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Metamath Proof Explorer

Theorem dvdsrneg

Description: An element divides its negative. (Contributed by Mario Carneiro, 1-Dec-2014)

Ref Expression
Hypotheses dvdsr.1 ⊢ 𝐵 = ( Base ‘ 𝑅 )
dvdsr.2 ⊢ ∥ = ( ∥r ‘ 𝑅 )
dvdsrneg.5 ⊢ 𝑁 = ( invg ‘ 𝑅 )
Assertion dvdsrneg ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐵 ) → 𝑋 ∥ ( 𝑁 ‘ 𝑋 ) )

Proof

Step Hyp Ref Expression
1 dvdsr.1 ⊢ 𝐵 = ( Base ‘ 𝑅 )
2 dvdsr.2 ⊢ ∥ = ( ∥r ‘ 𝑅 )
3 dvdsrneg.5 ⊢ 𝑁 = ( invg ‘ 𝑅 )
4 id ⊢ ( 𝑋 ∈ 𝐵 → 𝑋 ∈ 𝐵 )
5 ringgrp ⊢ ( 𝑅 ∈ Ring → 𝑅 ∈ Grp )
6 eqid ⊢ ( 1r ‘ 𝑅 ) = ( 1r ‘ 𝑅 )
7 1 6 ringidcl ⊢ ( 𝑅 ∈ Ring → ( 1r ‘ 𝑅 ) ∈ 𝐵 )
8 1 3 grpinvcl ⊢ ( ( 𝑅 ∈ Grp ∧ ( 1r ‘ 𝑅 ) ∈ 𝐵 ) → ( 𝑁 ‘ ( 1r ‘ 𝑅 ) ) ∈ 𝐵 )
9 5 7 8 syl2anc ⊢ ( 𝑅 ∈ Ring → ( 𝑁 ‘ ( 1r ‘ 𝑅 ) ) ∈ 𝐵 )
10 eqid ⊢ ( .r ‘ 𝑅 ) = ( .r ‘ 𝑅 )
11 1 2 10 dvdsrmul ⊢ ( ( 𝑋 ∈ 𝐵 ∧ ( 𝑁 ‘ ( 1r ‘ 𝑅 ) ) ∈ 𝐵 ) → 𝑋 ∥ ( ( 𝑁 ‘ ( 1r ‘ 𝑅 ) ) ( .r ‘ 𝑅 ) 𝑋 ) )
12 4 9 11 syl2anr ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐵 ) → 𝑋 ∥ ( ( 𝑁 ‘ ( 1r ‘ 𝑅 ) ) ( .r ‘ 𝑅 ) 𝑋 ) )
13 simpl ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐵 ) → 𝑅 ∈ Ring )
14 simpr ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐵 ) → 𝑋 ∈ 𝐵 )
15 1 10 6 3 13 14 ringnegl ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑁 ‘ ( 1r ‘ 𝑅 ) ) ( .r ‘ 𝑅 ) 𝑋 ) = ( 𝑁 ‘ 𝑋 ) )
16 12 15 breqtrd ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐵 ) → 𝑋 ∥ ( 𝑁 ‘ 𝑋 ) )