Metamath Proof Explorer


Theorem rngqiprngfulem1

Description: Lemma 1 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
Assertion rngqiprngfulem1 ⊢ φ → ∃ x ∈ B 1 Q = x ∼ ˙

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 eqid ⊢ Base Q = Base Q
12 eqid ⊢ 1 Q = 1 Q
13 11 12 ringidcl ⊢ Q ∈ Ring → 1 Q ∈ Base Q
14 10 13 syl ⊢ φ → 1 Q ∈ Base Q
15 9 a1i ⊢ φ → Q = R / 𝑠 ∼ ˙
16 5 a1i ⊢ φ → B = Base R
17 8 ovexi ⊢ ∼ ˙ ∈ V
18 17 a1i ⊢ φ → ∼ ˙ ∈ V
19 15 16 18 1 qusbas ⊢ φ → B / ∼ ˙ = Base Q
20 14 19 eleqtrrd ⊢ φ → 1 Q ∈ B / ∼ ˙
21 fvexd ⊢ φ → 1 Q ∈ V
22 elqsg ⊢ 1 Q ∈ V → 1 Q ∈ B / ∼ ˙ ↔ ∃ x ∈ B 1 Q = x ∼ ˙
23 21 22 syl ⊢ φ → 1 Q ∈ B / ∼ ˙ ↔ ∃ x ∈ B 1 Q = x ∼ ˙
24 20 23 mpbid ⊢ φ → ∃ x ∈ B 1 Q = x ∼ ˙