# Basic Logic Problem P(x,y)

## Homework Statement

I refer to part G of this little problem:

I don't see how to arrive at any conclusion, especially when I can't even see how ##z## comes into play. Assistance in interpreting the problem is appreciated!

## The Attempt at a Solution

I know that the answer for G is "False", which means I have to show that the following is true;
$$\exists x \in A, \ \forall y \in A, \exists z \in A, \ \neg P(x,y) \lor \neg P(y,z)$$

Here's how I have tried to interpret the problem.
$$\neg P(x,y) \lor \neg P(y,z)$$
is true when either or both predicates are false. Looking at a particular ##x## and all combinations of ##y## that come with it (##\exists x \in A, \ \forall y \in A##), none of them are able to make ##P(x,y)## false all the time.

So now I try to look at all ##y## (that are already paired with some ##x##) and try to find some ##z## for each of the ##y## that would cause ##p(z,y)## to always be false. However, I'm not sure where and how to apply the ##z## since it's not defined in the table.

Was this how I am supposed to interpret the problem? Apologies if what I have written is unintelligible as I am finding logic rather confusing.

#### Attachments

• 24.2 KB Views: 521

Related Calculus and Beyond Homework Help News on Phys.org
Delta2
Homework Helper
Gold Member
the table of truth of P(y,z) is the same as table of truth for P(x,y) with two differences: where x you put y and where y you put z.
that is it is
y 0 1 2
z \
0 T F T
1 T T F
2 T F F​

I see the table I tried to make using standard text editing doesn't look that good but I think you ll understand it. for example it is P(y,z)=T for y=0 and z=0, P(y,z)=F for y=1 and z=0, P(y,z)=T for y=2 and z=0...

WWCY
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

I refer to part G of this little problem:

View attachment 230498

I don't see how to arrive at any conclusion, especially when I can't even see how ##z## comes into play. Assistance in interpreting the problem is appreciated!

## The Attempt at a Solution

I know that the answer for G is "False", which means I have to show that the following is true;
##\ \exists x \in A, \ \forall y \in A, \exists z \in A, \ \neg P(x,y) \lor \neg P(y,z)##

Here's how I have tried to interpret the problem.
##\ \neg P(x,y) \lor \neg P(y,z)##
...
Don't be fixated on what variable name is used for a particular index (parameter) .

For example: If we set x = 0 and y = 1, then P(x,y) is True. However P(y,x) is False for the same choices of x and y, because this is P(1,0) .

WWCY and Delta2