Prove the distributive law for maximum and minimum operation Given $A=\{1,2,3,4\}\subseteq \mathbb{N}$ and define the operation on $A$ as below
\begin{eqnarray}
a\oplus b&=& \max(a,b)\\
a\otimes b&=& \min(a,b)
\end{eqnarray}
for all $a,b\in A$.
Prove $(a\oplus b)\otimes c=(a\otimes b)\oplus(a\otimes c)$ for all $a,b,c\in A$.
I only can prove it with tables, but it spent long time. If I using tables, I must check $4^3=64$ possible way.
Now I try to prove with definition of operations as below.
\begin{eqnarray}
(a\oplus b)\otimes c&=&\max(a,b)\otimes c\\
&=&\min(\max(a,b),c)
\end{eqnarray}
Now I get stuck here. I cannot make this form
\begin{eqnarray}
(a\otimes b)\oplus(a\otimes c) &=& \min(a,b)\oplus \min(a,c)\\
&=& \max(\min(a,b),\min(a,c)).
\end{eqnarray}
Anyone can give me hint how to prove this distributive law for this operations?

You have stated the distributive law incorrectly. You want $$(a\oplus b)\otimes c=(a\otimes c)\oplus(b\otimes c)\quad(*)$$
The easiest approach is to look at 6 cases: $a>b>c,a>c>b,b>a>c,b>c>a,c>a>b,c>b>a$.
If you look at $a>b>c$, then $(a\oplus b)\otimes c=a\otimes c=c$. But $(a\otimes b)\oplus(a\otimes c)=b\oplus c=b$.
It is straightforward to verify $(*)$ for each of the six cases. 
