Metamath Proof Explorer


Theorem rngqiprngghmlem1

Description: Lemma 1 for rngqiprngghm . (Contributed by AV, 25-Feb-2025)

Ref Expression
Hypotheses rng2idlring.r ⊢ φ → R ∈ Rng
rng2idlring.i ⊢ φ → I ∈ 2Ideal ⁡ R
rng2idlring.j ⊢ J = R ↾ 𝑠 I
rng2idlring.u ⊢ φ → J ∈ Ring
rng2idlring.b ⊢ B = Base R
rng2idlring.t ⊢ · ˙ = ⋅ R
rng2idlring.1 ⊢ 1 ˙ = 1 J
Assertion rngqiprngghmlem1 ⊢ φ ∧ A ∈ B → 1 ˙ · ˙ A ∈ Base J

Proof

Step Hyp Ref Expression
1 rng2idlring.r ⊢ φ → R ∈ Rng
2 rng2idlring.i ⊢ φ → I ∈ 2Ideal ⁡ R
3 rng2idlring.j ⊢ J = R ↾ 𝑠 I
4 rng2idlring.u ⊢ φ → J ∈ Ring
5 rng2idlring.b ⊢ B = Base R
6 rng2idlring.t ⊢ · ˙ = ⋅ R
7 rng2idlring.1 ⊢ 1 ˙ = 1 J
8 eqid ⊢ Base J = Base J
9 2 3 8 2idlelbas ⊢ φ → Base J ∈ LIdeal ⁡ R ∧ Base J ∈ LIdeal ⁡ opp r ⁡ R
10 9 simprd ⊢ φ → Base J ∈ LIdeal ⁡ opp r ⁡ R
11 ringrng ⊢ J ∈ Ring → J ∈ Rng
12 4 11 syl ⊢ φ → J ∈ Rng
13 3 12 eqeltrrid ⊢ φ → R ↾ 𝑠 I ∈ Rng
14 1 2 13 rng2idl0 ⊢ φ → 0 R ∈ I
15 2 3 8 2idlbas ⊢ φ → Base J = I
16 14 15 eleqtrrd ⊢ φ → 0 R ∈ Base J
17 1 10 16 3jca ⊢ φ → R ∈ Rng ∧ Base J ∈ LIdeal ⁡ opp r ⁡ R ∧ 0 R ∈ Base J
18 8 7 ringidcl ⊢ J ∈ Ring → 1 ˙ ∈ Base J
19 4 18 syl ⊢ φ → 1 ˙ ∈ Base J
20 19 anim1ci ⊢ φ ∧ A ∈ B → A ∈ B ∧ 1 ˙ ∈ Base J
21 eqid ⊢ 0 R = 0 R
22 eqid ⊢ LIdeal ⁡ opp r ⁡ R = LIdeal ⁡ opp r ⁡ R
23 21 5 6 22 rngridlmcl ⊢ R ∈ Rng ∧ Base J ∈ LIdeal ⁡ opp r ⁡ R ∧ 0 R ∈ Base J ∧ A ∈ B ∧ 1 ˙ ∈ Base J → 1 ˙ · ˙ A ∈ Base J
24 17 20 23 syl2an2r ⊢ φ ∧ A ∈ B → 1 ˙ · ˙ A ∈ Base J