A Unit Z 1 A