Metamath Proof Explorer


Theorem subrngpropd

Description: If two structures have the same ring components (properties), they have the same set of subrings. (Contributed by AV, 17-Feb-2025)

Ref Expression
Hypotheses subrngpropd.1 ⊢ φ → B = Base K
subrngpropd.2 ⊢ φ → B = Base L
subrngpropd.3 ⊢ φ ∧ x ∈ B ∧ y ∈ B → x + K y = x + L y
subrngpropd.4 ⊢ φ ∧ x ∈ B ∧ y ∈ B → x ⋅ K y = x ⋅ L y
Assertion subrngpropd ⊢ φ → SubRng ⁡ K = SubRng ⁡ L

Proof

Step Hyp Ref Expression
1 subrngpropd.1 ⊢ φ → B = Base K
2 subrngpropd.2 ⊢ φ → B = Base L
3 subrngpropd.3 ⊢ φ ∧ x ∈ B ∧ y ∈ B → x + K y = x + L y
4 subrngpropd.4 ⊢ φ ∧ x ∈ B ∧ y ∈ B → x ⋅ K y = x ⋅ L y
5 1 2 3 4 rngpropd ⊢ φ → K ∈ Rng ↔ L ∈ Rng
6 1 ineq2d ⊢ φ → s ∩ B = s ∩ Base K
7 eqid ⊢ K ↾ 𝑠 s = K ↾ 𝑠 s
8 eqid ⊢ Base K = Base K
9 7 8 ressbas ⊢ s ∈ V → s ∩ Base K = Base K ↾ 𝑠 s
10 9 elv ⊢ s ∩ Base K = Base K ↾ 𝑠 s
11 6 10 eqtrdi ⊢ φ → s ∩ B = Base K ↾ 𝑠 s
12 2 ineq2d ⊢ φ → s ∩ B = s ∩ Base L
13 eqid ⊢ L ↾ 𝑠 s = L ↾ 𝑠 s
14 eqid ⊢ Base L = Base L
15 13 14 ressbas ⊢ s ∈ V → s ∩ Base L = Base L ↾ 𝑠 s
16 15 elv ⊢ s ∩ Base L = Base L ↾ 𝑠 s
17 12 16 eqtrdi ⊢ φ → s ∩ B = Base L ↾ 𝑠 s
18 elinel2 ⊢ x ∈ s ∩ B → x ∈ B
19 elinel2 ⊢ y ∈ s ∩ B → y ∈ B
20 18 19 anim12i ⊢ x ∈ s ∩ B ∧ y ∈ s ∩ B → x ∈ B ∧ y ∈ B
21 eqid ⊢ + K = + K
22 7 21 ressplusg ⊢ s ∈ V → + K = + K ↾ 𝑠 s
23 22 elv ⊢ + K = + K ↾ 𝑠 s
24 23 oveqi ⊢ x + K y = x + K ↾ 𝑠 s y
25 eqid ⊢ + L = + L
26 13 25 ressplusg ⊢ s ∈ V → + L = + L ↾ 𝑠 s
27 26 elv ⊢ + L = + L ↾ 𝑠 s
28 27 oveqi ⊢ x + L y = x + L ↾ 𝑠 s y
29 3 24 28 3eqtr3g ⊢ φ ∧ x ∈ B ∧ y ∈ B → x + K ↾ 𝑠 s y = x + L ↾ 𝑠 s y
30 20 29 sylan2 ⊢ φ ∧ x ∈ s ∩ B ∧ y ∈ s ∩ B → x + K ↾ 𝑠 s y = x + L ↾ 𝑠 s y
31 eqid ⊢ ⋅ K = ⋅ K
32 7 31 ressmulr ⊢ s ∈ V → ⋅ K = ⋅ K ↾ 𝑠 s
33 32 elv ⊢ ⋅ K = ⋅ K ↾ 𝑠 s
34 33 oveqi ⊢ x ⋅ K y = x ⋅ K ↾ 𝑠 s y
35 eqid ⊢ ⋅ L = ⋅ L
36 13 35 ressmulr ⊢ s ∈ V → ⋅ L = ⋅ L ↾ 𝑠 s
37 36 elv ⊢ ⋅ L = ⋅ L ↾ 𝑠 s
38 37 oveqi ⊢ x ⋅ L y = x ⋅ L ↾ 𝑠 s y
39 4 34 38 3eqtr3g ⊢ φ ∧ x ∈ B ∧ y ∈ B → x ⋅ K ↾ 𝑠 s y = x ⋅ L ↾ 𝑠 s y
40 20 39 sylan2 ⊢ φ ∧ x ∈ s ∩ B ∧ y ∈ s ∩ B → x ⋅ K ↾ 𝑠 s y = x ⋅ L ↾ 𝑠 s y
41 11 17 30 40 rngpropd ⊢ φ → K ↾ 𝑠 s ∈ Rng ↔ L ↾ 𝑠 s ∈ Rng
42 1 2 eqtr3d ⊢ φ → Base K = Base L
43 42 sseq2d ⊢ φ → s ⊆ Base K ↔ s ⊆ Base L
44 5 41 43 3anbi123d ⊢ φ → K ∈ Rng ∧ K ↾ 𝑠 s ∈ Rng ∧ s ⊆ Base K ↔ L ∈ Rng ∧ L ↾ 𝑠 s ∈ Rng ∧ s ⊆ Base L
45 8 issubrng ⊢ s ∈ SubRng ⁡ K ↔ K ∈ Rng ∧ K ↾ 𝑠 s ∈ Rng ∧ s ⊆ Base K
46 14 issubrng ⊢ s ∈ SubRng ⁡ L ↔ L ∈ Rng ∧ L ↾ 𝑠 s ∈ Rng ∧ s ⊆ Base L
47 44 45 46 3bitr4g ⊢ φ → s ∈ SubRng ⁡ K ↔ s ∈ SubRng ⁡ L
48 47 eqrdv ⊢ φ → SubRng ⁡ K = SubRng ⁡ L