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Prove 1 == 0 in Wolfram Language

Posted 7 years ago

Consider the following code:

step1 = Exists[x, x \[Element] Complexes, x^2 + 1 == 0];
    Resolve[step1]
    Reduce[step1, x]
    FullSimplify[step1, x]
    FindInstance[step1, x, Complexes]

giving

True
True
True
{{x -> 0}}

Telling us that 1==0 and True==False.

Inactivate[Evaluate@step1[[3]], Equal] /. x -> 0
Activate[%]

giving

1 == 0
False

Voila!

8 Replies
POSTED BY: Gianluca Gorni

Paradox: Nothing can make False true, but False makes everything true. Go figure.

POSTED BY: Daniel Lichtblau

The variable x inside the quantifier is local to the quantifier. It is not the same as the global variable x specified in FindInstance. The statement

Exists[x, x \[Element] Complexes, x^2 + 1 == 0]

does not have any free variables. It is a true sentence, so the input to FindInstance is equivalent to FindInstance[True, x]. True is satisfied by any value of x, in particular by x->0.

A quantifier variable is bound to the nearest enclosing quantifier. For instance, the formula form below contains four different variables named x, one free, three bound to different quantifiers. Variables specified in the second argument of FindInstance are interpreted as free variables of its first argument.

In[1]:= form = x==17 && ForAll[x, x>2, x^2>4 && Exists[x, x>0, x<1 && ForAll[x,
x<0, x^2>0]]];

In[2]:= FindInstance[form, x]
Out[2]= {{x -> 17}}
POSTED BY: Adam Strzebonski

I don't even understand how to interpret the intended semantics of

 FindInstance[
  Exists[x, x \[Element] Complexes, x^2 + 1 == 0], x, Complexes]

but certainly the output {{x -> 0}} is bizarre.

Hope you've reported this to WRI tech support!

POSTED BY: Murray Eisenberg
In[3]:= FindInstance[step1, x, Booleans]

Out[3]= {}
POSTED BY: Frank Kampas
In[1]:= FindInstance[x^2 + 1 == 0, x]

Out[1]= {{x -> -I}}
POSTED BY: Frank Kampas

I know what the correct mathematical answer is, but I was practicing using Exists and FindInstance, which should also work.

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