Proof: let $A$ a ring, then $(-a) \cdot (-b) = a \cdot b $ $\forall a,b \in A$ I must prove this property:
Property: let $A$ a be ring, then $(-a) \cdot (-b) = a \cdot b $, $\forall a,b \in A$.
Proof: let $a \in A$ and $b \in A$, by hypothesis $A$ is a ring then $a \cdot 0=0$ and $b + (-b) = (-b) + b= 0$ and $(-b) \in A $, therefore $a \cdot (b + (-b))=0$, but by hypothesis $\cdot$ is distributive then $a \cdot (b + (-b))=a \cdot b + a \cdot (-b) =0$, therefore $-(a \cdot (-b))=a \cdot b$, but in a ring is true that $-(c \cdot d)=(-c) \cdot d$, $\forall c,d \in A$, therefore we have $(-a) \cdot (-b)=-(a \cdot (-b))=a \cdot b$.
It is correct?
Thanks in advance!!
 A: In light of MathGem's answer, let's look at your statement again. You said, "in a ring it is true $-(a\cdot b)=(-a)\cdot b$." What you're saying is that $$\tag 1 a\cdot b+a\cdot (-b)=0$$
which is true. 
Thus, you in turn claim that in a ring
$$a\cdot b+(-a)\cdot b=0$$
which is true. Then, it follows
$$\tag 2 (-a)\cdot (-b)+a\cdot (-b)=0$$
But since inverses are unique $(1),(2)$ imply $a\cdot b=(-a)\cdot(-b)$.

There is a nice proof, that goes as follows:
Look at the expression $$a\cdot b+a\cdot (-b)+(-a)\cdot(-b)$$ and use the distributive laws in two different ways. One would be the above is$$a\cdot b+a\cdot (-b)+(-a)\cdot(-b)= \\a\cdot (b+(-b))+(-a)\cdot(-b)=\\a\cdot 0+(-a)\cdot (-b)=\\ 0+(-a)\cdot (-b)=\\(-a)\cdot (-b)$$
What is the other?
A: There is an error. The statement "but in a ring is true..." begs the principle. Instead, let $\rm\:b = -c\:$ in what you just proved. More explicitly, you have proved $\rm\:x(-\color{#c00}y) = -(xy)\:$ for all $\rm\:x,y.\:$ Thus, applying this twice we deduce $\rm\ (-a)(-\color{#c00}b) = -((-a)b) = -(b(-\color{#c00}a)) = -(-(ba)) = ba = ab.$
A more conceptual way to view the proof is that $\rm\:(-a)(-b)\:$ and $\rm\:ab\:$ are both inverses of $\rm\:(-a)b,\:$ hence they are equal by uniqueness of inverses.
Equivalently, evaluate $\ \rm\overline{(-a)(-b)\! \,+\,} \overline{ \underline {(-a)b}} \underline{\,+\,ab_{\phantom{_{1}}}}\:$in two ways, over or underlined first.
See also prior posts on this Law of Signs in Rings.
