Metamath Proof Explorer


Theorem subrngid

Description: Every non-unital ring is a subring of itself. (Contributed by AV, 14-Feb-2025)

Ref Expression
Hypothesis subrngss.1 ⊢ 𝐵 = ( Base ‘ 𝑅 )
Assertion subrngid ( 𝑅 ∈ Rng → 𝐵 ∈ ( SubRng ‘ 𝑅 ) )

Proof

Step Hyp Ref Expression
1 subrngss.1 ⊢ 𝐵 = ( Base ‘ 𝑅 )
2 id ⊢ ( 𝑅 ∈ Rng → 𝑅 ∈ Rng )
3 1 ressid ⊢ ( 𝑅 ∈ Rng → ( 𝑅 ↾s 𝐵 ) = 𝑅 )
4 3 2 eqeltrd ⊢ ( 𝑅 ∈ Rng → ( 𝑅 ↾s 𝐵 ) ∈ Rng )
5 ssidd ⊢ ( 𝑅 ∈ Rng → 𝐵 ⊆ 𝐵 )
6 1 issubrng ⊢ ( 𝐵 ∈ ( SubRng ‘ 𝑅 ) ↔ ( 𝑅 ∈ Rng ∧ ( 𝑅 ↾s 𝐵 ) ∈ Rng ∧ 𝐵 ⊆ 𝐵 ) )
7 2 4 5 6 syl3anbrc ⊢ ( 𝑅 ∈ Rng → 𝐵 ∈ ( SubRng ‘ 𝑅 ) )