It is well known that for the following statement: $$∀y<b∃x₁,…,x_{m}[F(a,y,x₁,…,x_{m})=0]....................(1)$$ where $a$ is the parameter(s) of the polynomial $F$, the Chinese remainder theorem method results in the following system of Diophantine conditions solvable in the unknowns $q,w,z₀,…,z_{m}$ provided that (1) holds: $$F(a,z_0,z_1,…,z_m)\equiv 0 \pmod {\binom {q}b}$$
$$z_0=q$$
$b!×( b+w+B(a,b,w)) !$ divide $q+1$
$\binom {q}b $ divide $(\binom {z_1}w )$
$\binom {q}b $ divide $(\binom {z_2}w )$
....
$\binom {q}b $ divide $(\binom {z_m}w )$
where the polynomial $B(a,b,w)$ is obtained from $F(a,y,x₁,…,x_{m})$ by changing the signs of all its negative coefficients and systematically replacing $y$ by $b$ and $x₁,…,x_{m}$ by $w$.
The statement (1) is hold for all $y<b$, i.e., $0≤y<b$. Now, my question is: How one can transform the statement:
$$∀c<y<b∃x₁,…,x_{m}[F(a,y,x₁,…,x_{m})=0]....................(2)$$
to a system of Diophantine conditions where $c>0$.
We can rewrite your modified version as follows: $$\forall y<b\exists x_1,...,x_m[y=0\vee y=1\vee ... \vee y=c \vee F(a,y,x_1,...,x_m)=0].$$ Note that this really is equivalent to your original statement. To see this, think in terms of counterexamples (two statements are equivalent if they have the same potential counterexamples after all): a $y$ which is a counterexample to the above statement cannot be $\le c$ since then one of the "$y=i$"-disjuncts would automatically be true, and this corresponds to the "up front" requirement $y>c$ in your original statement.
And the above in turn can be rewritten as $$\forall y<b\exists x_1,...,x_m[G(a,y,x_1,...,x_m)=0]$$ where $G(a,y,x_1,...,x_m)=[\prod_{0\le i\le c} (y-i)]\cdot F(a,y,x_1,...,x_m)$.