Confused About the K-MAP Minimization
Assuming your initial table implies what I think it does, then I completely agree with your k-maps:
$$\begin{array}{rl}
\begin{smallmatrix}\begin{array}{r|cccc}
Y_1&amp;\overline{I_1}\:\overline{I_0}&amp;\overline{I_1}\: I_0&amp;I_1 \:I_0&amp;I_1 \:\overline{I_0}\\
\hline
\overline{I_3}\:\overline{I_2}&amp;x&amp;0&amp;0&amp;0\\
\overline{I_3}\:I_2 &amp;1&amp;0&amp;0&amp;0\\
I_3\: I_2 &amp;1&amp;0&amp;0&amp;1\\
I_3\:\overline{I_2} &amp;1&amp;0&amp;0&amp;1
\end{array}\end{smallmatrix}
&amp;
\begin{smallmatrix}\begin{array}{r|cccc}
Y_2&amp;\overline{I_1}\:\overline{I_0}&amp;\overline{I_1}\: I_0&amp;I_1 \:I_0&amp;I_1 \:\overline{I_0}\\
\hline
\overline{I_3}\:\overline{I_2}&amp;x&amp;0&amp;0&amp;1\\
\overline{I_3}\:I_2&amp;0&amp;0&amp;0&amp;1\\
I_3\: I_2&amp;1&amp;0&amp;0&amp;1\\
I_3\:\overline{I_2}&amp;1&amp;0&amp;0&amp;1
\end{array}\end{smallmatrix}
\end{array}$$
I also agree with your circled areas:
.....................
The green area for \$Y_1\$ is \$\overline{I_1}\:\overline{I_0}\$ and the orange area for \$Y_1\$ is \$I_3\:\overline{I_0}\$. So \$Y_1=\overline{I_1}\:\overline{I_0}+I_3\:\overline{I_0}\$.
The green area for \$Y_2\$ is \$I_1\:\overline{I_0}\$ and the orange area for \$Y_2\$ is \$I_3\:\overline{I_0}\$. So \$Y_2=I_1\:\overline{I_0}+I_3\:\overline{I_0}\$.
These are not the same.
I think you did very well, except that you missed just one small detail about the condition of \$I_1\$ between the two k-maps. That's all.