system for $N=4$ given by (\[N=4a\])-(\[N=4d\]) has four consistent reductions; two of them are local and the others are nonlocal. To obtain one-soliton solution of the reduced equations we use Type 1 and Type 2 approaches given in [@GurPek2].

Local Reductions for N=4
------------------------

Here we present two different local reductions of the coupled HI-mKdV system (\[coupledmKdVN\]) for $N=4$.

**i.** **$v_4=a_1v_2$, $v_3=b_1v_1$**, $a_1$ and $b_1$ are constants. The system (\[N=4a\])-(\[N=4d\]) is reduced to a system of two equations without any additional condition by this reduction. The reduced system is $$\begin{aligned}
&\mu v_{1,t}+6[(c_{12}+c_{14}a_1+c_{23}b_1+c_{34}a_1b_1)v_1v_2+c_{24}a_1v_2^2+c_{13}b_1v_1^2]v_{1,x}+v_{1,xxx}=0\\
&\mu v_{2,t}+6[(c_{12}+c_{14}a_1+c_{23}b_1+c_{34}a_1b_1)v_1v_2+c_{24}a_1v_2^2+c_{13}b_1v_1^2]v_{2,x}+v_{2,xxx}=0.\label{localiN=4}\end{aligned}$$ If we use the Type 1 approach, the constraints for the parameters of one-soliton solution of the system (\[localiN=4\]) are found from the following equalities: $$\begin{aligned}
\displaystyle
&v_4=a_1v_2 \Rightarrow \frac{g_4}{f}=a_1\frac{g_2}{f} \Rightarrow g_4=a_1g_2,\\
&v_3=b_1v_1 \Rightarrow \frac{g_3}{f}=b_1\frac{g_1}{f} \Rightarrow g_3=b_1g_1.\end{aligned}$$ To satisfy the above equalities one of the cases for the parameters is $$1)\, k_4=k_2, \quad 2)\, k_3=k_1,\quad 3)\, e^{\delta_4}=a_1e^{\delta_2},\quad 4)\, e^{\delta_3}=b_1e^{\delta_1}.$$ Under these constraints, one-soliton solution of the system (\[localiN=4\]) becomes $$\displaystyle
v_1=\frac{g_1}{f},\quad v_2=\frac{g_2}{f}$$ where $$\begin{aligned}
&g_1=e^{k_1x-\frac{k_1^3}{\mu}t+\delta_1}+\frac{a_1c_{24}(k_1-k_2)^2}{4k_2^2(k_1+k_2)^2}e^{(k_1+2k_2)x-\frac{(k_1^3+2k_2^3)}{\mu}t+\delta_1+2\delta_2}\\
&g_2=e^{k_2x-\frac{k_2^3}{\mu}t+\delta_1}+\frac{b_1c_{13}(k_1-k_2)^2}{4k_1^2(k_1+k_2)^2}e^{(2k_1+k_2)x-\frac{(2k_1^3+k_2^3)}{\mu}t+2\delta_1+\delta_2}\end{aligned}$$ and $$\begin{aligned}
f=&1+\frac{1}{(k_1+k_2)^2}[c_{12}+a_1c_{14}+b_1c_{23}+a_1b_1c_{34}]e^{(k_1+k_2)x-\frac{(k_1^3+k_2^3)}{\mu}t+\delta_1+\delta_2}\nonumber\\
&+\frac{b_1c_{13}}{4k_1^2}e^{2k_1x-\frac{2k_1^3}{\mu}t+2\delta_1}
+\frac{a_1c_{24}}{4k_2^2}e^{2k_2x-\frac{2k_2^3}{\mu}t+2\delta_2}\nonumber\\
&+\frac{a_1b_1c_{13}c_{24}(k_1-k_2)^4}{16k_1^2k_2^2(k_1+k_2)^4}e^{(2k_1+2k_2)x-\frac{(2k_1^3+2k_2^3)}{\mu}t+2\delta_1+2\delta_2}.\end{aligned}$$

Consider the following example.\
**Example 1.** Choose the parameters as $\mu=3, k_1=\frac{1}{2}, k_2=\frac{3}{2}$, and $a_1=b_1=e^{\delta_1}=e^{\delta_2}=c_{ij}=1$, $1\leq i< j\leq 4$. Hence the pair of the 