Metamath Proof Explorer


Theorem ascl0

Description: The scalar 0 embedded into a left module corresponds to the 0 of the left module if the left module is also a ring. (Contributed by AV, 31-Jul-2019)

Ref Expression
Hypotheses ascl0.a A=algScW
ascl0.f F=ScalarW
ascl0.l φWLMod
ascl0.r φWRing
Assertion ascl0 φA0F=0W

Proof

Step Hyp Ref Expression
1 ascl0.a A=algScW
2 ascl0.f F=ScalarW
3 ascl0.l φWLMod
4 ascl0.r φWRing
5 2 lmodfgrp WLModFGrp
6 3 5 syl φFGrp
7 eqid BaseF=BaseF
8 eqid 0F=0F
9 7 8 grpidcl FGrp0FBaseF
10 6 9 syl φ0FBaseF
11 eqid W=W
12 eqid 1W=1W
13 1 2 7 11 12 asclval 0FBaseFA0F=0FW1W
14 10 13 syl φA0F=0FW1W
15 eqid BaseW=BaseW
16 15 12 ringidcl WRing1WBaseW
17 4 16 syl φ1WBaseW
18 eqid 0W=0W
19 15 2 11 8 18 lmod0vs WLMod1WBaseW0FW1W=0W
20 3 17 19 syl2anc φ0FW1W=0W
21 14 20 eqtrd φA0F=0W