# Yang-Mills constraints and Poisson brackets

Let's have constraints for Yang-Mills theory: $$\varphi_{a} = \partial_{i}\pi^{i}_{a} - f_{abc}\pi^{b}_{i}A^{c}_{i}.$$ I have read the statement that $$\tag 1 [\varphi_{a}(\mathbf x), \varphi_{b}(\mathbf y )] = f_{abc}\varphi^{c}(\mathbf x) \delta (\mathbf x - \mathbf y).$$ $(1)$ can be computed by using canonical Poisson brackets $$[A_{a}^{i}(\mathbf x ), \pi_{b}^{j}(\mathbf y )] = \delta_{ab}\delta^{ij}\delta (\mathbf x - \mathbf y ), \quad [\pi_{a}^{i}(\mathbf x ), \pi_{b}^{j}(\mathbf y )] = [A_{a}^{i}(\mathbf x ), A_{b}^{j}(\mathbf y )] = 0.$$ But I can't get $(1)$. The problem is in two summands which have the form $$\tag 2 -\varepsilon_{klm}[\partial_{i}\pi^{i}_{a}(\mathbf x ), A_{m}^{j}(\mathbf y )]\pi_{l}^{j}(\mathbf y) = -\varepsilon_{kla}\pi_{l}^{i}(\mathbf y)\partial_{i}^{\mathbf x}\delta (x - y) = -\varepsilon_{kla}\pi_{l}^{i}(\mathbf y)\partial_{i}^{\mathbf y}\delta(\mathbf x - \mathbf y)$$ For getting $(1)$ I need to move derivative away from delta-function, but it can be done (in my opinion) only if there also is integration of $(2)$ over $x, y$. Where did I make the mistake? How to get $(1)$?

$$\tag{A} \{\partial_x+\partial_y\}\delta (x-y)~=~ 0,$$
$$\tag{B} \{f(x)-f(y)\}~\delta (x-y)~=~ 0,$$
$$\tag{C} \{f(y)-f(x)\}~\partial_x \delta (x-y)~=~ \delta (x-y) ~f^{\prime}(x),$$