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 ) ) |
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 ) ) |