Metamath Proof Explorer


Theorem pzriprnglem13

Description: Lemma 13 for pzriprng : Q is a unital ring. (Contributed by AV, 23-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 pzriprnglem13 ⊢ Q ∈ Ring

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 1 pzriprnglem1 ⊢ R ∈ Rng
8 1 2 3 pzriprnglem8 ⊢ I ∈ 2Ideal ⁡ R
9 1 2 pzriprnglem4 ⊢ I ∈ SubGrp ⁡ R
10 5 oveq2i ⊢ R / 𝑠 ∼ ˙ = R / 𝑠 R ~ QG I
11 6 10 eqtri ⊢ Q = R / 𝑠 R ~ QG I
12 eqid ⊢ 2Ideal ⁡ R = 2Ideal ⁡ R
13 11 12 qus2idrng ⊢ R ∈ Rng ∧ I ∈ 2Ideal ⁡ R ∧ I ∈ SubGrp ⁡ R → Q ∈ Rng
14 7 8 9 13 mp3an ⊢ Q ∈ Rng
15 1z ⊢ 1 ∈ ℤ
16 zex ⊢ ℤ ∈ V
17 snex ⊢ 1 ∈ V
18 16 17 xpex ⊢ ℤ × 1 ∈ V
19 18 snid ⊢ ℤ × 1 ∈ ℤ × 1
20 sneq ⊢ y = 1 → y = 1
21 20 xpeq2d ⊢ y = 1 → ℤ × y = ℤ × 1
22 21 sneqd ⊢ y = 1 → ℤ × y = ℤ × 1
23 22 eleq2d ⊢ y = 1 → ℤ × 1 ∈ ℤ × y ↔ ℤ × 1 ∈ ℤ × 1
24 23 rspcev ⊢ 1 ∈ ℤ ∧ ℤ × 1 ∈ ℤ × 1 → ∃ y ∈ ℤ ℤ × 1 ∈ ℤ × y
25 15 19 24 mp2an ⊢ ∃ y ∈ ℤ ℤ × 1 ∈ ℤ × y
26 eliun ⊢ ℤ × 1 ∈ ⋃ y ∈ ℤ ℤ × y ↔ ∃ y ∈ ℤ ℤ × 1 ∈ ℤ × y
27 25 26 mpbir ⊢ ℤ × 1 ∈ ⋃ y ∈ ℤ ℤ × y
28 1 2 3 4 5 6 pzriprnglem11 ⊢ Base Q = ⋃ y ∈ ℤ ℤ × y
29 27 28 eleqtrri ⊢ ℤ × 1 ∈ Base Q
30 oveq1 ⊢ i = ℤ × 1 → i ⋅ Q x = ℤ × 1 ⋅ Q x
31 30 eqeq1d ⊢ i = ℤ × 1 → i ⋅ Q x = x ↔ ℤ × 1 ⋅ Q x = x
32 31 ovanraleqv ⊢ i = ℤ × 1 → ∀ x ∈ Base Q i ⋅ Q x = x ∧ x ⋅ Q i = x ↔ ∀ x ∈ Base Q ℤ × 1 ⋅ Q x = x ∧ x ⋅ Q ℤ × 1 = x
33 id ⊢ ℤ × 1 ∈ Base Q → ℤ × 1 ∈ Base Q
34 1 2 3 4 5 6 pzriprnglem12 ⊢ x ∈ Base Q → ℤ × 1 ⋅ Q x = x ∧ x ⋅ Q ℤ × 1 = x
35 34 a1i ⊢ ℤ × 1 ∈ Base Q → x ∈ Base Q → ℤ × 1 ⋅ Q x = x ∧ x ⋅ Q ℤ × 1 = x
36 35 ralrimiv ⊢ ℤ × 1 ∈ Base Q → ∀ x ∈ Base Q ℤ × 1 ⋅ Q x = x ∧ x ⋅ Q ℤ × 1 = x
37 32 33 36 rspcedvdw ⊢ ℤ × 1 ∈ Base Q → ∃ i ∈ Base Q ∀ x ∈ Base Q i ⋅ Q x = x ∧ x ⋅ Q i = x
38 29 37 ax-mp ⊢ ∃ i ∈ Base Q ∀ x ∈ Base Q i ⋅ Q x = x ∧ x ⋅ Q i = x
39 eqid ⊢ Base Q = Base Q
40 eqid ⊢ ⋅ Q = ⋅ Q
41 39 40 isringrng ⊢ Q ∈ Ring ↔ Q ∈ Rng ∧ ∃ i ∈ Base Q ∀ x ∈ Base Q i ⋅ Q x = x ∧ x ⋅ Q i = x
42 14 38 41 mpbir2an ⊢ Q ∈ Ring