If $\{v_1+v_2, v_2+v_3, v_1+v_3\}$ are linearly independent then $\{v_1, v_2, v_3\}$ are linearly independent Problem. Prove that

for $v_1, v_2, v_3 \in \mathbb{R}^3$, if $\{v_1+v_2, v_2+v_3, v_1+v_3\}$ are linearly independent then $\{v_1, v_2, v_3\}$ are linearly independent.


What I tried: 
Let $m,n,p \in \mathbb{R}$ be such that
$$mv_1+nv_2+pv_3 = 0\;(\star)$$
From the hypothesis we know that if $a,b,c \in \mathbb{R}$ with $a(v_1+v_2)+b(v_2+v_3)+c(v_1+v_3) = 0$, then $a=b=c=0$.
First, every element $\begin{pmatrix}m\\n\\p \end{pmatrix} \in \mathbb{R}^3$ can be unique written in terms of $A = \biggl\{\begin{pmatrix}1\\0\\1 \end{pmatrix},\begin{pmatrix}1\\1\\0 \end{pmatrix},\begin{pmatrix}0\\1\\1 \end{pmatrix}\biggr\}$ because $A$ is a basis in $\mathbb{R}^3$, so we can let $\begin{cases} m=a+b \\ n=b+c \\ p=a+c \end{cases}$. So, from
$$
\begin{align}
(\star) \implies (a+b)v_1 + (b+c)v_2+(a+c)v_3=0 \\
\iff av_1+bv_1+bv_2+cv_2+av_3+cv_3=0 \\ 
\iff a(v_1+v_3)+b(v_1+v_2)+c(v_2+v_3)=0 \\
\implies a=b=c=0 \implies m=n=p=0
\end{align}$$
$\implies \{v_1, v_2, v_3\}$ are linearly independent 
Please correct me if I am wrong or not. Thanks!

Another approach using determinants:
Let's notice that we can show the tuple as a matrix
$$ \left\{ v_{1}+v_{2},v_{2}+v_{3},v_{1}+v_{3}\right\} =\left[\begin{array}{ccc}
v_{1} & 0 & v_{1}\\
v_{2} & v_{2} & 0\\
0 & v_{3} & v_{3}
\end{array}\right] $$
Because the columns are independent we know that the determinant is non zero. $$ \left|\begin{array}{ccc}
v_{1} & 0 & v_{1}\\
v_{2} & v_{2} & 0\\
0 & v_{3} & v_{3}
\end{array}\right|=v_{1}v_{2}v_{3}+v_{1}v_{2}v_{3}=2v_{1}v_{2}v_{3}\ne0 $$
Let us now work with $ A=\left[\begin{array}{ccc}
v_{1} & 0 & 0\\
0 & v_{2} & 0\\
0 & 0 & v_{3}
\end{array}\right]\Rightarrow\left|A\right|=\left|\begin{array}{ccc}
v_{1} & 0 & 0\\
0 & v_{2} & 0\\
0 & 0 & v_{3}
\end{array}\right|=v_{1}v_{2}v_{3}\ne0 $
