Poisson brackets: prove that they are canonical invariants I need a clarification about Poisson brackets. 
I'm studying on Goldstein's Classical Mechanics (1 ed.).
Goldstein proves that Poisson brackets are canonical invariants for any functions F and G. 
But there is a step that I can't understand.
After some steps, he says that: 
$$ \tag{1} [F, G]_q,_p = \sum_k ( \frac { \partial G}{\partial Q_k} [F,Q_k]_q, _p +\frac {\partial G}{\partial P_k}[F, P_k]_q, _p)$$ 
After other steps, he writes:
$$ \tag{2}[F,Q_k]= - \frac {\partial F}{\partial P_k}$$
and $$ \tag{3}[P_k, F]_q, _p = \sum_j \frac {\partial F}{\partial Q_j} [P_k, Q_j] + \sum_j \frac {\partial F}{\partial P_j}[P_k, P_j]$$
$$\Rightarrow \tag {4} [F,P_k]=\frac {\partial F}{\partial Q_k}$$ 
and now he replaces these relations in the first expression I have written, obtaining:
$$\tag {5}[F, G]_q, _p=[F, G]_Q, _P$$
Why does he obtain in the second last step $\frac {\partial F}{\partial Q_k}$ and not $-\frac {\partial F}{\partial Q_k}$? $[P_k, F]=-[F, P_k]$ isn't it?
EDIT: 
Golstein starts from (1) and substituites $Q_k$ to $F$ and $F$ to $G$ and so he obtains (2). 
Then he substitutes $P_k$ to $F$ and $F$ to $G$ and obtains (3). 
Immediately after he writes (4), that according to me is opposite to (3). 
And so I have thought to a printing error. I have tried to substitute $-P_k$ to $F$ and I have obtained 
$$[-P_k, F]=\frac {\partial F}{\partial Q_k}$$
Then, as $[-P_k, F]=[F,P_k]$, 
I can say that $$ [F,P_k]=\frac {\partial F}{\partial Q_k}$$
And so I can obtain (5).
Could you confirm that my argumentation is correct?
 A: Maybe it will be interesting to you to see another proof of this fact.
Firstly, define $2n$ vector $\textbf{x}=(q_1,...,q_n,p_1,...,p_n)^T$ and $2n\times2n$ matrix $J$
$$J=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$
With this notation we can write Hamiltonian equation in this way:
$$\dot{\textbf{x}}=J\frac{\partial H}{\partial \textbf{x}}$$.
Write transformation in the form $x_i \rightarrow  y_i(x)$ where $i=1,...,2n$.
Then we have
$$\dot{y_i}=\frac{\partial y_i}{\partial x_j}\dot{x_j}=\frac{\partial y_i}{\partial x_j}J_{jk}\frac{\partial H}{\partial x_k}=\frac{\partial y_i}{\partial x_j}J_{jk}\frac{\partial H}{\partial y_l}\frac{\partial y_l}{\partial x_k}$$
If we define $I_{ij}=\frac{\partial y_i}{\partial x_j}$ (Jacobian) then
$$\textbf{y}=IJI^T\frac{\partial H}{\partial \textbf{y}}$$
So Hamilton's equation are left invariant if
$$IJI^T=J$$
If this holds, the Jacobian $I$ is said to be symplectic. And a change variables with a 
symplectic Jacobian is said to be canonical transformation.
In this notation we can define Poisson bracket as $$\left\{f,g \right\}=\frac{\partial f}{\partial x_i}J_{ij}\frac{\partial g}{\partial x_j}$$
Now theorem.
Theorem:
The poisson bracket is invariant under canonical transformations.
Proof: if $x_i \rightarrow  y_i(x)$- is canonical transformation, we have $\frac{\partial f}{\partial x_i}=\frac{\partial f}{\partial y_k}I_{ki}$, so
$$\left\{f,g \right\}=\frac{\partial f}{\partial y_k}I_{ki}J_{ij}I_{lj}\frac{\partial g}{\partial y_l}=\frac{\partial f}{\partial y_k}J_{kl}\frac{\partial g}{\partial y_l}$$
$\blacksquare$
A: $[F,P_k]=\frac{\partial F}{\partial Q_i}\frac{\partial P_k}{\partial P_i}-\frac{\partial F}{\partial P_i}\frac{\partial P_k}{\partial Q_i}$.
Second term is zero and first term is
$\frac{\partial F}{\partial Q_i }\delta_{ik}=\frac{\partial F}{\partial Q_k}$
