Let $L$ be a differential operator.
$$$$
We suppose that $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x=f_j}$ and we assume that $\phi$ can be written as $Lx=f \land \psi$, where $\psi$ doesn't contain any $x$.
We prove this by induction on the number of equations, $n$.
Base case: For $n=1$ we have one equation, so it is of the form $Lx=f$.
Inductive hypothesis: We suppose that it holds for $n=k$, i.e., if $\phi$ contains $k$ equations, then we can reduce it into the form $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$
Inductive step: We will show that it holds for $n=k+1$, i.e., if we have $k+1$ equations we can reduce this system into the form $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$ From the inductive hypothesis we know that we can reduce the first $k$ equations into the above form. So we have two equations that contain $x$ and its derivatives. We add these two equations and we get a new differential equation $Lx=f$. So the initial system is equivalent to the new differential equation $Lx=f$.
$$$$
$$$$
We suppose that $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x \neq g_j}$ and we assume that $\phi$ can be written as $Lx \neq g \land \psi$, where $\psi$ doesn't contain any $x$.
We prove this by induction on the number of inequations, $n$.
Base case: For $n=1$ we have one inequation, so it is of the form $Lx \neq g$.
Inductive hypothesis: We suppose that it holds for $n=k$, i.e., if $\phi$ contains $k$ inequations, then we can reduce it into the form $$Lx \neq g \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$
Inductive step: We will show that it holds for $n=k+1$, i.e., if we have $k+1$ inequations we can reduce this system into the form $$Lx \neq g \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$ From the inductive hypothesis we know that we can reduce the first $k$ inequations into the above form. So we have two inequations that contain $x$ and its derivatives, $Lx \neq g \land L_{k+1}x \neq g_{k+1} \land \psi$. These inequations are equivalent to $L_{k+1}x = f_{k+1}+a \ \ , \ \ Lx \neq f+b$, where $a, b \neq 0$. Now we can add these two equations and we get $\tilde{L}x=L_{k+1}x+Lx=f_{k+1}+a+f+b, a, b \neq 0$. Then we have $\tilde{L}x \neq f_{k+1}+f$.
Is the last part correct? But if $a=-b$? Do we maybe have to suppose that $a,b >0$ ?
$$$$
$$$$
Is everything correct?
Could I improve something at the formulation?
You can continue as following: take the first k equations and create $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$ Now you can set new $\phi$ which contain exactly two equations, the new one and the extra one from the previous $\phi$. The size of new $\phi$ is less then $k$ so you fine.
Alternatively you split it into two parts of sizes $\lceil k/2\rceil$ and $\lfloor k/2\rfloor$ and proceed similarly.