Metamath Proof Explorer


Theorem zlmodzxzldeplem3

Description: Lemma 3 for zlmodzxzldep . (Contributed by AV, 24-May-2019) (Revised by AV, 10-Jun-2019)

Ref Expression
Hypotheses zlmodzxzldep.z ⊢ Z = ℤ ring freeLMod 0 1
zlmodzxzldep.a ⊢ A = 0 3 1 6
zlmodzxzldep.b ⊢ B = 0 2 1 4
zlmodzxzldeplem.f ⊢ F = A 2 B − 3
Assertion zlmodzxzldeplem3 ⊢ F linC ⁡ Z A B = 0 Z

Proof

Step Hyp Ref Expression
1 zlmodzxzldep.z ⊢ Z = ℤ ring freeLMod 0 1
2 zlmodzxzldep.a ⊢ A = 0 3 1 6
3 zlmodzxzldep.b ⊢ B = 0 2 1 4
4 zlmodzxzldeplem.f ⊢ F = A 2 B − 3
5 ovex ⊢ ℤ ring freeLMod 0 1 ∈ V
6 1 5 eqeltri ⊢ Z ∈ V
7 1 2 3 4 zlmodzxzldeplem1 ⊢ F ∈ ℤ A B
8 1 zlmodzxzlmod ⊢ Z ∈ LMod ∧ ℤ ring = Scalar ⁡ Z
9 simpr ⊢ Z ∈ LMod ∧ ℤ ring = Scalar ⁡ Z → ℤ ring = Scalar ⁡ Z
10 9 eqcomd ⊢ Z ∈ LMod ∧ ℤ ring = Scalar ⁡ Z → Scalar ⁡ Z = ℤ ring
11 8 10 ax-mp ⊢ Scalar ⁡ Z = ℤ ring
12 11 fveq2i ⊢ Base Scalar ⁡ Z = Base ℤ ring
13 zringbas ⊢ ℤ = Base ℤ ring
14 13 eqcomi ⊢ Base ℤ ring = ℤ
15 12 14 eqtri ⊢ Base Scalar ⁡ Z = ℤ
16 15 oveq1i ⊢ Base Scalar ⁡ Z A B = ℤ A B
17 7 16 eleqtrri ⊢ F ∈ Base Scalar ⁡ Z A B
18 3z ⊢ 3 ∈ ℤ
19 6nn ⊢ 6 ∈ ℕ
20 19 nnzi ⊢ 6 ∈ ℤ
21 1 zlmodzxzel ⊢ 3 ∈ ℤ ∧ 6 ∈ ℤ → 0 3 1 6 ∈ Base Z
22 18 20 21 mp2an ⊢ 0 3 1 6 ∈ Base Z
23 2z ⊢ 2 ∈ ℤ
24 4z ⊢ 4 ∈ ℤ
25 1 zlmodzxzel ⊢ 2 ∈ ℤ ∧ 4 ∈ ℤ → 0 2 1 4 ∈ Base Z
26 23 24 25 mp2an ⊢ 0 2 1 4 ∈ Base Z
27 2 eleq1i ⊢ A ∈ Base Z ↔ 0 3 1 6 ∈ Base Z
28 3 eleq1i ⊢ B ∈ Base Z ↔ 0 2 1 4 ∈ Base Z
29 27 28 anbi12i ⊢ A ∈ Base Z ∧ B ∈ Base Z ↔ 0 3 1 6 ∈ Base Z ∧ 0 2 1 4 ∈ Base Z
30 22 26 29 mpbir2an ⊢ A ∈ Base Z ∧ B ∈ Base Z
31 prelpwi ⊢ A ∈ Base Z ∧ B ∈ Base Z → A B ∈ 𝒫 Base Z
32 30 31 ax-mp ⊢ A B ∈ 𝒫 Base Z
33 lincval ⊢ Z ∈ V ∧ F ∈ Base Scalar ⁡ Z A B ∧ A B ∈ 𝒫 Base Z → F linC ⁡ Z A B = ∑ Z x ∈ A B F ⁡ x ⋅ Z x
34 6 17 32 33 mp3an ⊢ F linC ⁡ Z A B = ∑ Z x ∈ A B F ⁡ x ⋅ Z x
35 lmodcmn ⊢ Z ∈ LMod → Z ∈ CMnd
36 35 adantr ⊢ Z ∈ LMod ∧ ℤ ring = Scalar ⁡ Z → Z ∈ CMnd
37 8 36 ax-mp ⊢ Z ∈ CMnd
38 prex ⊢ 0 3 1 6 ∈ V
39 2 38 eqeltri ⊢ A ∈ V
40 prex ⊢ 0 2 1 4 ∈ V
41 3 40 eqeltri ⊢ B ∈ V
42 1 2 3 zlmodzxzldeplem ⊢ A ≠ B
43 39 41 42 3pm3.2i ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B
44 8 simpli ⊢ Z ∈ LMod
45 elmapi ⊢ F ∈ ℤ A B → F : A B ⟶ ℤ
46 39 prid1 ⊢ A ∈ A B
