Is the induced drag independent of wing span? I am a new aviation student and I was reading about induced drag the other day. I know that it is produced as a result of the tip vortices and that the greater the aspect ratio of an airplane the less the induced drag force. But when it came to the equation of the force, it is equal to:

$D_i = \frac{1}{2}\rho V^2 S \frac{C_L^2}{\pi  AR \epsilon}$

If we substitute aspect ratio $AR$ with span/chord $\frac{b}{c}$  and plan area $S$ as $b\cdot c$, the span term will be cancelled and the induced drag will be affected by the chord length only. 

It kind of contradicts the effect of aspect ratio on the induced drag force, doesn't it?
Aspect ratio $AR$ can be written as $\frac{b}{c}$ which is equal to $\frac{b^2}{S}$. 

Before we start substituting, note that $C_L$ also depends on the wing surface area $S$.

$L = \frac{1}{2}\rho V^2 C_L S$

or

$C_L =  \frac{L }{\frac{1}{2} \rho V^2 S}$

Substituting all this in the induced drag formula yields:

$D_i = \frac{1}{2}\rho V^2 S \frac{C_L^2}{\pi  AR \epsilon} =
\frac{L^2}{\frac{1}{2}\rho V^2 S \pi AR \epsilon} = \frac{L^2}{\frac{1}{2}\rho V^2   \pi b^2 \epsilon}$

This shows that the induced drag is proportional to the inverse of the square of the wingspan.