# show maximum number of arc disjoint directed $s-t$ paths is equal to max flow

Given: $$D = (V,A)$$ a directed graph and $$s,t \in V$$
Problem: Show that the maximum number of pairwise arc-disjoint directed $$s-t$$ paths is equal to the maximum value of the flow $$f$$ where $$f$$ is subject to capacity function $$c$$.
Approach: first show "$$\leq$$" and then show ''$$\geq$$''

"$$\leq$$":
$$\begin{array}{l} \text{Take } c =1 \text{ for all arcs}. & (1) \\ \text{ Let }P_1,..,P_k \text{ be arc-disjoint directed } s-t \text{ paths.} &(2) \\ \text{Then } f = \chi ^{A(P_1)} +...+ \chi ^{A(P_n)} \text{ is an } s-t \text{ flow and } 0\leq f \leq c & (3) \\ [f \leq c \text{ since } P_1,..,P_k \text{ are arc-disjoint] } &(4) \\ \text{ Moreover, } \text{value(f)} = k &(5) \\ \end{array}$$

This is how it was written down by my teacher. But I don't understand how he concludes this less of equal inequality. In line 2 you begin with $$k$$- paths and then in line (5) you conclude that value(f) = k , but why does "$$\leq$$" hold ? 