I got the next problem:
Let $A$ be a Lie algebra, prove that if the bracket associates $([[x,y],z]=[x,[y,z]]$) then the bracket is zero $([x,y]=0)$.
Can't get the result using the properties (alternating, Jacobi identity, anticommutativity), i think that the result is false. Any suggestions? Thanks.
The Lie algebra $N$ formed by strictly upper triangular $3\times 3$ matrices satisfies
$$[z, [x, y]]= 0\ \ \ \forall x, y, z\in N.$$
But $[x, y] \neq 0$.