# First order predicate formula "normal" mathematical meaning

Just having a last minute pre-exam panic, would be really grateful if someone could help me understand this question (I don't know why but maybe tiredness is stopping me from understanding??)

I'm asked to let $\phi$ be the following $\mathcal{L}$ sentence: $$\phi:\forall x\forall y\exists w\forall z((x\dot=y\land \lnot f(z)\dot =w)\lor (\lnot f(x) \dot=f(y) \land \lnot w \dot= f(z))$$

and express in ordinary mathematical terms what $\phi$ says about $f$.

I don't really understand: I get that given $x$ and $y$ we there is a $w$ such that for any $z$ either $x=y$ and $f(z)≠w$ or $f(x)≠f(y)$ and $f(z)≠w$.

But what actually does this tell me about $f$???

I assume the $\dot=$ is simply $=$?

$$\forall x\forall y\exists w\forall z((x=y\land \lnot f(z) =w)\lor (\lnot f(x) =f(y) \land \lnot w = f(z))$$

which is equivalent to (since $\neg f(z)=w$ is equivalent to $\neg w = f(z)$)

$$\forall x\forall y\exists w\forall z((x=y\land \lnot f(z)=w)\lor (\lnot f(x) =f(y) \land \lnot f(z) = w)$$

which by Distribution is equivalent to:

$$\forall x\forall y\exists w\forall z((x=y \lor \lnot f(x) =f(y)) \land \lnot w = f(z))$$

and by the Prenex Laws that is equivalent to:

$$\forall x\forall y((x=y \lor \lnot f(x) =f(y)) \land \exists w\forall z \lnot w = f(z))$$

and that to:

$$\forall x\forall y(x=y \lor \lnot f(x) =f(y)) \land \exists w\forall z \lnot w = f(z)$$

So, you really have two parts to your statement:

$$\forall x\forall y(x=y \lor \lnot f(x) =f(y))$$

which is equivalent to:

$$\forall x\forall y(f(x) =f(y) \rightarrow x=y )$$

and that says that $f$ is injective, and the second part is:

$$\exists w\forall z \lnot w = f(z)$$

and that says that $f$ is not surjective (or onto)

So: $\phi$ says that $f$ is injective but not surjective.

• yes the $\dot=$ is just the notation used at my university, it basically means $=$. Thanks very much for your answer. May 29 '17 at 21:37
• @elDin0 You're welcome! :) May 29 '17 at 21:40