Checking for linear independence Is the following proof correct?
Theorem. If $v_1,v_2,v_3,v_4$ is a linearly independent list then 
$$v_1-v_2,v_2-v_3,v_3-v_4,v_4$$
is also a linearly independent list.
Proof. Assume that $v_1,v_@,v_3,v_4$ is a linearly independent list, Consider now the following equation.
$$0=0(v_1-v_2)+0(v_2-v_3)+0(v_3-v_4)+0v_4\tag{1}$$
Let $a_1,a_2,a_3$ and $a_4$ be arbitrary scalars in $\mathbf{F}$ and assume that the following equation holds
$$0=a_1(v_1-v_2)+a_2(v_2-v_3)+a_3(v_3-v_4)+a_4v_4\tag{2}$$
After some algebraic manipulation we arrive at the following equation.
$$0=a_1v_1+(a_2-a_1)v_2+(a_3-a_2)v_3+(a_4-a_3)v_4\tag{3}$$
Since the list $v_1,v_2,v_3,v_4$ is linearly independent it follows that given any vector in $span(v_1,v_2,v_3,v_4)$ the choice of scalars is unique and since
$$0=0v_1+0v_2+0v_3+0v_4\tag{4}$$
It follows that all the scalars in $(3)$ must be $0$, consequently the only way to produce the $0$ vector as a linear combination of the vectors in the list $v_1-v_2,v_2-v_3,v_3-v_4,v_4$ is that indicated in $(1)$.
$\blacksquare$
Here $\mathbf{F}$ is either $\mathbb{C}$ or $\mathbb{R}$.
 A: Equation (1) should be omitted: it's an obvious fact that has no consequence on the rest.
The final argument is too fast: from linear independence of $v_1,v_2,v_3,v_4$ you deduce
\begin{cases}
a_1=0 \\
a_2-a_1=0 \\
a_3-a_2=0 \\
a_4-a_3=0
\end{cases}
and, from this, $a_1=a_2=a_3=a_4=0$. This should be mentioned, although easy.
A different approach is to consider the coordinates of the new vectors with respect to the original ones and so the matrix
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 1
\end{bmatrix}
A standard Gaussian elimination leads to the reduced row echelon form
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
which proves that the coordinate vectors are linearly independent and so also the vectors are.
A: It's valid. A note:
Before $(2)$ when you say "...and assume that the following equation holds..." it would be better to perhaps phrase this as "We wish to solve ... for $a_j$." Saying the former means that you're assuming solutions exist - but you're not sure! This is a tiny point, and is perhaps reflected in my writing style.
We may also streamline a bit by saying the following after $(3)$.

As $\{v_1, v_2, v_3, v_4\}$ is a linearly independent set, we must have that $a_1 = 0, \ a_2-a_1=0, \ a_3 - a_2 = 0,$ and $a_4-a_3=0.$ Back substitution yields that each $a_j = 0$, and so the original set is also a linearly independent set. $\square$

Again, these are just small suggestions, but nonetheless your proof is correct.
