= Solution
The five maximal four-cell groups give the <disjunctive normal form>
$$
\boxed{z=a_1\neg b_1\lor a_1a_0\lor a_1\neg b_0\lor a_0\neg b_1\lor\neg b_1\neg b_0.}
$$
For example, the group $a_1\neg b_0$ wraps between columns $00$ and $10$ in the bottom two rows. No group of eight cells is entirely one. The five terms are precisely the <prime implicants>, and each is essential: respectively, the cells $(a,b)=(2,1),(3,3),(2,2),(1,1),(0,0)$ are covered by no other <prime implicant>. Any <disjunctive normal form> can have its terms enlarged to <prime implicants> without increasing its size. Therefore all five displayed terms are necessary, proving minimality in number of products; each needs two <literals>, also giving ten <literals> in a minimal cover.
For an implementation with <AND gates> and <NOT gates>, call these five products $T_1,\ldots,T_5$. Form the shared complements $\neg b_1,\neg b_0$, calculate each $T_i$ with an <AND gate>, invert every $T_i$, and use
$$
z=\neg\bigl(\neg T_1\land\neg T_2\land\neg T_3\land\neg T_4\land\neg T_5\bigr).
$$
A five-input conjunction can be replaced by four two-input <AND gates>. This direct implementation uses nine two-input <AND gates> and eight <NOT gates>. That is a valid realization of the minimal <disjunctive normal form>, rather than a claim that this particular gate count is globally optimal among all factored circuits.
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