47 ffvelcdm ⊢ F : A B ⟶ ℤ ∧ A ∈ A B → F ⁡ A ∈ ℤ
48 46 47 mpan2 ⊢ F : A B ⟶ ℤ → F ⁡ A ∈ ℤ
49 7 45 48 mp2b ⊢ F ⁡ A ∈ ℤ
50 8 9 ax-mp ⊢ ℤ ring = Scalar ⁡ Z
51 50 eqcomi ⊢ Scalar ⁡ Z = ℤ ring
52 51 fveq2i ⊢ Base Scalar ⁡ Z = Base ℤ ring
53 52 14 eqtri ⊢ Base Scalar ⁡ Z = ℤ
54 49 53 eleqtrri ⊢ F ⁡ A ∈ Base Scalar ⁡ Z
55 2 22 eqeltri ⊢ A ∈ Base Z
56 eqid ⊢ Base Z = Base Z
57 eqid ⊢ Scalar ⁡ Z = Scalar ⁡ Z
58 eqid ⊢ ⋅ Z = ⋅ Z
59 eqid ⊢ Base Scalar ⁡ Z = Base Scalar ⁡ Z
60 56 57 58 59 lmodvscl ⊢ Z ∈ LMod ∧ F ⁡ A ∈ Base Scalar ⁡ Z ∧ A ∈ Base Z → F ⁡ A ⋅ Z A ∈ Base Z
61 44 54 55 60 mp3an ⊢ F ⁡ A ⋅ Z A ∈ Base Z
62 41 prid2 ⊢ B ∈ A B
63 ffvelcdm ⊢ F : A B ⟶ ℤ ∧ B ∈ A B → F ⁡ B ∈ ℤ
64 62 63 mpan2 ⊢ F : A B ⟶ ℤ → F ⁡ B ∈ ℤ
65 7 45 64 mp2b ⊢ F ⁡ B ∈ ℤ
66 65 53 eleqtrri ⊢ F ⁡ B ∈ Base Scalar ⁡ Z
67 3 26 eqeltri ⊢ B ∈ Base Z
68 56 57 58 59 lmodvscl ⊢ Z ∈ LMod ∧ F ⁡ B ∈ Base Scalar ⁡ Z ∧ B ∈ Base Z → F ⁡ B ⋅ Z B ∈ Base Z
69 44 66 67 68 mp3an ⊢ F ⁡ B ⋅ Z B ∈ Base Z
70 61 69 pm3.2i ⊢ F ⁡ A ⋅ Z A ∈ Base Z ∧ F ⁡ B ⋅ Z B ∈ Base Z
71 eqid ⊢ + Z = + Z
72 fveq2 ⊢ x = A → F ⁡ x = F ⁡ A
73 id ⊢ x = A → x = A
74 72 73 oveq12d ⊢ x = A → F ⁡ x ⋅ Z x = F ⁡ A ⋅ Z A
75 fveq2 ⊢ x = B → F ⁡ x = F ⁡ B
76 id ⊢ x = B → x = B
77 75 76 oveq12d ⊢ x = B → F ⁡ x ⋅ Z x = F ⁡ B ⋅ Z B
78 56 71 74 77 gsumpr ⊢ Z ∈ CMnd ∧ A ∈ V ∧ B ∈ V ∧ A ≠ B ∧ F ⁡ A ⋅ Z A ∈ Base Z ∧ F ⁡ B ⋅ Z B ∈ Base Z → ∑ Z x ∈ A B F ⁡ x ⋅ Z x = F ⁡ A ⋅ Z A + Z F ⁡ B ⋅ Z B
79 37 43 70 78 mp3an ⊢ ∑ Z x ∈ A B F ⁡ x ⋅ Z x = F ⁡ A ⋅ Z A + Z F ⁡ B ⋅ Z B
80 4 fveq1i ⊢ F ⁡ A = A 2 B − 3 ⁡ A
81 2ex ⊢ 2 ∈ V
82 39 81 fvpr1 ⊢ A ≠ B → A 2 B − 3 ⁡ A = 2
83 42 82 ax-mp ⊢ A 2 B − 3 ⁡ A = 2
84 80 83 eqtri ⊢ F ⁡ A = 2
85 84 oveq1i ⊢ F ⁡ A ⋅ Z A = 2 ⋅ Z A
86 4 fveq1i ⊢ F ⁡ B = A 2 B − 3 ⁡ B
87 negex ⊢ − 3 ∈ V
88 41 87 fvpr2 ⊢ A ≠ B → A 2 B − 3 ⁡ B = − 3
89 42 88 ax-mp ⊢ A 2 B − 3 ⁡ B = − 3
90 86 89 eqtri ⊢ F ⁡ B = − 3
91 90 oveq1i ⊢ F ⁡ B ⋅ Z B = -3 ⋅ Z B
92 85 91 oveq12i ⊢ F ⁡ A ⋅ Z A + Z F ⁡ B ⋅ Z B = 2 ⋅ Z A + Z -3 ⋅ Z B
93 eqid ⊢ 0 0 1 0 = 0 0 1 0
94 1 93 zlmodzxz0 ⊢ 0 0 1 0 = 0 Z
95 94 eqcomi ⊢ 0 Z = 0 0 1 0
96 1 2 3 95 71 58 zlmodzxzequap ⊢ 2 ⋅ Z A + Z -3 ⋅ Z B = 0 Z
97 92 96 eqtri ⊢ F ⁡ A ⋅ Z A + Z F ⁡ B ⋅ Z B = 0 Z
98 34 79 97 3eqtri ⊢ F linC ⁡ Z A B = 0 Z