If $x\perp \mathrm{span}\{r_1,\dots,r_p\}$, can we prove $x\notin\mathrm{span}\{v_1,\dots,v_p\}$? Notations: For a scalar $a\in\mathbb{R}$, denote 
$$\mathrm{sgn}(a)=\left\{
          \begin{array}{l l}
            1  & \mbox{if } a>0\\
            0  & \mbox{if } a=0\\
            -1 & \mbox{if } a<0
          \end{array}.\right.$$
For a vector $r\in\mathbb{R}^n$, $\mathrm{sgn}(r)$ is element-wise.
Now we have a set of vectors $r_1, \dots, r_p\in\mathbb{R}^n$, $p<n$. The corresponding sign vectors are $v_1,\dots,v_p\in\mathbb{R}^n$ satisfying
$$\mathrm{sgn}(r_i)=v_i$$
which implies that the elements of $v_i$ can only be $1$, $-1$ or $0$.
Question: Given a nonzero vector $x\in\mathbb{R}^n$ satisfying $x\perp \mathrm{span}\{r_1,\dots,r_p\}$, can we prove $x\notin\mathrm{span}\{v_1,\dots,v_p\}$? If not, can you given any counterexamples?
 A: Counterexample: we have $x=v_1+v_2+v_3$ when
$$
(r_1,r_2,r_3,x,v_1,v_2,v_3)=\begin{pmatrix}
1&1&-3&1&1&1&-1\\
1&-2&2&1&1&-1&1\\
-2&1&1&1&-1&1&1\\
0&0&0&0&0&0&0
\end{pmatrix}.
$$
Edit. Here is a better example:
$$
(r_1,r_2,r_3,x,v_1,v_2,v_3)=\begin{pmatrix}
1&1&-5&1&1&1&-1\\
1&-5&1&1&1&-1&1\\
-5&1&1&1&-1&1&1\\
1&1&1&3&1&1&1
\end{pmatrix}.
$$
The key is, each $v_i$ contains a negative entry, but the entries of $x=v_1+v_2+v_3$ are all positive. So we have a great liberty to pick some $r_1,r_2,r_3$ that are orthogonal to $x$.
A: The following uses $p=n$ thus strictly speaking is incorrect. It may serve as an example as to why that restriction is necessary for the question though.
Using row vectors, look at
$$\{r_1,\dots,r_p\} = \pmatrix{1 & 7 & 1 \\ -1 & 1 & 1 \\ -2 & -6 & 0}$$
$$ x = \pmatrix{3 & -1 & 4 }$$
Then
$$\{v_1,\dots,v_p\} = \pmatrix{1 & 1 & 1 \\ -1 & 1 & 1 \\ -1 & -1 & 0}$$
The $r$ do not have full span and $x$ is perpendicular to them. After applying the sgn(), the $v$ set has full span, as may be seen with the simple row operation to upper triangular from adding the first row to the other rows:
$$\pmatrix{1 & 1 & 1\\ 0 & 2 & 2 \\ 0 & 0 & 1 \\}$$
Thus a counterexample is given.
It may be more difficult to find a counterexample if given the additional restriction that each $r$ is independent, but I believe it could be given in that case as well.
