Basic Logic Problem P(x,y)

[WWCY]

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!

Homework Equations

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.

 

[Delta2]

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...

 

[SammyS]

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!

Homework Equations

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) .