About the pointwise infimum of a continuous piecewise linear function I am reading the logic paper "Interpolation in fuzzy logic" by Matthias Baaz and Helmut Veith
https://doi.org/10.1007/s001530050151
On page 479 it is claimed that "It suffices to show that continuous piecewise linear functions are closed under taking suprema and infima of a single coordinate $y$; geometrically, this means taking the upper boundary of the union of orthogonal projections of all polytopes on the hyperplane y = 0, [15]."
Unfortunately, reference [15] is non helpful at all because it is said that
[15] is "Hryniv, O.: Personal Communication."
Thus, I am wondering how to prove that if $f: [0,1]^{n+1} \longmapsto [0,1]$ is a continuous piecewise linear function then the function $\bigwedge f: [0,1]^n \longmapsto [0,1]$ defined by
$$
(\bigwedge f) (\vec{x}):= \inf \{ f(y,\vec{x}): y \in [0,1]  \}
$$
is also a continuous piecewise linear function. Can anybody suggest a way to prove this statement?
Indeed, I am more interested in a similar statement: if $f$ is a continuous piecewise linear function with rational coefficients then $\bigwedge f$ is also continuous piecewise linear with rational coefficients. I suspect that a proof of the previous result can give me hints about how to prove this statement.
 A: I'll denote elements of $[0,1]^{n+1}$ by $\vec z$ and elements of $[0,1]^{n}$ by $\vec x$.
First, $\bigwedge f$ is continuous. For each linear piece $f_i$, there is $\lambda_i$ such that $|f_i(\vec z)-f_i(\vec z_0)|<\lambda_i|\vec z-\vec z_0|$.  Then by dividing the segment between $\vec{z}$ and $\vec{z_0}$ into parts according to the linear pieces, we can conclude that $|f(\vec z)-f(\vec z_0)|<\lambda|\vec z-\vec z_0|$ with $\lambda=\max \lambda_i$.
For $\vec x,\vec x_0\in[0,1]^n$, let $\hat y$ and $\hat y_0$ be the respective values of $y$ and $y_0$ at which the infima of $f(y,\vec x)$ and $f(y_0,\vec x_0)$ are attained. If $f(\hat y,\vec x)\ge f(\hat y_0,\vec x_0)$, then
$$
\begin{eqnarray}
\left|(\bigwedge f) (\vec{x})-
(\bigwedge f) (\vec{x}_0)\right|
&=&
f(\hat y,\vec x)-f(\hat y_0,\vec x_0)
\\
&=&
(f(\hat y,\vec x)-f(\hat y_0,\vec x))+(f(\hat y_0,\vec x)-f(\hat y_0,\vec x_0))
\\
&\le&
f(\hat y_0,\vec x)-f(\hat y_0,\vec x_0)
\\
&\le&
\lambda\left|\vec x-\vec x_0\right|\;,
\end{eqnarray}
$$
and likewise, if $f(\hat y,\vec x)\le f(\hat y_0,\vec x_0)$, then
$$
\begin{eqnarray}
\left|(\bigwedge f) (\vec{x})-
(\bigwedge f) (\vec{x}_0)\right|
&=&
f(\hat y_0,\vec x_0)-f(\hat y,\vec x)
\\
&=&
(f(\hat y_0,\vec x_0)-f(\hat y,\vec x_0))+(f(\hat y,\vec x_0)-f(\hat y,\vec x))
\\
&\le&
f(\hat y,\vec x_0)-f(\hat y,\vec x)
\\
&\le&
\lambda\left|\vec x-\vec x_0\right|\;.
\end{eqnarray}
$$
Thus $\bigwedge f$ is continuous. To see that it is piecewise linear, orthogonally project all the points defining the pieces of $f$ to $[0,1]^n$ and triangulate the resulting set of points. For each piece $f_i$ of $f$, let $D_i$ be the orthogonal projection of the domain of $f_i$ to $[0,1]^n$. Then for each simplex of the triangulation and each piece $f_i$, the simplex is either entirely inside or entirely outside $D_i$, and the infimum $\bigwedge f_i$ of $f_i$ with respect to $y$ is a linear function on the simplex (since it is attained at the boundary of the domain of $f_i$, that boundary is a linear function of $\vec x$ within the simplex, and $f_i$ is linear). Thus, on each simplex, $\bigwedge f$ is the infimum with respect to $i$ of the infima $\bigwedge f_i$ of all pieces $f_i$ for which $D_i$ contains the simplex. For each pair of these $\bigwedge f_i$, find the hyperplane on which they are equal and, if the hyperplane intersects the simplex, subdivide the simplex accordingly. Then on each resulting polytope, $\bigwedge f$ is defined by one of the $\bigwedge f_i$ alone, which is a linear function on the simplex, and therefore on the polytope. Thus $\bigwedge f$ is piecewise linear.
