I have a deterministic function $f(x_1, x_2, ..., x_n)$ that takes $n$ arguments.

Given a set of arguments $X = (x_i)$, I can compute $U_X = \{ i \in [1, n] : x_i \text{ was read during the evaluation of } f(X) \}$

Would it be valid to use the set $K_X = \{(i, x_i): i \in U_X\}$ as a memoization key for $f(X)$?

In particular, I am worried that there may exist $X=(x_i)$ and $Y=(y_i)$ such that:

$$ \tag1 U_X \subset U_Y $$ $$ \tag2 \forall i \in U_X, x_i = y_i $$ $$ \tag3 f(X) \neq f(Y) $$

In my case, the consequence of the existence of such $X$ and $Y$ would be that $K_X$ would be used as a memoization key for $f(Y)$, and would thus return the wrong result.

My intuition says that, with $f$ being a deterministic function of its arguments, there should not even exist $X$ and $Y$ (with $U_X$ a strict subset of $U_Y$) such that both $(1)$ and $(2)$ hold (much less all three!), but I would like a demonstration of it (and, if it turns out to be trivial, at least pointers to the formalism that makes it trivial).