Metamath Proof Explorer


Theorem subrngint

Description: The intersection of a nonempty collection of subrings is a subring. (Contributed by AV, 15-Feb-2025)

Ref Expression
Assertion subrngint ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) → 𝑆 ∈ ( SubRng ‘ 𝑅 ) )

Proof

Step Hyp Ref Expression
1 subrngsubg ( 𝑟 ∈ ( SubRng ‘ 𝑅 ) → 𝑟 ∈ ( SubGrp ‘ 𝑅 ) )
2 1 ssriv ( SubRng ‘ 𝑅 ) ⊆ ( SubGrp ‘ 𝑅 )
3 sstr ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ ( SubRng ‘ 𝑅 ) ⊆ ( SubGrp ‘ 𝑅 ) ) → 𝑆 ⊆ ( SubGrp ‘ 𝑅 ) )
4 2 3 mpan2 ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) → 𝑆 ⊆ ( SubGrp ‘ 𝑅 ) )
5 subgint ( ( 𝑆 ⊆ ( SubGrp ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) → 𝑆 ∈ ( SubGrp ‘ 𝑅 ) )
6 4 5 sylan ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) → 𝑆 ∈ ( SubGrp ‘ 𝑅 ) )
7 ssel2 ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑟𝑆 ) → 𝑟 ∈ ( SubRng ‘ 𝑅 ) )
8 7 ad4ant14 ( ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) ∧ 𝑟𝑆 ) → 𝑟 ∈ ( SubRng ‘ 𝑅 ) )
9 simprl ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) → 𝑥 𝑆 )
10 elinti ( 𝑥 𝑆 → ( 𝑟𝑆𝑥𝑟 ) )
11 10 imp ( ( 𝑥 𝑆𝑟𝑆 ) → 𝑥𝑟 )
12 9 11 sylan ( ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) ∧ 𝑟𝑆 ) → 𝑥𝑟 )
13 simprr ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) → 𝑦 𝑆 )
14 elinti ( 𝑦 𝑆 → ( 𝑟𝑆𝑦𝑟 ) )
15 14 imp ( ( 𝑦 𝑆𝑟𝑆 ) → 𝑦𝑟 )
16 13 15 sylan ( ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) ∧ 𝑟𝑆 ) → 𝑦𝑟 )
17 eqid ( .r𝑅 ) = ( .r𝑅 )
18 17 subrngmcl ( ( 𝑟 ∈ ( SubRng ‘ 𝑅 ) ∧ 𝑥𝑟𝑦𝑟 ) → ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑟 )
19 8 12 16 18 syl3anc ( ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) ∧ 𝑟𝑆 ) → ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑟 )
20 19 ralrimiva ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) → ∀ 𝑟𝑆 ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑟 )
21 ovex ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ V
22 21 elint2 ( ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑆 ↔ ∀ 𝑟𝑆 ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑟 )
23 20 22 sylibr ( ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) ∧ ( 𝑥 𝑆𝑦 𝑆 ) ) → ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑆 )
24 23 ralrimivva ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) → ∀ 𝑥 𝑆𝑦 𝑆 ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑆 )
25 ssn0 ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) → ( SubRng ‘ 𝑅 ) ≠ ∅ )
26 n0 ( ( SubRng ‘ 𝑅 ) ≠ ∅ ↔ ∃ 𝑟 𝑟 ∈ ( SubRng ‘ 𝑅 ) )
27 subrngrcl ( 𝑟 ∈ ( SubRng ‘ 𝑅 ) → 𝑅 ∈ Rng )
28 27 exlimiv ( ∃ 𝑟 𝑟 ∈ ( SubRng ‘ 𝑅 ) → 𝑅 ∈ Rng )
29 26 28 sylbi ( ( SubRng ‘ 𝑅 ) ≠ ∅ → 𝑅 ∈ Rng )
30 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
31 30 17 issubrng2 ( 𝑅 ∈ Rng → ( 𝑆 ∈ ( SubRng ‘ 𝑅 ) ↔ ( 𝑆 ∈ ( SubGrp ‘ 𝑅 ) ∧ ∀ 𝑥 𝑆𝑦 𝑆 ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑆 ) ) )
32 25 29 31 3syl ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) → ( 𝑆 ∈ ( SubRng ‘ 𝑅 ) ↔ ( 𝑆 ∈ ( SubGrp ‘ 𝑅 ) ∧ ∀ 𝑥 𝑆𝑦 𝑆 ( 𝑥 ( .r𝑅 ) 𝑦 ) ∈ 𝑆 ) ) )
33 6 24 32 mpbir2and ( ( 𝑆 ⊆ ( SubRng ‘ 𝑅 ) ∧ 𝑆 ≠ ∅ ) → 𝑆 ∈ ( SubRng ‘ 𝑅 ) )