5
$\begingroup$

I am new to Mathematica and to Math. I do not study math in English, so please bear with me as I try to state my questions..

This is what I am trying to "solve":

$$∃!x : A(x) ⇔ ∃x : ¬A(x)$$

In words: "Is 'there is exactly one x for which the assertion A is true' equivalent to 'there is at least one x for which assertion A is not true' ?"

Referred to that I have questions:

  • Is this something you would normally do with Mathematica?
  • In the program I opened a "New Notebook" and typed in:

    Equivalent[!Exists[x, a[x]], Exists[x, !a[x] ]] 
    

    and it - surprise! ;) - did not work. What have I done wrong ?

  • How would the answer look like ?
$\endgroup$
2
  • 2
    $\begingroup$ As far as I remember math quantifiers, "there i exactly one x..." is not one of them. They are either "For all x..." or "There is at least one x...". What You are looking for is a uniqueness quantification, but there's no such built-in function in Mathematica. Besides, negation of Exists[ ] results in ForAll[ ]. $\endgroup$
    – Wojciech
    Commented Dec 13, 2013 at 15:07
  • 2
    $\begingroup$ @WojciechSitkiewicz I guess it could be written as Exists[x, a[x]] && ForAll[{x,y}, a[x] && a[y] , x==y] or so $\endgroup$
    – ssch
    Commented Dec 13, 2013 at 16:09

2 Answers 2

3
$\begingroup$

Based on ssch suggestion I have come up with a following code:

Equivalent[Exists[x, a[x]] && ForAll[{x, y}, a[x] && a[y], x == y], 
Exists[x, ! a[x]]] // TautologyQ

False

$\endgroup$
3
$\begingroup$

Actually, ∃! is a standard for "there exists a unique ..." . However, the proposition is false. Consider A(x) where x is greater than 5. There exists lots of x for which x<=5. And surely there is not a unique x, for x>5.

To express ∃! in Mathematica one needs

Exists[x, A[x]] && ForAll[{a, b}, A[a] && A[b] \[Implies] a == b]

Then Mathematica can resolve

In[98]:= Equivalent[
  Exists[x, A[x]] && ForAll[{a, b}, A[a] && A[b] \[Implies] a == b], 
  Exists[x, Not[A[x]]]] // TautologyQ

Out[98]= False 

After doing this, I see that the full answer has already been provided.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.