# How to prove the transitivity of the comparator?

• The solution can't be correct without the transitivity:

If AB > BA and BC > CB, then we have AC > CA.

But how to prove it?

• What I have is:

``````1)   AB > BA
2)   BC > CB
``````

from 2) we have

``````3)   BAC > CAB
``````

from 1) we have

``````4)  BAC > CAB > CBA
``````

from 1) we also have

``````5) ABC > BAC > CAB > CBA
``````

from 5 and "take out" the B we have

``````6) AC > CA
``````

Glad you asked this, even after showing this property, I'm still not sure why this "sorting" solution works.

EDIT: Just realized I didn't prove that AB > BA => ACB > BCA. I believe that inequality is true. Don't want to prove it at the moment.

EDIT2 : Realized what I said above is wrong too. My intuition is really not helping me for this problem.

• that's my concern too, I don't know why someone just take it for granted, although it seems correct, but I can't prove it

1. BAC > CAB let's take A 999, B70078, C7007, this assertion is not correct

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.