1
$\begingroup$

The statements are:

  1. $\forall x \exists y(fx = y)$
  2. $\forall x \exists x(fx = x)$
  3. $\exists x(fx = x)$

I know that two statements $\alpha, \beta$ are equivalent iff $I \vDash (\alpha)\equiv (\beta)$ for every interpretation $I.$

It might be wise trying to prove $I \vDash (\alpha) \rightarrow (\beta)$ and $I \vDash (\alpha) \leftarrow (\beta)$ seperately.

But how do I do that?

Obviously, the first statement doesn't imply the second, whereas the second statement does indeed imply the first. The second statement describes a constant function, whereas the first statement describes a function in general. Of course every constant function is a function, but not the other way around. But I don't think that this is a proper answer.

$\endgroup$
5
  • $\begingroup$ I know that it's obvious which of these statements are equivalent, but I guess the goal of this excercise is to prove it with different interpretations, not with finding some counterexample on $\Bbb N$ or something like that. $\endgroup$
    – Julian
    Commented May 10, 2018 at 10:08
  • $\begingroup$ A counterexample is enough to show that you cannot prove the equivalence. $\endgroup$ Commented May 10, 2018 at 10:11
  • $\begingroup$ Yes, but I thought that I have to work with the different rules for interpretations here. They look like this: de.wikiversity.org/wiki/Mathematische_Logik/Modellbeziehung/… $\endgroup$
    – Julian
    Commented May 10, 2018 at 10:12
  • $\begingroup$ Or would this only for some sort of syntactical rules for interpretation? $\endgroup$
    – Julian
    Commented May 10, 2018 at 10:16
  • $\begingroup$ The second statement does not describe a constant function but simply a function that has at least one fixed point. A constant endofunction will satisfy this, but will so will many other non-constant functions. If you were working in a multisorted context, then constant functions between different sorts would not necessarily satisfy this. $\endgroup$ Commented May 10, 2018 at 17:27

1 Answer 1

4
$\begingroup$

2 and 3 are cleary equivalent, because there is no $x$ free into $∃x(fx=x)$.

Using the quantifier axiom $\forall x \varphi \to \varphi(t/x)$, we have that $\forall x \varphi \vdash \varphi$.

And from Generalization Theorem we have that :

if $x$ is not free in $\varphi$, then $\varphi \vdash \forall x \varphi$.

For a formal proof of $\vdash \varphi \leftrightarrow \forall x \varphi$, when $x$ is not free in $\varphi$, we need the specification of the proof system : Natural Deduction, etc.

A "semantical" proof needs the details regarding the semantical specifications: e.g. $M \vDash \forall x \varphi$ iff $M \vDash \varphi[m/x]$, for every $m \in M$.

Thus, $M \vDash \forall x \exists x (fx=x)$ iff $M \vDash \exists x (fx=x)[m/x]$, for every $m \in M$.

But $x$ is not free in $\exists x (fx=x)$ and thus $\exists x (fx=x)[m/x] = \exists x (fx=x)$.


1 and 3 are not: we can find a counterexample in $\mathbb N$ interpreting $f$ as the successor function.

We have that : $\mathbb N \vDash ∀x∃y(s(x)=y)$ but clearly $\mathbb N \nvDash ∃x(s(x)=x)$.

$\endgroup$
5
  • $\begingroup$ But I could find a counterexample for 2 and 3 too. Define $f$ as a function on $\Bbb N$ such that $f(0) = 0$ and $f(n) = n+1$ for every $n \ge 1$. In this case, the second statement wouldn't hold, but the third statement would. $\endgroup$
    – Julian
    Commented May 10, 2018 at 10:23
  • $\begingroup$ Besides that, we never had the theorem that you mentioned, so I am not allowed to use it. $\endgroup$
    – Julian
    Commented May 10, 2018 at 10:25
  • 1
    $\begingroup$ @Julian - if $f(0)=0$ we have that $\exists x (fx=x)$ is true in that model. This is not a counterexample: but a "positive" example is not a proof. $\endgroup$ Commented May 10, 2018 at 10:56
  • $\begingroup$ But isn't this what's it about? Proving that it doesn't hold in a certain model? $\endgroup$
    – Julian
    Commented May 10, 2018 at 11:33
  • $\begingroup$ I mean, that's what you did when you used a counterexample by assuming that $f$ is a successfor function on $\Bbb N$. $\endgroup$
    – Julian
    Commented May 10, 2018 at 11:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .