Complement property of DES
I think I got it:
I'm going to show that after round i the result will be $\overline{L_i}$ and $\overline{R_i}$ when the input was $\overline{L_{i-1}}$ and $\overline{R_{i-1}}$ and using $\overline k$ as key.

$L_i=R_{i-1} \implies \overline{L_i} = \overline{R_{i-1}}$

$\overline{R_i} = \overline{L_{i-1}} \oplus f(\overline{R_{i-1}}, \overline{k_i}) = \overline{L_{i-1}} \oplus f(R_{i-1}, {k_i}) = \overline{L_{i-1} \oplus f(R_{i-1}, k_i)}$.

$f(\overline R, \overline k)=f(R, k)$ because within the fuction the complement of expansion(R) and $\overline k$ are xored which eleminates the complement.

That way round the final result will be $\overline c$ when using $\overline m$ as input and $\overline k$ as key. The initial permutation does not change that.