Reducing $f$ modulo $g_i(x_i)={\prod }_{s\in S_i}(x_i-s)$ lowers each ${\mathrm{deg}}_{x_i}$ below $\mid S_i\mid$, preserves the values on the grid, and preserves any top-degree coefficient whose exponents stay below the grid sizes

Set $r^{(0)}:=f$. For each $i=1,\dots ,n$, view $r^{(i-1)}$ as a polynomial in $x_i$ with coefficients in $F[x_1,\dots ,x_{i-1},x_{i+1},\dots ,x_n]$. Since $g_i$ is monic of degree $\mid S_i\mid$, [F2] gives $r^{(i-1)}=q_i{g}_i+r^{(i)}$ with $r^{(i)}=0$ or ${\mathrm{deg}}_{x_i}{r}^{(i)}<\mid S_i\mid$. Because $g_i$ involves only $x_i$, this subtraction leaves all previously obtained bounds ${\mathrm{deg}}_{x_j}<\mid S_j\mid$ for $j<i$ unchanged; and every nonleading term of $g_i$ has smaller $x_i$-degree than its leading term, so the division step introduces no monomial of larger total degree. Therefore $\tilde{f}:=r^{(n)}$ satisfies part 1.

Let $a=(a_1,\dots ,a_n)\in S_1\times \cdots \times S_n$. For each $i$, step 1.1 gives $r^{(i-1)}-r^{(i)}=q_i{g}_i$, and [F3] gives $g_i(a_i)=0$ because $a_i\in S_i$. Hence $r^{(i-1)}(a)=r^{(i)}(a)$ for every $i$, so chaining these equalities yields $\tilde{f}(a)=f(a)$.

In the situation of part 3, fix a stage $i$. Any monomial of total degree $\mathrm{deg}f$ in $q_i{g}_i$ must use the leading term $x_i^{\mid S_i\mid }$ of $g_i$, because every other term of $g_i$ has smaller total degree. So every top-degree monomial of $q_i{g}_i$ has $x_i$-exponent at least $\mid S_i\mid$. Since the target monomial has exponent $t_i<\mid S_i\mid$ for every $i$, no stage changes its coefficient. Therefore $[x_1^{t_1}\cdots x_n^{t_n}]\tilde{f}=[x_1^{t_1}\cdots x_n^{t_n}]f$.