Metamath Proof Explorer


Theorem pzriprnglem11

Description: Lemma 11 for pzriprng : The base set of the quotient of R and J . (Contributed by AV, 22-Mar-2025)

Ref Expression
Hypotheses pzriprng.r ⊢ R = ℤ ring × 𝑠 ℤ ring
pzriprng.i ⊢ I = ℤ × 0
pzriprng.j ⊢ J = R ↾ 𝑠 I
pzriprng.1 ⊢ 1 ˙ = 1 J
pzriprng.g ⊢ ∼ ˙ = R ~ QG I
pzriprng.q ⊢ Q = R / 𝑠 ∼ ˙
Assertion pzriprnglem11 ⊢ Base Q = ⋃ r ∈ ℤ ℤ × r

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 pzriprng.1 ⊢ 1 ˙ = 1 J
5 pzriprng.g ⊢ ∼ ˙ = R ~ QG I
6 pzriprng.q ⊢ Q = R / 𝑠 ∼ ˙
7 df-qs ⊢ ℤ × ℤ / ∼ ˙ = e | ∃ p ∈ ℤ × ℤ e = p ∼ ˙
8 6 a1i ⊢ ∼ ˙ = R ~ QG I → Q = R / 𝑠 ∼ ˙
9 1 pzriprnglem2 ⊢ Base R = ℤ × ℤ
10 9 eqcomi ⊢ ℤ × ℤ = Base R
11 10 a1i ⊢ ∼ ˙ = R ~ QG I → ℤ × ℤ = Base R
12 ovexd ⊢ ∼ ˙ = R ~ QG I → R ~ QG I ∈ V
13 5 12 eqeltrid ⊢ ∼ ˙ = R ~ QG I → ∼ ˙ ∈ V
14 1 pzriprnglem1 ⊢ R ∈ Rng
15 14 a1i ⊢ ∼ ˙ = R ~ QG I → R ∈ Rng
16 8 11 13 15 qusbas ⊢ ∼ ˙ = R ~ QG I → ℤ × ℤ / ∼ ˙ = Base Q
17 5 16 ax-mp ⊢ ℤ × ℤ / ∼ ˙ = Base Q
18 nfcv ⊢ Ⅎ _ s e | e = p ∼ ˙
19 nfcv ⊢ Ⅎ _ r e | e = p ∼ ˙
20 nfcv ⊢ Ⅎ _ p e | e = s r ∼ ˙
21 eceq1 ⊢ p = s r → p ∼ ˙ = s r ∼ ˙
22 21 eqeq2d ⊢ p = s r → e = p ∼ ˙ ↔ e = s r ∼ ˙
23 22 abbidv ⊢ p = s r → e | e = p ∼ ˙ = e | e = s r ∼ ˙
24 18 19 20 23 iunxpf ⊢ ⋃ p ∈ ℤ × ℤ e | e = p ∼ ˙ = ⋃ s ∈ ℤ ⋃ r ∈ ℤ e | e = s r ∼ ˙
25 iunab ⊢ ⋃ p ∈ ℤ × ℤ e | e = p ∼ ˙ = e | ∃ p ∈ ℤ × ℤ e = p ∼ ˙
26 iuncom ⊢ ⋃ s ∈ ℤ ⋃ r ∈ ℤ e | e = s r ∼ ˙ = ⋃ r ∈ ℤ ⋃ s ∈ ℤ e | e = s r ∼ ˙
27 df-sn ⊢ s r ∼ ˙ = e | e = s r ∼ ˙
28 27 eqcomi ⊢ e | e = s r ∼ ˙ = s r ∼ ˙
29 28 a1i ⊢ s ∈ ℤ → e | e = s r ∼ ˙ = s r ∼ ˙
