# What is the meaning of the statement $\forall x \exists y \forall z (z \neq y \iff f(x) \neq z)$?

I need to understand the meaning of the FOL statement below.

$$\forall x \exists y \forall z (z \neq y \iff f(x) \neq z)$$

Does this imply that $$x$$, $$y$$, and $$z$$ cannot be the same or $$f(x)$$ has no value?

The statement is

"for all $$x$$, there exists a value of $$y$$ such that for all $$z$$,
$$z\neq y$$ if and only if $$z \neq f(x)$$".

This can be simplified: \begin{align} & & \forall x \exists y \forall z (z\neq y \iff z \neq f(x))\\ &\implies & \forall x \exists y \forall z (z=y \iff z = f(x))\\ &\implies & \forall x \exists y \forall z (y = f(x))\\ &\implies & \forall x \exists y (y = f(x))\\ \end{align}

If we denote the set of all values of $$x$$ by $$X$$ and the set of all values of $$y$$ by $$Y$$, then this tells us that the function $$f$$ maps every $$x$$ in $$X$$ to a $$y$$ in $$Y$$. That is, $$f: X \to Y$$.

• Thanks a lot Philip, I got that for each value of x there is a corresponding y mapping, will there be a case such that- 1. x,y,z cannot be same, 2. for each argument x, f(x) is unique, 3) y indicates that f(x) has no value ..
– ammu
Apr 20, 2019 at 4:54
• No to all. The first two are more restrictive than the given statement. For 1, let f be the identity mapping. For 2, f could be a constant function. I'm assuming all variables share the same domain. 3 is in direct contradiction with the statement, which says that f(x) always has a value. Apr 20, 2019 at 5:08
• @ammu Don't forget to accept answers by clicking the check mark if they solved your problem. Apr 20, 2019 at 18:45