Metamath Proof Explorer


Theorem lidlsubcl

Description: An ideal is closed under subtraction. (Contributed by Stefan O'Rear, 28-Mar-2015) (Proof shortened by OpenAI, 25-Mar-2020)

Ref Expression
Hypotheses lidlcl.u 𝑈 = ( LIdeal ‘ 𝑅 )
lidlsubcl.m = ( -g𝑅 )
Assertion lidlsubcl ( ( ( 𝑅 ∈ Ring ∧ 𝐼𝑈 ) ∧ ( 𝑋𝐼𝑌𝐼 ) ) → ( 𝑋 𝑌 ) ∈ 𝐼 )

Proof

Step Hyp Ref Expression
1 lidlcl.u 𝑈 = ( LIdeal ‘ 𝑅 )
2 lidlsubcl.m = ( -g𝑅 )
3 1 lidlsubg ( ( 𝑅 ∈ Ring ∧ 𝐼𝑈 ) → 𝐼 ∈ ( SubGrp ‘ 𝑅 ) )
4 3 3adant3 ( ( 𝑅 ∈ Ring ∧ 𝐼𝑈 ∧ ( 𝑋𝐼𝑌𝐼 ) ) → 𝐼 ∈ ( SubGrp ‘ 𝑅 ) )
5 simp3l ( ( 𝑅 ∈ Ring ∧ 𝐼𝑈 ∧ ( 𝑋𝐼𝑌𝐼 ) ) → 𝑋𝐼 )
6 simp3r ( ( 𝑅 ∈ Ring ∧ 𝐼𝑈 ∧ ( 𝑋𝐼𝑌𝐼 ) ) → 𝑌𝐼 )
7 2 subgsubcl ( ( 𝐼 ∈ ( SubGrp ‘ 𝑅 ) ∧ 𝑋𝐼𝑌𝐼 ) → ( 𝑋 𝑌 ) ∈ 𝐼 )
8 4 5 6 7 syl3anc ( ( 𝑅 ∈ Ring ∧ 𝐼𝑈 ∧ ( 𝑋𝐼𝑌𝐼 ) ) → ( 𝑋 𝑌 ) ∈ 𝐼 )
9 8 3expa ( ( ( 𝑅 ∈ Ring ∧ 𝐼𝑈 ) ∧ ( 𝑋𝐼𝑌𝐼 ) ) → ( 𝑋 𝑌 ) ∈ 𝐼 )