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 = ( 𝑅s 𝐴 )
subrng0.2 0 = ( 0g𝑅 )
Assertion subrng0 ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 0 = ( 0g𝑆 ) )

Proof

Step Hyp Ref Expression
1 subrng0.1 𝑆 = ( 𝑅s 𝐴 )
2 subrng0.2 0 = ( 0g𝑅 )
3 subrngsubg ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝐴 ∈ ( SubGrp ‘ 𝑅 ) )
4 1 2 subg0 ( 𝐴 ∈ ( SubGrp ‘ 𝑅 ) → 0 = ( 0g𝑆 ) )
5 3 4 syl ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 0 = ( 0g𝑆 ) )