First Order Logic : Tableau

Nov 2012
2
0
Germany
This formulae gives me some troubles.

formulae.png

Im getting this far before im in trouble:

formulae2.png

How do I push the negation inwards ?

Appreciate all hints or solutions!

 

Plato

MHF Helper
Aug 2006
22,507
8,664
This formulae gives me some troubles.
View attachment 25545

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 2012
2
0
Germany
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?
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.

\(\displaystyle \forall y p(A,y) , \neg ( \forall y \exists x p(x,y) ) \)
 
Last edited: