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**Amarylli$****Guest**

Hi all,

"\forall" x : N p is false but "\exists" x : N p is true.

**Amarylli$****Guest**

Amarylli$ wrote:

Hi all,

\forall x : N p is false but \exists x : N p is true.

for this, i need to define a pair of predicates p and q involving free variable x

how can i go abt doin tht?

thanks!

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm not familiar with the notation N p

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Amarylli$****Guest**

well..what i understand is that i have to create expression somewhat like below:

p <-> x+ y > x^2

so x and y here are the free variables.. hence in the question above..i need to create such expression that it satisfies the logic

and N are natural numbers in this context.

- predicate

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm going to guess that N p means p(x). If this is the case, all you must do is find a statement that is true for some x, but not all.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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