Metamath Proof Explorer


Theorem subrng0

Description: A subring always has the same additive identity. (Contributed by AV, 14-Feb-2025)

Ref Expression
Hypotheses subrng0.1
|- S = ( R |`s A )
subrng0.2
|- .0. = ( 0g ` R )
Assertion subrng0
|- ( A e. ( SubRng ` R ) -> .0. = ( 0g ` S ) )

Proof

Step Hyp Ref Expression
1 subrng0.1
 |-  S = ( R |`s A )
2 subrng0.2
 |-  .0. = ( 0g ` R )
3 subrngsubg
 |-  ( A e. ( SubRng ` R ) -> A e. ( SubGrp ` R ) )
4 1 2 subg0
 |-  ( A e. ( SubGrp ` R ) -> .0. = ( 0g ` S ) )
5 3 4 syl
 |-  ( A e. ( SubRng ` R ) -> .0. = ( 0g ` S ) )