Let $(X_k)_{k\in \mathbb N_0}$ be a simple random walk on $\mathbb Z$ and $M_k := \max \{X_1,\ldots,X_k\}$, the maximum process. Then $Y_k := M_k - X_k$ is a Markov chain. By $\sigma_0 = 0, \sigma_m := \inf\{ l> \sigma_{m-1} : Y_l \neq Y_{l-1} \}$ denote the $m$th time of movement of $(Y_k)_k$. So by defining $$\tilde{Y}_m := Y_{\sigma_m}$$ we remove the flat steps from $(Y_k)_k$. What I want to show is: $$\mathcal{L}((|X_k|)_{k\in\mathbb N_0}) = \mathcal{L}((\tilde Y_k)_{k\in\mathbb N_0}),$$ where $\mathcal L$ denotes the laws of the processes.
My attempt was to use sort of reflection principle for random walks, but I failed.
The Markov chains $\{|X_k|\}_{k=0}^{\infty}$ and $\{\tilde{Y}_m\}_{m=0}^{\infty}$ both have the same state space and the same transition probabilities. Specifically, if $|X_k|=0$ then $|X_{k+1}|=1$ with prob 1, and likewise if $\tilde{Y}_m=0$ then $\tilde{Y}_{m+1}=1$ with prob 1. Now if $|X_k|>0$ then on the next step it is equally likely to move either up or down, and the same holds for $\tilde{Y}_m$.