$\left\{ Z_{n},n\geq1\right\} $ is a martingale, which means that $\forall n$, it satisfies the below two conditions,
1) This is the uniformly bounded condition. $$ E\left[\left|Z_{n}\right|\right]<\infty $$
2) Standard martingalge condition. $$ E\left[\left.Z_{n+1}\right|Z_{1},\ldots,Z_{n}\right]=Z_{n} $$
Question:
How do we prove that $\left\{ Z_{m+k}-Z_{m},m\geq1,k=1,\ldots,n\right\} $ is a martingale?
Context:
This is an omitted step in the proof of the Martingale Convergence Theorem from Stochastic Processes, Second Edition, Sheldon Ross, Theorem 6.4.6, Page 315.
Steps Tried:
$$ \text{Let, }Y_{k}=Z_{m+k}-Z_{m} $$ $$ E\left[\left.Y_{k+1}\right|Y_{k}\ldots Y_{1}\right]=E\left[\left.Z_{m+k+1}-Z_{m}\right|Z_{m+k}-Z_{m},\ldots,Z_{m+1}-Z_{m}\right] $$
We'll use the tower property of conditional expectation, and first condition on a larger set of random variables. First since $Z_{m+k}-Z_m,\dots, Z_{m+1}-Z_m,Z_m$ and $Z_{m+k},\dots,Z_m$ determine the same events, we have
$$E [Z_{m+k+1} | Z_{m+k}-Z_m,\dots,Z_{m+1}-Z_m,Z_m]=E[Z_{m+k+1} |Z_{m+k},\dots,Z_m]=Z_{m+k}=(*)$$
By the tower property, to get the conditional expectation of $Z_{m+K+1}$ WRT to the smaller set of variables $Z_{m+k}-Z_m,\dots,Z_{m+1}-Z_m$, we take the conditional expectation of $(*)$ with respect to the latter set (tower property):
$$E[Z_{m+k+1} | Z_{m+k}-Z_m,\dots,Z_{m+1}-Z_m]=E [ Z_{m+k} |Z_{m+k}-Z_m,\dots,Z_{m+1}-Z_m],$$
But this gives
$$E[Z_{m+k+1} - Z_m | Z_{m+k}-Z_m,\dots,Z_{m+1}-Z_m ] = E [ Z_{m+k}-Z_m |Z_{m+k}-Z_m,\dots,Z_{m+1}-Z_m ] = Z_{m+k}-Z_m,$$
which is what we wanted to show.