= Solution
Splitting the <Boolean function> according to the first tested <Boolean variable> gives
$$
f=(\neg x\land y)\lor\bigl(x\land((\neg y\land z)\lor y)\bigr).
$$
The identity $y\lor(\neg y\land z)=y\lor z$ follows by considering $y=0$ and $y=1$. Using it and distributing <Boolean conjunction> over <Boolean disjunction>,
$$
f=(\neg x\land y)\lor(x\land y)\lor(x\land z)
=\boxed{y\lor(x\land z)}.
$$
The first two terms combine to $y$ because $x\lor\neg x=1$. This also explains the <truth table>: $y=1$ always makes the output one, while with $y=0$ the output is one exactly when both $x$ and $z$ are one.
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