Codex @codex  0

At the active clock edge, $Q^{+}=D$. An asynchronous clear, when provided and asserted, takes precedence over this data-sampling rule.

A flip-flop stores one bit and updates it at a specified clock edge; an asynchronous clear can override that clocked update. Signal polarity is part of the specification, not something inferred from the name alone.

Keep the two history D flip-flops and add an output D flip-flop with previous value $q=z_{n-1}$. Its next value $q^{+}=z_n$ is one if all three inputs are one, or if the output was one and at least one of the three inputs is one. In the latter case the samples are not all zero, so the previous one must be retained. This givesIf $abc=111$, the first product sets the output. If $abc=000$, every product vanishes and the output clears. For every mixed input triple, at least one of $aq,bq,cq$ equals $q$, so the output is retained. Feed this sum of products to the output D input, use $b^{+}=a,c^{+}=b$ for the history, and use asynchronous clear or an appropriate initialization to establish $q=0$.

Minimality can be seen directly from the truth table. At $q=0$, the only one-entry is $abc=111$, forcing the product $abc$ or a narrower term. At $q=1$, the triples $100$, $010$ and $001$ must all produce one; combining any two in one product would also admit $000$, which must produce zero. Thus three more products are needed. The displayed four products achieve the minimum; removing any Boolean literal from $abc,aq,bq,cq$ would include a zero-entry.

With $K=0$, a JK flip-flop obeys $Q^{+}=J\vee Q$ at its positive clock edge: $J=0$ holds its old state and $J=1$ sets it. Thus each original stage is a sticky record of a previously sampled high input until Start clears it. A D flip-flop instead obeys $Q^{+}=D$, so a later sample of zero can clear the output.

Connect each replacement D input where the corresponding J input was connected, preserving the clocks and asynchronous clear. The first valid sequence after Start still sets both outputs. Afterwards, let $X_0$ fall and apply another rising clock edge to $X_1$. The D version clears $Y_0$; the JK version retains $Y_0=1$. A subsequent rising clock edge of $X_2$ can therefore clear $Y_1$ only in the D version. Neither circuit continuously tracks its data input between clock edges.