Metamath Proof Explorer


Theorem rngqiprngfulem2

Description: Lemma 2 for rngqiprngfu (and lemma for rngqiprngu ). (Contributed by AV, 16-Mar-2025)

Ref Expression
Hypotheses rngqiprngfu.r ⊢ φ → R ∈ Rng
rngqiprngfu.i ⊢ φ → I ∈ 2Ideal ⁡ R
rngqiprngfu.j ⊢ J = R ↾ 𝑠 I
rngqiprngfu.u ⊢ φ → J ∈ Ring
rngqiprngfu.b ⊢ B = Base R
rngqiprngfu.t ⊢ · ˙ = ⋅ R
rngqiprngfu.1 ⊢ 1 ˙ = 1 J
rngqiprngfu.g ⊢ ∼ ˙ = R ~ QG I
rngqiprngfu.q ⊢ Q = R / 𝑠 ∼ ˙
rngqiprngfu.v ⊢ φ → Q ∈ Ring
rngqiprngfu.e ⊢ φ → E ∈ 1 Q
Assertion rngqiprngfulem2 ⊢ φ → E ∈ B

Proof

Step Hyp Ref Expression
1 rngqiprngfu.r ⊢ φ → R ∈ Rng
2 rngqiprngfu.i ⊢ φ → I ∈ 2Ideal ⁡ R
3 rngqiprngfu.j ⊢ J = R ↾ 𝑠 I
4 rngqiprngfu.u ⊢ φ → J ∈ Ring
5 rngqiprngfu.b ⊢ B = Base R
6 rngqiprngfu.t ⊢ · ˙ = ⋅ R
7 rngqiprngfu.1 ⊢ 1 ˙ = 1 J
8 rngqiprngfu.g ⊢ ∼ ˙ = R ~ QG I
9 rngqiprngfu.q ⊢ Q = R / 𝑠 ∼ ˙
10 rngqiprngfu.v ⊢ φ → Q ∈ Ring
11 rngqiprngfu.e ⊢ φ → E ∈ 1 Q
12 1 2 3 4 5 6 7 8 9 10 rngqiprngfulem1 ⊢ φ → ∃ x ∈ B 1 Q = x ∼ ˙
13 11 adantr ⊢ φ ∧ x ∈ B → E ∈ 1 Q
14 eleq2 ⊢ 1 Q = x ∼ ˙ → E ∈ 1 Q ↔ E ∈ x ∼ ˙
15 14 adantl ⊢ φ ∧ x ∈ B ∧ 1 Q = x ∼ ˙ → E ∈ 1 Q ↔ E ∈ x ∼ ˙
16 elecg ⊢ E ∈ 1 Q ∧ x ∈ B → E ∈ x ∼ ˙ ↔ x ∼ ˙ E
17 11 16 sylan ⊢ φ ∧ x ∈ B → E ∈ x ∼ ˙ ↔ x ∼ ˙ E
18 rngabl ⊢ R ∈ Rng → R ∈ Abel
19 1 18 syl ⊢ φ → R ∈ Abel
20 eqid ⊢ 2Ideal ⁡ R = 2Ideal ⁡ R
21 5 20 2idlss ⊢ I ∈ 2Ideal ⁡ R → I ⊆ B
22 2 21 syl ⊢ φ → I ⊆ B
23 19 22 jca ⊢ φ → R ∈ Abel ∧ I ⊆ B
24 23 adantr ⊢ φ ∧ x ∈ B → R ∈ Abel ∧ I ⊆ B
25 eqid ⊢ - R = - R
26 5 25 8 eqgabl ⊢ R ∈ Abel ∧ I ⊆ B → x ∼ ˙ E ↔ x ∈ B ∧ E ∈ B ∧ E - R x ∈ I
27 24 26 syl ⊢ φ ∧ x ∈ B → x ∼ ˙ E ↔ x ∈ B ∧ E ∈ B ∧ E - R x ∈ I
28 simp2 ⊢ x ∈ B ∧ E ∈ B ∧ E - R x ∈ I → E ∈ B
29 27 28 biimtrdi ⊢ φ ∧ x ∈ B → x ∼ ˙ E → E ∈ B
30 17 29 sylbid ⊢ φ ∧ x ∈ B → E ∈ x ∼ ˙ → E ∈ B
31 30 adantr ⊢ φ ∧ x ∈ B ∧ 1 Q = x ∼ ˙ → E ∈ x ∼ ˙ → E ∈ B
32 15 31 sylbid ⊢ φ ∧ x ∈ B ∧ 1 Q = x ∼ ˙ → E ∈ 1 Q → E ∈ B
33 32 ex ⊢ φ ∧ x ∈ B → 1 Q = x ∼ ˙ → E ∈ 1 Q → E ∈ B
34 13 33 mpid ⊢ φ ∧ x ∈ B → 1 Q = x ∼ ˙ → E ∈ B
35 34 rexlimdva ⊢ φ → ∃ x ∈ B 1 Q = x ∼ ˙ → E ∈ B
36 12 35 mpd ⊢ φ → E ∈ B