Metamath Proof Explorer


Theorem asclpropd

Description: If two structures have the same components (properties), one is an associative algebra iff the other one is. The last hypotheses on 1r can be discharged either by letting W = _V (if strong equality is known on .s ) or assuming K is a ring. (Contributed by Mario Carneiro, 5-Jul-2015)

Ref Expression
Hypotheses asclpropd.f ⊢ F = Scalar ⁡ K
asclpropd.g ⊢ G = Scalar ⁡ L
asclpropd.1 ⊢ φ → P = Base F
asclpropd.2 ⊢ φ → P = Base G
asclpropd.3 ⊢ φ ∧ x ∈ P ∧ y ∈ W → x ⋅ K y = x ⋅ L y
asclpropd.4 ⊢ φ → 1 K = 1 L
asclpropd.5 ⊢ φ → 1 K ∈ W
Assertion asclpropd ⊢ φ → algSc ⁡ K = algSc ⁡ L

Proof

Step Hyp Ref Expression
1 asclpropd.f ⊢ F = Scalar ⁡ K
2 asclpropd.g ⊢ G = Scalar ⁡ L
3 asclpropd.1 ⊢ φ → P = Base F
4 asclpropd.2 ⊢ φ → P = Base G
5 asclpropd.3 ⊢ φ ∧ x ∈ P ∧ y ∈ W → x ⋅ K y = x ⋅ L y
6 asclpropd.4 ⊢ φ → 1 K = 1 L
7 asclpropd.5 ⊢ φ → 1 K ∈ W
8 5 oveqrspc2v ⊢ φ ∧ z ∈ P ∧ 1 K ∈ W → z ⋅ K 1 K = z ⋅ L 1 K
9 8 anassrs ⊢ φ ∧ z ∈ P ∧ 1 K ∈ W → z ⋅ K 1 K = z ⋅ L 1 K
10 7 9 mpidan ⊢ φ ∧ z ∈ P → z ⋅ K 1 K = z ⋅ L 1 K
11 6 oveq2d ⊢ φ → z ⋅ L 1 K = z ⋅ L 1 L
12 11 adantr ⊢ φ ∧ z ∈ P → z ⋅ L 1 K = z ⋅ L 1 L
13 10 12 eqtrd ⊢ φ ∧ z ∈ P → z ⋅ K 1 K = z ⋅ L 1 L
14 13 mpteq2dva ⊢ φ → z ∈ P ⟼ z ⋅ K 1 K = z ∈ P ⟼ z ⋅ L 1 L
15 3 mpteq1d ⊢ φ → z ∈ P ⟼ z ⋅ K 1 K = z ∈ Base F ⟼ z ⋅ K 1 K
16 4 mpteq1d ⊢ φ → z ∈ P ⟼ z ⋅ L 1 L = z ∈ Base G ⟼ z ⋅ L 1 L
17 14 15 16 3eqtr3d ⊢ φ → z ∈ Base F ⟼ z ⋅ K 1 K = z ∈ Base G ⟼ z ⋅ L 1 L
18 eqid ⊢ algSc ⁡ K = algSc ⁡ K
19 eqid ⊢ Base F = Base F
20 eqid ⊢ ⋅ K = ⋅ K
21 eqid ⊢ 1 K = 1 K
22 18 1 19 20 21 asclfval ⊢ algSc ⁡ K = z ∈ Base F ⟼ z ⋅ K 1 K
23 eqid ⊢ algSc ⁡ L = algSc ⁡ L
24 eqid ⊢ Base G = Base G
25 eqid ⊢ ⋅ L = ⋅ L
26 eqid ⊢ 1 L = 1 L
27 23 2 24 25 26 asclfval ⊢ algSc ⁡ L = z ∈ Base G ⟼ z ⋅ L 1 L
28 17 22 27 3eqtr4g ⊢ φ → algSc ⁡ K = algSc ⁡ L