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Let $(G,.,e)$ be a torsion-free group with a total order that is compatible with the group operation, by which we mean that if $a,b \in G$ with $a \leq b$ then $ca \leq cb$ and $ac \leq bc$ for all $c\in G$. We can turn $G$ into an idempotent semiring by defining $a+b=max \left\lbrace a,b \right\rbrace$ for all $a,b \in G$. For any non-empty finite $L \subseteq G$, how to prove that

$a+bc=a$ and $a+bc=a$ for all $a,b \in L$, where $c=min\left\lbrace (minL)(maxL)^{-1},(maxL)^{-1}(minL)\right\rbrace$

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    If it's given to be finite, then you can write down the elements of $L$ as $a_1 \leq a_2 \leq \ldots \leq a_n$, and then see how the situation simplifies. – 2017-02-22
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    Are you sure you wanted to write $ (\min L)(\max L)^{-1},(\max L)^{-1}(\min L)$? Because it looks like the two numbers are identical... – 2017-02-22
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    This is false. Take $L$ to be any finite set containing $0$ and at least one negative and one positive number. Then take $b=0$ and $a < 0$ in $L$. – 2017-02-22
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    Why did this question change so radically? – 2017-02-25

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Given

$c = min\left\lbrace (minL)(maxL)^{-1},(maxL)^{-1}(minL)\right\rbrace \\ \\ \Rightarrow c \leq (maxL)^{-1}(minL) \\ \Rightarrow (max L)c \leq (max L)(maxL)^{-1}(minL) \\ \Rightarrow (max L) c \leq (minL) \\ \Rightarrow bc \leq (max L)c \leq (minL) \leq a \\ \Rightarrow max \left\lbrace a, bc \right\rbrace =a \\ \Rightarrow a+bc=a \\ \mbox{similarly we can prove that a+cb=a} $