Split network flow function

Let
a graph $$G = (V,E)$$
a network $$N = (G, s, t, c)$$
and an integral flow funtion $$f$$

The value of f $$v(f)=v_1+v_2+...+v_p$$, where $$v_i$$ is a flow leaving the source. I must prove that there are $$p$$ integral flows functions so that $$v(f_i)=v_i$$ and $$f = f_1+f_2+...+f_p$$.
So I have to split the initial flow into $$p$$ flows so that the sum of the new flows is the original flow and $$v(fi)=vi$$.

I followed those instructions manually on some examples, but I can't find an algorithm to fit on all cases.
Any ideas on how I should do it?

It suffices to prove the claim when $$v_i=1$$ for each $$i$$. This will allow us to split the flow $$f$$ into $$v(f)$$ flows with value $$1$$ each. Then given arbitrary natural $$v_i$$’s with $$\sum v_i=v(f)$$, we partition these $$v(f)$$ onto $$p$$ groups consisting of $$v_1,\dots, v_p$$ flows respectively, and next for each $$i$$ we merge flows of $$i$$-th group into one flow with value $$v_i$$.
We prove the possibility of splitting a flow $$f$$ into $$v(f)$$ flows of value $$1$$ by induction with respect to $$v(f)>0$$. If $$v(f)=1$$ then we put $$f_1=f$$. Assume that we have already proved the claim for $$v(f)=p$$. Let $$f$$ be any feasible integral $$s-t$$ flow on $$N$$ with $$v(f)=p+1$$ Start from $$s$$ and go along the flow $$f$$ (that is, along edges $$e$$ such that $$f(e)>0$$), subtracting $$1$$ from $$f$$ on each passed edge. Since both the graph $$G$$ and the number $$v(f)$$ are finite, we can subtract only finitely many $$1$$’s, so we shall stick at some step. Since $$\operatorname{netinflow}(f, v)= \operatorname{netoutlow} (f, v)$$ for all vertices $$v$$ in $$G$$ except $$s$$ and $$t$$, and $$\operatorname{netinflow}(f, s)=0$$, we can stick only at the vertex $$t$$. So we went a path from $$s$$ to $$t$$. This path endowed with the values which we subtracted, naturally generate an integral flow $$f’$$ from $$s$$ to to $$t$$ with $$v(f’)=1$$. Putting $$f’’=f-f’$$ we obtain a feasible integral $$s-t$$ flow on $$N$$ with $$v(f’)=p$$, which can be split into $$p$$ required flows by the induction hypothesis.