My professor defined a martingale sequence to be a sequence $X_0, \dots, X_i$ such that $\mathbb{E}[X_i \mid X_0, \dots, X_{i-1}] = X_{i-1}$, and claimed that this implies $\mathbb{E}[X_i \mid X_{i-1}] = X_{i-1}$ without a proof.
While intuitively clear, I still want to see a proof of this statement so I attempted one myself. If we have $\mathbb{E}_Y [\mathbb{E}[X \mid Y, Z]] := \sum_y \text{Pr}(Y = y)\mathbb{E}[X \mid Y, Z] = \mathbb{E}[X \mid Z]$ we would be done since then $\mathbb{E}[X_i \mid X_{i-1}] = \mathbb{E}_{X_0}\mathbb{E}_{X_1} \dots \mathbb{E}_{X_{i-2}}[\mathbb{E}[X_i \mid X_0, \dots, X_{i-1}]]$, the latter of which is just $\mathbb{E}_{X_0}\mathbb{E}_{X_1} \dots \mathbb{E}_{X_{i-2}}[X_{i-1}] = X_{i-1}$. I wasn't able to prove that $\mathbb{E}_Y[\mathbb{E}[X \mid Y, Z]] = \mathbb{E}[X \mid Z]$, while I do think this is a reasonable generalization of the easier case $\mathbb{E}[\mathbb{E}[X \mid Y]] = \mathbb{E}[X]$.
My thanks in advance, and I apologize if $\mathbb{E}_Y$ is not a standard notation, it basically means that take the expectation only over the random variable $Y$.
In general, if if you have a sub-$\sigma$-algebra $\mathcal{H} \subset \mathcal{A}$ of a$\sigma$-algebra $\mathcal{A}$ it holds that: $\mathbb{E}[\mathbb{E}[X,\mathcal{A}],\mathcal{H}]= \mathbb{E}[X,\mathcal{H}]$ (This is often called the tower rule). Here $\mathcal{A} = \sigma(X_1,\cdots,X_{j-1})$ and $\mathcal{H} = \sigma(X_{j-1})$. The result then follows: $\mathbb{E}[X_{j},\sigma(X_{j-1})] $ $= \mathbb{E}[\mathbb{E}[X_j,\sigma(X_1,\cdots,X_{j-1})],\sigma(X_{j-1})] $ $= \mathbb{E}[X_{j-1},\sigma(X_{j-1})] = X_{j-1}$.