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=ScalarK
asclpropd.g G=ScalarL
asclpropd.1 φP=BaseF
asclpropd.2 φP=BaseG
asclpropd.3 φxPyWxKy=xLy
asclpropd.4 φ1K=1L
asclpropd.5 φ1KW
Assertion asclpropd φalgScK=algScL

Proof

Step Hyp Ref Expression
1 asclpropd.f F=ScalarK
2 asclpropd.g G=ScalarL
3 asclpropd.1 φP=BaseF
4 asclpropd.2 φP=BaseG
5 asclpropd.3 φxPyWxKy=xLy
6 asclpropd.4 φ1K=1L
7 asclpropd.5 φ1KW
8 5 oveqrspc2v φzP1KWzK1K=zL1K
9 8 anassrs φzP1KWzK1K=zL1K
10 7 9 mpidan φzPzK1K=zL1K
11 6 oveq2d φzL1K=zL1L
12 11 adantr φzPzL1K=zL1L
13 10 12 eqtrd φzPzK1K=zL1L
14 13 mpteq2dva φzPzK1K=zPzL1L
15 3 mpteq1d φzPzK1K=zBaseFzK1K
16 4 mpteq1d φzPzL1L=zBaseGzL1L
17 14 15 16 3eqtr3d φzBaseFzK1K=zBaseGzL1L
18 eqid algScK=algScK
19 eqid BaseF=BaseF
20 eqid K=K
21 eqid 1K=1K
22 18 1 19 20 21 asclfval algScK=zBaseFzK1K
23 eqid algScL=algScL
24 eqid BaseG=BaseG
25 eqid L=L
26 eqid 1L=1L
27 23 2 24 25 26 asclfval algScL=zBaseGzL1L
28 17 22 27 3eqtr4g φalgScK=algScL