30 29 iuneq2i ⊢ ⋃ s ∈ ℤ e | e = s r ∼ ˙ = ⋃ s ∈ ℤ s r ∼ ˙
31 simpr ⊢ r ∈ ℤ ∧ s ∈ ℤ ∧ p = s r ∼ ˙ → p = s r ∼ ˙
32 1 2 3 4 5 pzriprnglem10 ⊢ s ∈ ℤ ∧ r ∈ ℤ → s r ∼ ˙ = ℤ × r
33 32 ancoms ⊢ r ∈ ℤ ∧ s ∈ ℤ → s r ∼ ˙ = ℤ × r
34 33 adantr ⊢ r ∈ ℤ ∧ s ∈ ℤ ∧ p = s r ∼ ˙ → s r ∼ ˙ = ℤ × r
35 31 34 eqtrd ⊢ r ∈ ℤ ∧ s ∈ ℤ ∧ p = s r ∼ ˙ → p = ℤ × r
36 35 ex ⊢ r ∈ ℤ ∧ s ∈ ℤ → p = s r ∼ ˙ → p = ℤ × r
37 36 rexlimdva ⊢ r ∈ ℤ → ∃ s ∈ ℤ p = s r ∼ ˙ → p = ℤ × r
38 0zd ⊢ r ∈ ℤ ∧ p = ℤ × r → 0 ∈ ℤ
39 simpr ⊢ r ∈ ℤ ∧ p = ℤ × r → p = ℤ × r
40 opeq1 ⊢ s = 0 → s r = 0 r
41 40 eceq1d ⊢ s = 0 → s r ∼ ˙ = 0 r ∼ ˙
42 39 41 eqeqan12d ⊢ r ∈ ℤ ∧ p = ℤ × r ∧ s = 0 → p = s r ∼ ˙ ↔ ℤ × r = 0 r ∼ ˙
43 0zd ⊢ r ∈ ℤ → 0 ∈ ℤ
44 1 2 3 4 5 pzriprnglem10 ⊢ 0 ∈ ℤ ∧ r ∈ ℤ → 0 r ∼ ˙ = ℤ × r
45 43 44 mpancom ⊢ r ∈ ℤ → 0 r ∼ ˙ = ℤ × r
46 45 eqcomd ⊢ r ∈ ℤ → ℤ × r = 0 r ∼ ˙
47 46 adantr ⊢ r ∈ ℤ ∧ p = ℤ × r → ℤ × r = 0 r ∼ ˙
48 38 42 47 rspcedvd ⊢ r ∈ ℤ ∧ p = ℤ × r → ∃ s ∈ ℤ p = s r ∼ ˙
49 48 ex ⊢ r ∈ ℤ → p = ℤ × r → ∃ s ∈ ℤ p = s r ∼ ˙
50 37 49 impbid ⊢ r ∈ ℤ → ∃ s ∈ ℤ p = s r ∼ ˙ ↔ p = ℤ × r
51 50 abbidv ⊢ r ∈ ℤ → p | ∃ s ∈ ℤ p = s r ∼ ˙ = p | p = ℤ × r
52 iunsn ⊢ ⋃ s ∈ ℤ s r ∼ ˙ = p | ∃ s ∈ ℤ p = s r ∼ ˙
53 df-sn ⊢ ℤ × r = p | p = ℤ × r
54 51 52 53 3eqtr4g ⊢ r ∈ ℤ → ⋃ s ∈ ℤ s r ∼ ˙ = ℤ × r
55 30 54 eqtrid ⊢ r ∈ ℤ → ⋃ s ∈ ℤ e | e = s r ∼ ˙ = ℤ × r
56 55 iuneq2i ⊢ ⋃ r ∈ ℤ ⋃ s ∈ ℤ e | e = s r ∼ ˙ = ⋃ r ∈ ℤ ℤ × r
57 26 56 eqtri ⊢ ⋃ s ∈ ℤ ⋃ r ∈ ℤ e | e = s r ∼ ˙ = ⋃ r ∈ ℤ ℤ × r
58 24 25 57 3eqtr3i ⊢ e | ∃ p ∈ ℤ × ℤ e = p ∼ ˙ = ⋃ r ∈ ℤ ℤ × r
59 7 17 58 3eqtr3i ⊢ Base Q = ⋃ r ∈ ℤ ℤ × r