ringmneg2

Description: Negation of a product in a ring. ( mulneg2 analog.) (Contributed by Jeff Madsen, 19-Jun-2010) (Revised by Mario Carneiro, 2-Jul-2014)

	Ref	Expression
Hypotheses	ringneglmul.b	$⊢ B = {Base}_{R}$
	ringneglmul.t	$⊢ \underset{˙}{\cdot} = \cdot_{R}$
	ringneglmul.n	$⊢ N = {inv}_{g} (R)$
	ringneglmul.r	$⊢ φ \to R \in Ring$
	ringneglmul.x	$⊢ φ \to X \in B$
	ringneglmul.y	$⊢ φ \to Y \in B$
Assertion	ringmneg2	$⊢ φ \to X \underset{˙}{\cdot} N (Y) = N (X \underset{˙}{\cdot} Y)$

Proof

Step	Hyp	Ref	Expression
1		ringneglmul.b	$⊢ B = {Base}_{R}$
2		ringneglmul.t	$⊢ \underset{˙}{\cdot} = \cdot_{R}$
3		ringneglmul.n	$⊢ N = {inv}_{g} (R)$
4		ringneglmul.r	$⊢ φ \to R \in Ring$
5		ringneglmul.x	$⊢ φ \to X \in B$
6		ringneglmul.y	$⊢ φ \to Y \in B$
7		ringgrp	$⊢ R \in Ring \to R \in Grp$
8	4 7	syl	$⊢ φ \to R \in Grp$
9		eqid	$⊢ 1_{R} = 1_{R}$
10	1 9	ringidcl	$⊢ R \in Ring \to 1_{R} \in B$
11	4 10	syl	$⊢ φ \to 1_{R} \in B$
12	1 3	grpinvcl	$⊢ (R \in Grp \land 1_{R} \in B) \to N (1_{R}) \in B$
13	8 11 12	syl2anc	$⊢ φ \to N (1_{R}) \in B$
14	1 2	ringass	$⊢ (R \in Ring \land (X \in B \land Y \in B \land N (1_{R}) \in B)) \to (X \underset{˙}{\cdot} Y) \underset{˙}{\cdot} N (1_{R}) = X \underset{˙}{\cdot} (Y \underset{˙}{\cdot} N (1_{R}))$
15	4 5 6 13 14	syl13anc	$⊢ φ \to (X \underset{˙}{\cdot} Y) \underset{˙}{\cdot} N (1_{R}) = X \underset{˙}{\cdot} (Y \underset{˙}{\cdot} N (1_{R}))$
16	1 2	ringcl	$⊢ (R \in Ring \land X \in B \land Y \in B) \to X \underset{˙}{\cdot} Y \in B$
17	4 5 6 16	syl3anc	$⊢ φ \to X \underset{˙}{\cdot} Y \in B$
18	1 2 9 3 4 17	rngnegr	$⊢ φ \to (X \underset{˙}{\cdot} Y) \underset{˙}{\cdot} N (1_{R}) = N (X \underset{˙}{\cdot} Y)$
19	1 2 9 3 4 6	rngnegr	$⊢ φ \to Y \underset{˙}{\cdot} N (1_{R}) = N (Y)$
20	19	oveq2d	$⊢ φ \to X \underset{˙}{\cdot} (Y \underset{˙}{\cdot} N (1_{R})) = X \underset{˙}{\cdot} N (Y)$
21	15 18 20	3eqtr3rd	$⊢ φ \to X \underset{˙}{\cdot} N (Y) = N (X \underset{˙}{\cdot} Y)$

Metamath Proof Explorer

Theorem ringmneg2

Proof