$d$ be an integer with $R^+(f,d) + R^+({g},d)<1.$ Then $$\begin{aligned}
\degthr(f\wedge {g}) \leq 2d.\end{aligned}$$

Consider rational functions $p_1(x)/q_1(x)$ and $p_2(y)/q_2(y)$ of degree at most $d$ such that $q_1$ and $q_2$ are positive on $X$ and $Y,$ respectively, and $$\begin{aligned}
   \sup_{X} \left| f(x) - \frac{p_1(x)}{q_1(x)}\right| + 
   \sup_{Y} \left| {g}(y) - \frac{p_2(y)}{q_2(y)}\right| < 1. \end{aligned}$$ Then $$\begin{aligned}
 f(x)\wedge {g}(y) \equiv \sign\{1 + f(x)+{g}(y)\}
 \equiv \sign{\left\{ 1 + \frac{p_1(x)}{q_1(x)} +
\frac{p_2(y)}{q_2(y)} \right\}}.\end{aligned}$$ Multiplying the last expression by the positive quantity $q_1(x)q_2(y)$ gives $f(x)\wedge {g}(y) \equiv \sign\{ q_1(x)q_2(y) + p_1(x)q_2(y)
+ p_2(y)q_1(x)\}.$

We will also need a converse to Theorem \[thm:rational-is-possible\], proved in .

\[thm:sherstov-degthr-R\] Let $f\colon X\to\moo$ and ${g}\colon Y\to\moo$ be given functions, where $X,Y\subset\Re^n$ are arbitrary finite sets. Assume that $f$ and $g$ are not identically false. Let $d=\degthr(f\wedge g).$ Then $$\begin{aligned}
R^+(f,4d) + R^+(g,2d) < 1.\end{aligned}$$

Symmetric functions. {#symmetric-functions. .unnumbered}
--------------------

Let $S_n$ denote the symmetric group on $n$ elements. For $\sigma\in S_n$ and $x\in\zoon$, we denote $\sigma
x=(x_{\sigma(1)},\ldots,x_{\sigma(n)})\in\zoon.$ For $x\in\zoon,$ we define $|x|= x_1+x_2+\cdots+x_n.$ A function $\phi\colon \zoon\to\Re$ is called *symmetric* if $\phi(x) = \phi(\sigma x)$ for every $x\in\zoon$ and every $\sigma\in S_n.$ Equivalently, $\phi$ is symmetric if $\phi(x)$ is uniquely determined by $|x|.$ Symmetric functions on $\zoon$ are intimately 