# Thread: Symbolic Logic statement ... need to know if correct or close to

1. ## Symbolic Logic statement ... need to know if correct or close to

I am to Write the sentence in words that corresponds to the following symbolic statement. Find the negation both symbolically and in words:
∀x ∃y ∋ y > x

For all x: P(x) is true and there is at least one y such that P(y) is true For every element P(y) > P(x)

There exist at least one x such that P(x) is true and For all y: P(y) is true For every element P(y)≤ P(x)

2. ## Re: Symbolic Logic statement ... need to know if correct or close to

Originally Posted by Odail
I am to Write the sentence in words that corresponds to the following symbolic statement. Find the negation both symbolically and in words:
∀x ∃y ∋ y > x

For all x: P(x) is true and there is at least one y such that P(y) is true For every element P(y) > P(x)

There exist at least one x such that P(x) is true and For all y: P(y) is true For every element P(y)≤ P(x)
I'd just write:

For all x there exists y such that y is greater than x.

negate:

there exists a y such that x is greater than y

3. ## Re: Symbolic Logic statement ... need to know if correct or close to

Thanks ... but could you explain wht the x was left out of the negation ?

4. ## Re: Symbolic Logic statement ... need to know if correct or close to

well you wouldn't want to word is: there exists a y such that for all x, x is greater than y [this sounds like theres a y smaller than all x]

could say

$\exists y \in Y$ and $x \in X \,\,s.t.\,\, x > y$

5. ## Re: Symbolic Logic statement ... need to know if correct or close to

ok kool thanks much appreciated

6. ## Re: Symbolic Logic statement ... need to know if correct or close to

Originally Posted by Odail
I am to Write the sentence in words that corresponds to the following symbolic statement. Find the negation both symbolically and in words:
∀x ∃y ∋ y > x
My browser shows this formula as $\forall\exists y\ni y>x$. This is not a well-formed formula because $\ni$ is a binary relation requiring two arguments. That is, $B\ni a$ (more commonly written as $a\in B$ means that $a$ is an element of a set $B$. In fact, I don't believe $\ni$ should be a part of your formula at all. You probably meant $\forall x\exists y\;y>x$.

Originally Posted by Odail
For all x: P(x) is true and there is at least one y such that P(y) is true For every element P(y) > P(x)
This does not make much sense because $P$ was not defined. Every object you mention must be properly introduced.

Originally Posted by Jonroberts74
I'd just write:

For all x there exists y such that y is greater than x.
I agree.

Originally Posted by Jonroberts74
negate:

there exists a y such that x is greater than y
This statement does not properly introduce $x$. It is not clear whether this holds for all $x$ or for some $x$. In addition, this sentence switches $x$ and $y$ compared to the original statement, which introduces an additional difficulty. Usually one constructs the negation of $\forall x\exists y\;P(x,y)$ as follows: $\exists x\forall y\;\neg P(x,y)$, i.e., the quantifiers change, but the variable names don't. Of course, the latter formula is equivalent to $\exists y\forall x\;\neg P(y,x)$, but again, this renaming introduces extra complexity. It seems that you got confused by this extra complexity because you renamed $x$ and $y$ that follow the quantifiers, but did not rename them in the rest of the formula. Finally, the negation of a strict inequality $y>x$ is a non-strict inequality.

Originally Posted by Jonroberts74
well you wouldn't want to word is: there exists a y such that for all x, x is greater than y [this sounds like theres a y smaller than all x]
Yes, this is incorrect. The correct negation is

There exists an y such that for all x, x is less than or equal to y,

or

There exists an x such that for all y, y is less than or equal to x.

That is, there exists a greatest element.

Originally Posted by Jonroberts74
could say

$\exists y \in Y$ and $x \in X \,\,s.t.\,\, x > y$
The original formula does not have a capital $Y$. Again, $x$ is not properly introduced by a quantifier here. It looks like $\exists$ pertains also to $x$, but this is not correct.