Metamath Proof Explorer


Theorem ringm2neg

Description: Double negation of a product in a ring. ( mul2neg analog.) (Contributed by Mario Carneiro, 4-Dec-2014) (Proof shortened by AV, 30-Mar-2025)

Ref Expression
Hypotheses ringneglmul.b
|- B = ( Base ` R )
ringneglmul.t
|- .x. = ( .r ` R )
ringneglmul.n
|- N = ( invg ` R )
ringneglmul.r
|- ( ph -> R e. Ring )
ringneglmul.x
|- ( ph -> X e. B )
ringneglmul.y
|- ( ph -> Y e. B )
Assertion ringm2neg
|- ( ph -> ( ( N ` X ) .x. ( N ` Y ) ) = ( X .x. Y ) )

Proof

Step Hyp Ref Expression
1 ringneglmul.b
 |-  B = ( Base ` R )
2 ringneglmul.t
 |-  .x. = ( .r ` R )
3 ringneglmul.n
 |-  N = ( invg ` R )
4 ringneglmul.r
 |-  ( ph -> R e. Ring )
5 ringneglmul.x
 |-  ( ph -> X e. B )
6 ringneglmul.y
 |-  ( ph -> Y e. B )
7 ringrng
 |-  ( R e. Ring -> R e. Rng )
8 4 7 syl
 |-  ( ph -> R e. Rng )
9 1 2 3 8 5 6 rngm2neg
 |-  ( ph -> ( ( N ` X ) .x. ( N ` Y ) ) = ( X .x. Y ) )