Consider a system described by Hamilton's equations $$\dot{q}_i=\frac{\partial H}{\partial p_i}=\{q_i,H\}, \quad \dot{p}_i=-\frac{\partial H}{\partial q_i}=\{p_i,H\}.\tag{1}$$ I want to prove that a time-independent transformation of the form $$q_i\to Q_i(q,p), \quad p_i\to P_i(q,p)\tag{2}$$ preserves Hamilton's equations provided $$\{Q_i,Q_j\}=0,\qquad \{Q_i,P_j\}=\delta_{ij},\qquad\{P_i,P_j\}=0, \tag{3}$$ i.e. if the transformation is canonical.
However, I find that $$\dot{Q}_i=\sum_k\left[\frac{\partial Q_i}{\partial q_k}\dot{q}_k+\frac{\partial Q_i}{\partial p_k}\dot{p}_k\right]=\{Q_i,H\},\tag{4}$$ and similarly, $$\dot{P}_i=\sum_k\left[\frac{\partial P_i}{\partial q_k}\dot{q}_k+\frac{\partial P_i}{\partial p_k}\dot{p}_k\right]=\{P_i,H\}\tag{5}$$ without using (3)!
Does it mean that any transformation of the form (2) is canonical?
But as far as I knew, only a subset of transformations of type (1) which satisfies (3) preserve the form of Hamilton's equations i.e. are canonical. But it turns out that I am able to reproduce Hamilton's equations in the transformed variables without using the restriction that (3), rather trivially. Did I make any mistakes?