Metamath Proof Explorer


Theorem dimpropd

Description: If two structures have the same components (properties), they have the same dimension. (Contributed by Thierry Arnoux, 18-May-2023)

Ref Expression
Hypotheses dimpropd.b1 ⊢ φ → B = Base K
dimpropd.b2 ⊢ φ → B = Base L
dimpropd.w ⊢ φ → B ⊆ W
dimpropd.p ⊢ φ ∧ x ∈ W ∧ y ∈ W → x + K y = x + L y
dimpropd.s1 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y ∈ W
dimpropd.s2 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y = x ⋅ L y
dimpropd.f ⊢ F = Scalar ⁡ K
dimpropd.g ⊢ G = Scalar ⁡ L
dimpropd.p1 ⊢ φ → P = Base F
dimpropd.p2 ⊢ φ → P = Base G
dimpropd.a ⊢ φ ∧ x ∈ P ∧ y ∈ P → x + F y = x + G y
dimpropd.v1 ⊢ φ → K ∈ LVec
dimpropd.v2 ⊢ φ → L ∈ LVec
Assertion dimpropd ⊢ φ → dim ⁡ K = dim ⁡ L

Proof

Step Hyp Ref Expression
1 dimpropd.b1 ⊢ φ → B = Base K
2 dimpropd.b2 ⊢ φ → B = Base L
3 dimpropd.w ⊢ φ → B ⊆ W
4 dimpropd.p ⊢ φ ∧ x ∈ W ∧ y ∈ W → x + K y = x + L y
5 dimpropd.s1 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y ∈ W
6 dimpropd.s2 ⊢ φ ∧ x ∈ P ∧ y ∈ B → x ⋅ K y = x ⋅ L y
7 dimpropd.f ⊢ F = Scalar ⁡ K
8 dimpropd.g ⊢ G = Scalar ⁡ L
9 dimpropd.p1 ⊢ φ → P = Base F
10 dimpropd.p2 ⊢ φ → P = Base G
11 dimpropd.a ⊢ φ ∧ x ∈ P ∧ y ∈ P → x + F y = x + G y
12 dimpropd.v1 ⊢ φ → K ∈ LVec
13 dimpropd.v2 ⊢ φ → L ∈ LVec
14 eqid ⊢ LBasis ⁡ K = LBasis ⁡ K
15 14 lbsex ⊢ K ∈ LVec → LBasis ⁡ K ≠ ∅
16 12 15 syl ⊢ φ → LBasis ⁡ K ≠ ∅
17 n0 ⊢ LBasis ⁡ K ≠ ∅ ↔ ∃ x x ∈ LBasis ⁡ K
18 16 17 sylib ⊢ φ → ∃ x x ∈ LBasis ⁡ K
19 14 dimval ⊢ K ∈ LVec ∧ x ∈ LBasis ⁡ K → dim ⁡ K = x
20 12 19 sylan ⊢ φ ∧ x ∈ LBasis ⁡ K → dim ⁡ K = x
21 1 2 3 4 5 6 7 8 9 10 11 12 13 lbspropd ⊢ φ → LBasis ⁡ K = LBasis ⁡ L
22 21 eleq2d ⊢ φ → x ∈ LBasis ⁡ K ↔ x ∈ LBasis ⁡ L
23 22 biimpa ⊢ φ ∧ x ∈ LBasis ⁡ K → x ∈ LBasis ⁡ L
24 eqid ⊢ LBasis ⁡ L = LBasis ⁡ L
25 24 dimval ⊢ L ∈ LVec ∧ x ∈ LBasis ⁡ L → dim ⁡ L = x
26 13 23 25 syl2an2r ⊢ φ ∧ x ∈ LBasis ⁡ K → dim ⁡ L = x
27 20 26 eqtr4d ⊢ φ ∧ x ∈ LBasis ⁡ K → dim ⁡ K = dim ⁡ L
28 18 27 exlimddv ⊢ φ → dim ⁡ K = dim ⁡ L