# Thread: First Order Logic : Tableau

1. ## First Order Logic : Tableau

This formulae gives me some troubles.

Im getting this far before im in trouble:

How do I push the negation inwards ?

Appreciate all hints or solutions!

2. ## Re: First Order Logic : Tableau

Originally Posted by Razoor
This formulae gives me some troubles.
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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?

3. ## Re: First Order Logic : Tableau

Originally Posted by Plato
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?
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.

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