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 ) )