Metamath Proof Explorer


Theorem subrg0

Description: A subring always has the same additive identity. (Contributed by Stefan O'Rear, 27-Nov-2014)

Ref Expression
Hypotheses subrg0.1 ⊢ 𝑆 = ( 𝑅 ↾s 𝐴 )
subrg0.2 ⊢ 0 = ( 0g ‘ 𝑅 )
Assertion subrg0 ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 0 = ( 0g ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 subrg0.1 ⊢ 𝑆 = ( 𝑅 ↾s 𝐴 )
2 subrg0.2 ⊢ 0 = ( 0g ‘ 𝑅 )
3 subrgsubg ⊢ ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝐴 ∈ ( SubGrp ‘ 𝑅 ) )
4 1 2 subg0 ⊢ ( 𝐴 ∈ ( SubGrp ‘ 𝑅 ) → 0 = ( 0g ‘ 𝑆 ) )
5 3 4 syl ⊢ ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 0 = ( 0g ‘ 𝑆 ) )