Metamath Proof Explorer


Theorem pzriprnglem6

Description: Lemma 6 for pzriprng : J has a ring unity. (Contributed by AV, 19-Mar-2025)

Ref Expression
Hypotheses pzriprng.r ⊢ R = ℤ ring × 𝑠 ℤ ring
pzriprng.i ⊢ I = ℤ × 0
pzriprng.j ⊢ J = R ↾ 𝑠 I
Assertion pzriprnglem6 ⊢ X ∈ I → 1 0 ⋅ J X = X ∧ X ⋅ J 1 0 = X

Proof

Step Hyp Ref Expression
1 pzriprng.r ⊢ R = ℤ ring × 𝑠 ℤ ring
2 pzriprng.i ⊢ I = ℤ × 0
3 pzriprng.j ⊢ J = R ↾ 𝑠 I
4 1 2 pzriprnglem3 ⊢ X ∈ I ↔ ∃ a ∈ ℤ X = a 0
5 1 2 pzriprnglem5 ⊢ I ∈ SubRng ⁡ R
6 eqid ⊢ ⋅ R = ⋅ R
7 3 6 ressmulr ⊢ I ∈ SubRng ⁡ R → ⋅ R = ⋅ J
8 7 eqcomd ⊢ I ∈ SubRng ⁡ R → ⋅ J = ⋅ R
9 5 8 ax-mp ⊢ ⋅ J = ⋅ R
10 9 oveqi ⊢ 1 0 ⋅ J a 0 = 1 0 ⋅ R a 0
11 10 a1i ⊢ a ∈ ℤ → 1 0 ⋅ J a 0 = 1 0 ⋅ R a 0
12 zringbas ⊢ ℤ = Base ℤ ring
13 zringring ⊢ ℤ ring ∈ Ring
14 13 a1i ⊢ a ∈ ℤ → ℤ ring ∈ Ring
15 1zzd ⊢ a ∈ ℤ → 1 ∈ ℤ
16 0z ⊢ 0 ∈ ℤ
17 16 a1i ⊢ a ∈ ℤ → 0 ∈ ℤ
18 id ⊢ a ∈ ℤ → a ∈ ℤ
19 zringmulr ⊢ × = ⋅ ℤ ring
20 19 oveqi ⊢ 1 ⁢ a = 1 ⋅ ℤ ring a
21 15 18 zmulcld ⊢ a ∈ ℤ → 1 ⁢ a ∈ ℤ
22 20 21 eqeltrrid ⊢ a ∈ ℤ → 1 ⋅ ℤ ring a ∈ ℤ
23 19 eqcomi ⊢ ⋅ ℤ ring = ×
24 23 oveqi ⊢ 0 ⋅ ℤ ring 0 = 0 ⋅ 0
25 id ⊢ 0 ∈ ℤ → 0 ∈ ℤ
26 25 25 zmulcld ⊢ 0 ∈ ℤ → 0 ⋅ 0 ∈ ℤ
27 16 26 ax-mp ⊢ 0 ⋅ 0 ∈ ℤ
28 24 27 eqeltri ⊢ 0 ⋅ ℤ ring 0 ∈ ℤ
29 28 a1i ⊢ a ∈ ℤ → 0 ⋅ ℤ ring 0 ∈ ℤ
30 eqid ⊢ ⋅ ℤ ring = ⋅ ℤ ring
31 1 12 12 14 14 15 17 18 17 22 29 30 30 6 xpsmul ⊢ a ∈ ℤ → 1 0 ⋅ R a 0 = 1 ⋅ ℤ ring a 0 ⋅ ℤ ring 0
32 zcn ⊢ a ∈ ℤ → a ∈ ℂ
33 32 mullidd ⊢ a ∈ ℤ → 1 ⁢ a = a
34 20 33 eqtr3id ⊢ a ∈ ℤ → 1 ⋅ ℤ ring a = a
35 0cn ⊢ 0 ∈ ℂ
36 35 mul02i ⊢ 0 ⋅ 0 = 0
37 24 36 eqtri ⊢ 0 ⋅ ℤ ring 0 = 0
38 37 a1i ⊢ a ∈ ℤ → 0 ⋅ ℤ ring 0 = 0
39 34 38 opeq12d ⊢ a ∈ ℤ → 1 ⋅ ℤ ring a 0 ⋅ ℤ ring 0 = a 0
40 11 31 39 3eqtrd ⊢ a ∈ ℤ → 1 0 ⋅ J a 0 = a 0
41 9 oveqi ⊢ a 0 ⋅ J 1 0 = a 0 ⋅ R 1 0
42 41 a1i ⊢ a ∈ ℤ → a 0 ⋅ J 1 0 = a 0 ⋅ R 1 0
43 19 oveqi ⊢ a ⋅ 1 = a ⋅ ℤ ring 1
44 18 15 zmulcld ⊢ a ∈ ℤ → a ⋅ 1 ∈ ℤ
45 43 44 eqeltrrid ⊢ a ∈ ℤ → a ⋅ ℤ ring 1 ∈ ℤ
46 1 12 12 14 14 18 17 15 17 45 29 30 30 6 xpsmul ⊢ a ∈ ℤ → a 0 ⋅ R 1 0 = a ⋅ ℤ ring 1 0 ⋅ ℤ ring 0
47 23 oveqi ⊢ a ⋅ ℤ ring 1 = a ⋅ 1
48 32 mulridd ⊢ a ∈ ℤ → a ⋅ 1 = a
49 47 48 eqtrid ⊢ a ∈ ℤ → a ⋅ ℤ ring 1 = a
50 49 38 opeq12d ⊢ a ∈ ℤ → a ⋅ ℤ ring 1 0 ⋅ ℤ ring 0 = a 0
51 42 46 50 3eqtrd ⊢ a ∈ ℤ → a 0 ⋅ J 1 0 = a 0
52 40 51 jca ⊢ a ∈ ℤ → 1 0 ⋅ J a 0 = a 0 ∧ a 0 ⋅ J 1 0 = a 0
53 oveq2 ⊢ X = a 0 → 1 0 ⋅ J X = 1 0 ⋅ J a 0
54 id ⊢ X = a 0 → X = a 0
55 53 54 eqeq12d ⊢ X = a 0 → 1 0 ⋅ J X = X ↔ 1 0 ⋅ J a 0 = a 0
56 oveq1 ⊢ X = a 0 → X ⋅ J 1 0 = a 0 ⋅ J 1 0
57 56 54 eqeq12d ⊢ X = a 0 → X ⋅ J 1 0 = X ↔ a 0 ⋅ J 1 0 = a 0
58 55 57 anbi12d ⊢ X = a 0 → 1 0 ⋅ J X = X ∧ X ⋅ J 1 0 = X ↔ 1 0 ⋅ J a 0 = a 0 ∧ a 0 ⋅ J 1 0 = a 0
59 52 58 syl5ibrcom ⊢ a ∈ ℤ → X = a 0 → 1 0 ⋅ J X = X ∧ X ⋅ J 1 0 = X
60 59 rexlimiv ⊢ ∃ a ∈ ℤ X = a 0 → 1 0 ⋅ J X = X ∧ X ⋅ J 1 0 = X
61 4 60 sylbi ⊢ X ∈ I → 1 0 ⋅ J X = X ∧ X ⋅ J 1 0 = X