= Solution
\b[No.] In the standard <successor structure>, every element other than the named zero has a predecessor:
$$
\forall x\bigl(x=0\ \vee\ \exists y\,s(y)=x\bigr).
$$
The proposed structure fails this <first-order sentence> at the integer $-1$. Its nonnegative elements map to positive elements, while every negative element maps to an integer at most $-2$; hence no element maps to $-1$. This gives a direct first-order distinction, without treating the negative branch as a two-sided successor chain.
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