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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?

Ongky Denny Wijaya
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  • Probably the easiest is to notice that with the association $n = {0, 1, \ldots, n-1}$ you have that $a\oplus b = a\cup b$ and $a\otimes b = a\cap b$. Then you can just use De Morgan's. – nbritten Nov 28 '19 at 14:03

1 Answers1

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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.

almagest
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