First Order Logic : Tableau

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November 4th 2012, 02:41 PM
Razoor

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First 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!
November 4th 2012, 04:43 PM
Plato

Re: First Order Logic : Tableau

Quote:

Originally Posted by Razoor

This formulae gives me some troubles.
Attachment 25545[COLOR=#000000]

I find your answer puzzling.
Lets say we are dealing with $\mathbb{N}$ and $p(m,n)$ means $m\le n$ .
Then $(\exists x)(\forall y)[p(x,y)]$ 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?
November 5th 2012, 01:52 AM
Razoor

Re: First Order Logic : Tableau

Quote:

Originally Posted by Plato

I find your answer puzzling.
Lets say we are dealing with $\mathbb{N}$ and $p(m,n)$ means $m\le n$ .
Then $(\exists x)(\forall y)[p(x,y)]$ 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.

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

"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:

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

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

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