列表的最大值
我有自然数列表,其最大值不是零。我也有反对陈述,索引nth kl 0的值大于最大值。我如何证明这个错误的语句?因为列表中存在的所有值都小于最大值(对于所有NL,N< = List Max LT; = list Max l )。
I have list of natural numbers and its maximum value is not zero. I also have contra statement that value at index nth k l 0 is greater than maximum value.How I can prove that this wrong statement?Because all values present in the list are smaller than maximum value (for all n l ,n<=list max l).
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确实有点复杂。代码中的搜索命令向您显示如何找到所需的引理。
如您所见,您甚至不需要列表的最大值大于0的前提。
It is indeed a bit intricate. The Search commands in the code show you how to find the required lemmas.
As you see you don't even need the premise that the max of your list is larger than 0.
您也可以尝试应用置换以证明更“直接”的属性。
这是一个使用(有限)元组的变体,而不是具有
MathComp
的列表,以给出一个想法:You could also try to apply contraposition to prove a more "direct" property.
Here is a variant using (finite) tuples instead of lists with
mathcomp
, to give an idea: