Predicate logic with multiple quantifiers

[brynjolf23]

Hello everyone. This is my first post on this forum. Thank you for taking the time to help me with my question.
I have no idea where to start. :(

Question 1:
Find an example of a predicate P(x,y) where the domain of x and y are D such that
$\forall x \in D, \exists y \in D, P(x,y)$ is true but $\exists y \in D, \forall x \in D, P(x,y)$ is false.

Let D={1,2,3}

Question 2:
Is it possible to find a predicate P(x,y) such that:
$\exists y \in D, \forall x \in D, P(x,y)$ is true but $\forall x \in D, \exists y \in D, P(x,y)$ is false

Let D={1,2,3}

Question 3:
Is there any method of finding a suitable predicate? or do i just have to guess my way through?

 

[Greg]

Hi brynjolf23 and welcome to MHB!

I've placed the appropriate delimiters around your LaTeX code. They are (for this site)

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Thanks and enjoy the forum!

 

[castor28]

Hello everyone. This is my first post on this forum. Thank you for taking the time to help me with my question.
I have no idea where to start. :(

Question 1:
Find an example of a predicate P(x,y) where the domain of x and y are D such that
$\forall x \in D, \exists y \in D, P(x,y)$ is true but $\exists y \in D, \forall x \in D, P(x,y)$ is false.

Let D={1,2,3}

Question 2:
Is it possible to find a predicate P(x,y) such that:
$\exists y \in D, \forall x \in D, P(x,y)$ is true but $\forall x \in D, \exists y \in D, P(x,y)$ is false

Let D={1,2,3}

Question 3:
Is there any method of finding a suitable predicate? or do i just have to guess my way through?

Hi,

To be able to help you, we need to know the context of your questions. That is why we ask you to show any work you have done (even if it is wrong); at least, you should tell us what you are studying. This is specially important in formal logic, since there are several methods of deduction.

In this case, to get you started, I can offer an informal description. Both statements are about finding pairs $(x,y)$ that make $P(x,y)$ true. In the first statement, $\forall x \exists y\, P(x,y)$ you can find, for each $x$, at least one $y$ that makes $P(x,y)$ true. Note that each $y$ may depend on $x$.

On the other hand, in the statement $\exists y\forall x\, P(x,y)$, there must be a single $y$ that works for all $x$.