Let $X_t=(X^1,\dots,X^d$) be a d-dimensional semimartingale. Then the formula for Ito's lemma I have found in several places, including wikipedia, is:
$f(X_T)-f(X_0)=\sum_{i=1}^d\int_0^T f_{i}(X_{s-})\,dX^i_s + \frac{1}{2}\sum_{i,j=1}^d \int_0^T f_{i,j}(X_{s-})\,d[X^i,X^j]_s\\ \qquad+ \sum_{s\le T} \left(\Delta f(X_s)-\sum_{i=1}^df_{i}(X_{s-})\,\Delta X^i_s -\frac{1}{2}\sum_{i,j=1}^d f_{i,j}(X_{s-})\,\Delta X^i_s \, \Delta X^j_s\right)$
for a twice continuously differentiable function $f$.
Now, I try to apply this to the function $f(t,B,N^1,N^2)$ where $B_t=\sigma W_t$ a Brownian motion and $N^1_t$ and $N^2_t$ independent poisson processes.
Thus, I think it looks like this:
$df=f_tdt+f_BdB+f_{N^1}dN^1+f_{N^2}dN^2+\frac{1}{2}(f_{BB}d[B]+f_{N^1N^1}d[N^1]+f_{N^2N^2}d[N^2])+\Delta f-f_{N^1}dN^1-f_{N^2}dN^2-\frac{1}{2}(f_{N^1N^1}dN^1+f_{N^2N^2}dN^2)$
because the mixed terms become zero due to independencies and $\Delta B_t=0$ as well as $\Delta t =0$ and $[t]=0$ while $\Delta N=1$
So simplifying further yields: $df=f_tdt+f_B\sigma dW +\frac{1}{2}f_{BB}\sigma^2dt+\Delta f$
Is this correct or did I make a mistake somewhere? I am especially unsure if the mixed terms really all disappear and whether $\sum_{s\le T}(\sum_{i=1}^df_{i}(X_{s-}\,\Delta )X^i_s)$ indeed becomes $f_{N^1}dN^1-f_{N^2}dN^2$. If not, I don't know how to interpret this term when writing the whole thing as $df$.