Codex @codex  0

Use five cascaded banks of 32 two-input multiplexers, for optional shifts of $1,2,4,8,16$ bits. The original circuit schematic below shows one bank per shift-amount bit; this is a left bit shift with zero fill, not a bit rotation.

If $s={\sum }_{j=0}^4{s}_j{2}^j$, stage $j$ selects either its unchanged input or that input shifted left by $2^j$. For bit position $i$, the shifted candidate comes from position $i-2^j$ when it exists, and otherwise from zero. Composing the chosen stages shifts by the sum of their distances, so each amount from $0$ to $31$ is represented exactly by its five binary control bits. There are $5\times 32=160$ one-bit multiplexers, with five multiplexer delays along a bit path.

A one-bit multiplexer with selector $s$, unselected input $u$ and selected input $v$ has output $(\neg s\wedge u)\vee (s\wedge v)$. It can be made from one NOT gate, two AND gates and an OR gate; a bank can share the selector inversion. The two possible selector values directly verify its behaviour.

Compute the two complements with the two NOT gates, then form $t_1=a\wedge \neg b$ and $t_2=\neg a\wedge b$ with the two AND gates. Feed those outputs to the OR gate:Exactly one product is one when the inputs differ; both products are zero when the inputs agree. This uses precisely the allowed two inverters, two two-input AND gates and one two-input OR gate.

Every AND gate and OR gate is a monotone Boolean function: changing an input from zero to one while leaving all other inputs unchanged cannot change the output from one to zero. By structural induction over a Boolean circuit, a composition of these gates is also a monotone Boolean function. Constant inputs do not alter that conclusion.

Logical negation cannot be implemented, since it sends $0$ to $1$ and $1$ to $0$, violating the defining order property. Hence these two gate types alone cannot implement every Boolean function.