I have a simple question that I believe to know the answer to but want to be sure that I can apply the tower property without loss.
Let $X=(X_{j})_{j=1}^{n}$ be a vector valued random variable distributed according to the distribution $P$ and let $f$ be some function of $X$. Then my question is if $X^{1}=(X_{j})_{j=1}^{k}$ and $X^{2}=(X_{j})_{j=1}^{k'}$ where $k'<k<n$ (i.e. $X^{1}$ is a subvector of $X$ and $X^{2}$ is a subvector of $X^{1}$), then does the following equality (that looks like the tower property) hold; $$ \mathbb{E}_{P}[\mathbb{E}_{P}[f(X)\;|\;X^{1}=x^{1}]\;|\;X^{2}=x^{2}]=\mathbb{E}_{P}[f(X)\;|\;X^{2}=x^{2}] $$ whenever $x^{2}=(x^{1}_{j})_{j=1}^{k'}$. Namely, $x^{2}$ is the subvector of $x^{1}$ consisting of the first $k'$ terms.
This looks familiar to the tower property for martingales, and I want to know if it is true even when the R.V.'s are not necessarily independent.
My attempt to prove it would be to say let $X^{1}$ be the degenerate random variable equal to $x^{1}$ and let $X^{2}=(x^{1}_{j})_{j=1}^{k'}$ be a projection of $X^{1}$. Then $X^{2}$ is a function of $X^{1}$ and therefore $$ \mathbb{E}[\mathbb{E}[f(X)|X^{1}]|X^{2}]=\mathbb{E}[f(X)|X^{2}] $$
Does this make sense?