Show that the autocovariance function depends on $s$ and $t$ only through their difference $\left|s-t\right|$

Consider the time series $$x_t = \beta_1 + \beta_2t + w_t,$$ where $$\beta_1, \beta_2$$ are known constants and $$w_t$$ is a white noise process with variance $$\sigma^2_w$$.

I want to show that the process $$y_t = x_t - x_{t-1}$$ is stationary. I have the following definition that I can use

Def: A stationary time series $$x_t$$ is a finite variance process such that

• i) the mean value function, $$\mu_t$$, is constant and does not depend on time $$t$$, and
• ii) the autocovariance function $$\gamma(s,t)$$ depends on $$s$$ and $$t$$ only through their difference $$\left|s -t\right|$$.

What I've tried: We have $$y_t = x_t - x_{t-1} = \beta_2 + w_t - w_{t-1}$$. Therefore $$\mu_t =\operatorname{E}[y_t] = \beta_2$$, which is constant and does not depend on time $$t$$, so that the first condition is satisfied. If I look at the autocovariance function I get \begin{align} \gamma(s, t) = \operatorname{cov}(y_s, y_t) &= \operatorname{E}[(y_s - \mu_s)(y_t - \mu_t)] = \operatorname{E}[w_sw_t - w_sw_{t-1} - w_{s-1}w_t + w_{s-1}w_{t-1}]\\ &= \operatorname{E}[w_s w_t] - \operatorname{E}[w_s w_{t-1}] - \operatorname{E}[w_{s-1}w_t] + \operatorname{E}[w_{s-1}w_{t-1}] \end{align} If I look at this result then it makes a lot of sense to conclude that the autocovariance only depends on $$s$$ and $$t$$ through their difference. Since $$w_t$$ is i.i.d. the autocovariance function should be the same for different $$s,t$$ as long as their difference was the same. Unfortunately I'm not sure I should explain this in more detail and make a clear conclusion.

Question: How do I show that the autocovariance function depends on $$s$$ and $$t$$ only through their difference $$\left|s-t\right|$$?

We use the fact that $$(w_s,w_t)$$ has the same distribution as $$(w_0,w_{t-s})$$ (which is the same as that of a couple of independent centered random variables with variance $$\sigma_w^2$$). Therefore, $$\mathbb E\left[w_sw_t\right]=\mathbb E\left[w_0w_{t-s}\right].$$ Do a similar reasoning for the other three terms in the expression of $$\gamma(s,t)$$.