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