Metamath Proof Explorer


Theorem cmnpropd

Description: If two structures have the same group components (properties), one is a commutative monoid iff the other one is. (Contributed by Mario Carneiro, 6-Jan-2015)

Ref Expression
Hypotheses ablpropd.1 φ B = Base K
ablpropd.2 φ B = Base L
ablpropd.3 φ x B y B x + K y = x + L y
Assertion cmnpropd φ K CMnd L CMnd

Proof

Step Hyp Ref Expression
1 ablpropd.1 φ B = Base K
2 ablpropd.2 φ B = Base L
3 ablpropd.3 φ x B y B x + K y = x + L y
4 1 2 3 mndpropd φ K Mnd L Mnd
5 3 oveqrspc2v φ u B v B u + K v = u + L v
6 3 oveqrspc2v φ v B u B v + K u = v + L u
7 6 ancom2s φ u B v B v + K u = v + L u
8 5 7 eqeq12d φ u B v B u + K v = v + K u u + L v = v + L u
9 8 2ralbidva φ u B v B u + K v = v + K u u B v B u + L v = v + L u
10 1 raleqdv φ v B u + K v = v + K u v Base K u + K v = v + K u
11 1 10 raleqbidv φ u B v B u + K v = v + K u u Base K v Base K u + K v = v + K u
12 2 raleqdv φ v B u + L v = v + L u v Base L u + L v = v + L u
13 2 12 raleqbidv φ u B v B u + L v = v + L u u Base L v Base L u + L v = v + L u
14 9 11 13 3bitr3d φ u Base K v Base K u + K v = v + K u u Base L v Base L u + L v = v + L u
15 4 14 anbi12d φ K Mnd u Base K v Base K u + K v = v + K u L Mnd u Base L v Base L u + L v = v + L u
16 eqid Base K = Base K
17 eqid + K = + K
18 16 17 iscmn K CMnd K Mnd u Base K v Base K u + K v = v + K u
19 eqid Base L = Base L
20 eqid + L = + L
21 19 20 iscmn L CMnd L Mnd u Base L v Base L u + L v = v + L u
22 15 18 21 3bitr4g φ K CMnd L CMnd