Metamath Proof Explorer


Theorem lsppropd

Description: If two structures have the same components (properties), they have the same span function. (Contributed by Mario Carneiro, 9-Feb-2015) (Revised by Mario Carneiro, 14-Jun-2015) (Revised by AV, 24-Apr-2024)

Ref Expression
Hypotheses lsspropd.b1 ⊢ φ → B = Base K
lsspropd.b2 ⊢ φ → B = Base L
lsspropd.w ⊢ φ → B ⊆ W
lsspropd.p ⊢ φ ∧ x ∈ W ∧ y ∈ W → x + K y = x + L y
lsspropd.s1 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y ∈ W
lsspropd.s2 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y = x ⋅ L y
lsspropd.p1 ⊢ φ → P = Base Scalar ⁡ K
lsspropd.p2 ⊢ φ → P = Base Scalar ⁡ L
lsppropd.v1 ⊢ φ → K ∈ X
lsppropd.v2 ⊢ φ → L ∈ Y
Assertion lsppropd ⊢ φ → LSpan ⁡ K = LSpan ⁡ L

Proof

Step Hyp Ref Expression
1 lsspropd.b1 ⊢ φ → B = Base K
2 lsspropd.b2 ⊢ φ → B = Base L
3 lsspropd.w ⊢ φ → B ⊆ W
4 lsspropd.p ⊢ φ ∧ x ∈ W ∧ y ∈ W → x + K y = x + L y
5 lsspropd.s1 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y ∈ W
6 lsspropd.s2 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y = x ⋅ L y
7 lsspropd.p1 ⊢ φ → P = Base Scalar ⁡ K
8 lsspropd.p2 ⊢ φ → P = Base Scalar ⁡ L
9 lsppropd.v1 ⊢ φ → K ∈ X
10 lsppropd.v2 ⊢ φ → L ∈ Y
11 1 2 eqtr3d ⊢ φ → Base K = Base L
12 11 pweqd ⊢ φ → 𝒫 Base K = 𝒫 Base L
13 1 2 3 4 5 6 7 8 lsspropd ⊢ φ → LSubSp ⁡ K = LSubSp ⁡ L
14 13 rabeqdv ⊢ φ → t ∈ LSubSp ⁡ K | s ⊆ t = t ∈ LSubSp ⁡ L | s ⊆ t
15 14 inteqd ⊢ φ → ⋂ t ∈ LSubSp ⁡ K | s ⊆ t = ⋂ t ∈ LSubSp ⁡ L | s ⊆ t
16 12 15 mpteq12dv ⊢ φ → s ∈ 𝒫 Base K ⟼ ⋂ t ∈ LSubSp ⁡ K | s ⊆ t = s ∈ 𝒫 Base L ⟼ ⋂ t ∈ LSubSp ⁡ L | s ⊆ t
17 eqid ⊢ Base K = Base K
18 eqid ⊢ LSubSp ⁡ K = LSubSp ⁡ K
19 eqid ⊢ LSpan ⁡ K = LSpan ⁡ K
20 17 18 19 lspfval ⊢ K ∈ X → LSpan ⁡ K = s ∈ 𝒫 Base K ⟼ ⋂ t ∈ LSubSp ⁡ K | s ⊆ t
21 9 20 syl ⊢ φ → LSpan ⁡ K = s ∈ 𝒫 Base K ⟼ ⋂ t ∈ LSubSp ⁡ K | s ⊆ t
22 eqid ⊢ Base L = Base L
23 eqid ⊢ LSubSp ⁡ L = LSubSp ⁡ L
24 eqid ⊢ LSpan ⁡ L = LSpan ⁡ L
25 22 23 24 lspfval ⊢ L ∈ Y → LSpan ⁡ L = s ∈ 𝒫 Base L ⟼ ⋂ t ∈ LSubSp ⁡ L | s ⊆ t
26 10 25 syl ⊢ φ → LSpan ⁡ L = s ∈ 𝒫 Base L ⟼ ⋂ t ∈ LSubSp ⁡ L | s ⊆ t
27 16 21 26 3eqtr4d ⊢ φ → LSpan ⁡ K = LSpan ⁡ L