# Edge choosability(edge list coloring) of cycles

I have 2 cycles with 6 length as shown below.

I want to show that the above graph is 4-edge-choosable. I don't know where to start. It's known that every cycle of even length is 2-edge-choosable, therefore, we can color one of these cycles(suppose we are coloring red one) at first and at each vertice of that cycle we'll have 2 permissible colors to color other cycle(in our case blue one). But what can we do when for example permissible colors at vertex $6$ are colors that are choosen for edges $23$ and $34$, i.e.

We have colored edges of red cycle with these colors:

$$16 \rightarrow \alpha_1 \\ 23 \rightarrow \alpha_2 \\ ...\\ 61 \rightarrow \alpha_6$$

Suppose that list of colors for $63$ edge is $L =$ { $\beta_1, \beta_2, \beta_3, \beta_4$} and now for $63$ edge we have
$L' = L\setminus${$\alpha_5, \alpha_6$} 2 permissible colors when we are looking from the vertex $6$ and from the other side we have $L'' = L\setminus${$\alpha_2, \alpha_3$} 2 permissible colors when we are looking from the vertex $3$. What if $L' =${$\alpha_2, \alpha_3$} and $L'' =${$\alpha_5, \alpha_6$}. How can we color edges of first(red) cycle so that we shouldn't get the latter situation? Or maybe color them dynamically, by changing edge colors when starting second cycle coloring?