Metamath Proof Explorer


Theorem srgisid

Description: In a semiring, the only left-absorbing element is the additive identity. Remark in Golan p. 1. (Contributed by Thierry Arnoux, 1-May-2018)

Ref Expression
Hypotheses srgz.b
|- B = ( Base ` R )
srgz.t
|- .x. = ( .r ` R )
srgz.z
|- .0. = ( 0g ` R )
srgisid.1
|- ( ph -> R e. SRing )
srgisid.2
|- ( ph -> Z e. B )
srgisid.3
|- ( ( ph /\ x e. B ) -> ( Z .x. x ) = Z )
Assertion srgisid
|- ( ph -> Z = .0. )

Proof

Step Hyp Ref Expression
1 srgz.b
 |-  B = ( Base ` R )
2 srgz.t
 |-  .x. = ( .r ` R )
3 srgz.z
 |-  .0. = ( 0g ` R )
4 srgisid.1
 |-  ( ph -> R e. SRing )
5 srgisid.2
 |-  ( ph -> Z e. B )
6 srgisid.3
 |-  ( ( ph /\ x e. B ) -> ( Z .x. x ) = Z )
7 6 ralrimiva
 |-  ( ph -> A. x e. B ( Z .x. x ) = Z )
8 1 3 srg0cl
 |-  ( R e. SRing -> .0. e. B )
9 oveq2
 |-  ( x = .0. -> ( Z .x. x ) = ( Z .x. .0. ) )
10 9 eqeq1d
 |-  ( x = .0. -> ( ( Z .x. x ) = Z <-> ( Z .x. .0. ) = Z ) )
11 10 rspcv
 |-  ( .0. e. B -> ( A. x e. B ( Z .x. x ) = Z -> ( Z .x. .0. ) = Z ) )
12 4 8 11 3syl
 |-  ( ph -> ( A. x e. B ( Z .x. x ) = Z -> ( Z .x. .0. ) = Z ) )
13 7 12 mpd
 |-  ( ph -> ( Z .x. .0. ) = Z )
14 1 2 3 srgrz
 |-  ( ( R e. SRing /\ Z e. B ) -> ( Z .x. .0. ) = .0. )
15 4 5 14 syl2anc
 |-  ( ph -> ( Z .x. .0. ) = .0. )
16 13 15 eqtr3d
 |-  ( ph -> Z = .0. )