= Solution
Let $(L^+,G^+)$ denote the state after the next active <clock edge>. In the <truth table>, an entry is the two-bit next-state code, and `–` marks a don't-care value for the unused current state `11`:
|| Current $LG$
|| $ab=00$
|| $ab=01$
|| $ab=10$
|| $ab=11$
| $00$ (`Eq`)
| $00$
| $10$
| $01$
| $00$
| $10$ (`Lt`)
| $10$
| $10$
| $01$
| $10$
| $01$ (`Gt`)
| $01$
| $10$
| $01$
| $01$
| $11$ (unused)
| $--$
| $--$
| $--$
| $--$
For each input column, `01` forces $L^+G^+=10$, `10` forces $01$, and the two equal-input columns copy the current code. Thus this compact <truth table> specifies all twelve valid combinations of state and input, or twenty-four individual next-state <bits>:
$$
\boxed{ab=01\Rightarrow(L^+,G^+)=(1,0),\quad
ab=10\Rightarrow(L^+,G^+)=(0,1).}
$$
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