Write formula using quantifiers

**MachinePL1993** · November 12th 2012, 09:29 AM

Hello there.
I need to write a following formula using quantifiers, but it has to be in prenex normal form.

The formula is : function f is not weakly decreasing and not weakly increasing.

I got to here:

$\exists x \exists y(x \le y \wedge f(x)>f(y)) \wedge \exists x\exists y(x \le y \wedge f(x)<f(y))$

by negating :

$(\forall x \forall y(x \le y \wedge f(x)\le f(y)) \vee (\forall x \forall y(x \le y \wedge f(x)\le f(y))$

But I have no idea how to proceed, since my solution is not in prenex normal form.

Please help.

**Plato** · November 12th 2012, 10:13 AM

How are those terms, weakly decreasing and weakly increasing defined?

**MachinePL1993** · November 12th 2012, 10:39 AM

Weakly increasing- $(\forall x \forall y(x \le y \wedge f(x)\le f(y))$

Weakly decreasing- $(\forall x \forall y(x \le y \wedge f(x) \geq f(y))$

I came up with this:

$\exists x \exists y\exists a \exists b((x \le y \wedge (f(x)>f(y))) \wedge (a \le b \wedge (f(a)<f(b)))$

But is it correct? If so then why and how can I derive it from those two formulas above?

**Plato** · November 12th 2012, 11:29 AM

Originally Posted by MachinePL1993

Weakly increasing- $(\forall x \forall y(x \le y \wedge f(x)\le f(y))$

Weakly decreasing- $(\forall x \forall y(x \le y \wedge f(x) \geq f(y))$

I came up with this:
$\exists x \exists y\exists a \exists b((x \le y \wedge (f(x)>f(y))) \wedge (a \le b \wedge (f(a)<f(b)))$
But is it correct? If so then why and how can I derive it from those two formulas above?

Are you sure that it is not an implication?
Weakly increasing- $(\forall x \forall y(x \le y \Rightarrow f(x)\le f(y))$
Weakly decreasing- $(\forall x \forall y(x \le y \Rightarrow f(x) \geq f(y))$
Then the negation you have would be correct.

**MachinePL1993** · November 12th 2012, 11:50 AM

Yes, I'm sorry, there should be an implication.

Weakly increasing- $(\forall x \forall y(x \le y \rightarrow f(x)\le f(y))$

Weakly decreasing- $(\forall x \forall y(x \le y \rightarrow f(x) \geq f(y))$

But now I have to bring it to prenex normal form. Is there some other way to arrive to a proper formula without converting my negation from above using an algorithm to prenex normal form like I did two messages prior? Is my answer from above even correct?

**Plato** · November 12th 2012, 12:06 PM

Originally Posted by MachinePL1993

But now I have to bring it to prenex normal form. Is there some other way to arrive to a proper formula without converting my negation from above using an algorithm to prenex normal form like I did two messages prior? Is my answer from above even correct?

Yes, I would accept that. But I don't know the level of rigor expected of you.

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