This formulae gives me some troubles.

Attachment 25545

Im getting this far before im in trouble:

Attachment 25546

How do I push the negation inwards ?

Appreciate all hints or solutions!

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- Nov 4th 2012, 02:41 PMRazoorFirst Order Logic : Tableau
This formulae gives me some troubles.

Attachment 25545

Im getting this far before im in trouble:

Attachment 25546

How do I push the negation inwards ?

Appreciate all hints or solutions!

- Nov 4th 2012, 04:43 PMPlatoRe: First Order Logic : Tableau
I find your answer puzzling.

Lets say we are dealing with $\displaystyle \mathbb{N}$ and $\displaystyle p(m,n)$ means $\displaystyle m\le n$.

Then $\displaystyle (\exists x)(\forall y)[p(x,y)]$ says "some natural number precedes every natural number". Does that imply that "every natural is preceded by some natural number"?

Can you use IE, UI, UG & EU in a valid way to get what you need? - Nov 5th 2012, 01:52 AMRazoorRe: First Order Logic : Tableau
Im not quite sure what you mean with your last question. But this is how i understand the sentence.

$\displaystyle \exists x \forall y p(x,y) $

"There exist an X where all Y is in the same function p(X,Y) which implies that for all Y is there an X in the same function p(X,Y)"

First of all i need to remove the imply arrow by negating the right side and split up the left and right part like this:

$\displaystyle \exists x \forall y p(x,y) , \neg ( \forall y \exists x p(x,y) ) $

You can then remove $\displaystyle (\exists x)$ by introducing a new constant for X.

$\displaystyle \forall y p(A,y) , \neg ( \forall y \exists x p(x,y) ) $