In describing d'Alembert's principle, the lecture note I was provided with states that the total force $\mathbb F_l$ acting on a particle can be taken as,
$$\mathbb F_l=F_l+\sum_mf_{ml}+C_l,$$
where $F_l$ is the sum of the applied forces on $l^{th}$ particle, $f_{ml}$ being the internal force on $l^{th}$ particle due to an $m^{th}$ particle, and $C_l$ denoting the constraint forces. However considering the law of action and reaction it further states that $f_{ml}+f_{lm}=0$, which I have no trouble understanding. But considering a virtual displacement $\delta \bf{r}_l$ on $l^{th}$ particle, in the next line it concludes that the virtual work $\delta W$ done should be,
$$\delta W~=~\sum_{l=1}^N(F_l+C_l)\centerdot\delta\textbf{r}_l,$$
ignoring $f_{ml}$ terms. But if we take those terms into account shouldn't it be
$$\delta W~=~\sum_{l=1}^N(F_l+C_l)\centerdot\delta\textbf{r}_l+\sum_l\sum_mf_{ml}\centerdot\delta\textbf{r}_l.$$
In other words, I don't see how comes $$\sum_l\sum_mf_{ml}\centerdot\delta\textbf{r}_l~\stackrel?=~0.$$
For the purpose of illustrating my problem, consider a system of two particles for which one can expand the above double summation and write
$$f_{11}\centerdot\delta\textbf{r}_1+f_{21}\centerdot\delta\textbf{r}_1+f_{12}\centerdot\delta\textbf{r}_2+f_{22}\centerdot\delta\textbf{r}_2.$$
How can this add up to zero in general? Even if I assume $f_{11}=0$ and $f_{22}=0$, I am left with $$f_{21}\centerdot\delta\textbf{r}_1+f_{12}\centerdot\delta\textbf{r}_2.$$
Do I have to assume $\delta\textbf{r}_1=\delta\textbf{r}_2$ i.e. that the virtual displacements of the particles correspond to merely a displacement of the system? Or have I missed out on something?