Answers.Body: <p>The statement is  </p>
  
  <p>"for all <span class="math-container">$x$</span>, there exists a value of <span class="math-container">$y$</span> such that for all <span class="math-container">$z$</span>,<br>
  <span class="math-container">$z\neq y$</span> if and only if <span class="math-container">$z \neq f(x)$</span>".</p>
  
  <p>This can be simplified:
  <span class="math-container">$$\begin{align}
  &amp; &amp; \forall x \exists y \forall z (z\neq y \iff z \neq f(x))\\
  &amp;\implies &amp; \forall x \exists y \forall z (z=y \iff z = f(x))\\
  &amp;\implies &amp; \forall x \exists y \forall z (y = f(x))\\
  &amp;\implies &amp; \forall x \exists y (y = f(x))\\
  \end{align}$$</span></p>
  
  <p>If we denote the set of all values of <span class="math-container">$x$</span> by <span class="math-container">$X$</span> and the set of all values of <span class="math-container">$y$</span> by <span class="math-container">$Y$</span>, then this tells us that the function <span class="math-container">$f$</span> maps every <span class="math-container">$x$</span> in <span class="math-container">$X$</span> to a <span class="math-container">$y$</span> in <span class="math-container">$Y$</span>.  That is, <span class="math-container">$f: X \to Y$</span>.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-04-20T04:54:34.667Z
Answers.Text: 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 ..
Answers.UserDisplayName: ammu
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-04-20T05:08:00.297Z
Answers.Text: 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.
Answers.UserDisplayName: Philip Raeisghasem
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-04-20T18:45:17.917Z
Answers.Text: @ammu Don't forget to accept answers by clicking the check mark if they solved your problem.
Answers.UserDisplayName: Philip Raeisghasem
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-04-14T19:57:34.443Z
Answers.LastActivityDate: 2019-04-14T19:57:34.443Z
Answers.UserDisplayName: Philip Raeisghasem
Body: <p>I need to understand the meaning of the FOL statement below.</p>
  <p><span class="math-container">$$
  \forall x \exists y \forall z (z \neq y \iff f(x) \neq z)
  $$</span></p>
  <p>Does this imply that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, and <span class="math-container">$z$</span> cannot be the same or <span class="math-container">$f(x)$</span> has no value?</p>
  
Tags: <logic>
Title: What is the meaning of the statement $\forall x \exists y \forall z (z \neq y \iff f(x) \neq z)$?
UserDisplayName: ammu
ThreadText: # What is the meaning of the statement $\forall x \exists y \forall z (z \neq y \iff f(x) \neq z)$?
  
  (Asked by: username_0 on 2019-04-14)
  
  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?
  
  **Tags:** logic
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2019-04-14. Score: 4 | Accepted answer
  
  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, $f: X \to Y$.
  
  > **Comments:**
  > **username_0**: 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 ..  
  > **username_1**: 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.  
  > **username_1**: @ammu Don't forget to accept answers by clicking the check mark if they solved your problem.  
  
  ---
  
  