The Simplify command is having trouble verifying a compound inequality, even while it correctly handles the two corresponding single inequalities.

I'm starting with some simple global assumptions:

$Assumptions = Element[k, Integers] && i == k + 1;

Under these assumptions, the following uses of Simplify all return True, as expected:

In[2]:= Simplify[i == k + 1]
Out[2]= True

In[3]:= Simplify[k < k + 1 < k + 2]
Out[3]= True

In[4]:= Simplify[k < i && i < k + 2]
Out[4]= True

However, combining the two inequalities from the last input above into one compound inequality does not yield the expected output:

In[5]:= Simplify[k < i < k + 2]
Out[5]= k < i < 2 + k
(*NOT True as expected*)

Why does Simplifynot recognize that this compound inequality is true?

(Note: I realize it seems silly to check if k<i<k+2 since I'm already assuming i=k+1. However, I was having problems while dealing with more complex code, and this behavior of Simplify seemed to be at the root of it.)


1 Answer 1


Something I've noted about Mathematica, is it is best to be complete in defining variable domains. Also define i as an integer, which you might think it implicitly is from your definition i == k + 1:

$Assumptions = {i, k} ∈ Integers && i == k + 1;

Then try your code

Simplify[k < i < k + 2]
(* True *)

Curious to see if that fixes your bigger problem.

  • 1
    $\begingroup$ Thanks! That does fix the issue I was having! $\endgroup$
    – Tim Wagner
    Commented Dec 30, 2019 at 4:04

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