Want to negate the following statements

**issacnewton** · July 10th 2011, 09:53 AM

Hi

I have to negate the following statements and then express again
in English. I need to know if I am making any mistakes.

a)Everyone who is majoring in math has a friend who needs
help with his homework.

b)Everyone has a roommate who dislikes everyone.

c)There is someone in the freshman class who doesn't
have a roommate.

d)Everyone likes someone,but no one likes everyone.

My answers are as follows--------

a) Let P(x)= x is majoring in math
Q(x)=x needs help with homework
M(x,y)=x and y are friends

The statement would be

$\exists x (P(x))\rightarrow \exists y (M(x,y)\wedge Q(y))$

So the negated statement would be

$\neg \left[\exists x (P(x))\rightarrow \exists y (M(x,y)\wedge Q(y))\right]$

$\neg \left[\neg(\exists x (P(x)) \vee \exists y (M(x,y)\wedge Q(y))\right]$

$\neg \left[\forall x \neg P(x) \vee \exists y (M(x,y)\wedge Q(y))\right]$

$\neg (\forall x \neg P(x)) \wedge \neg \exists y (M(x,y)\wedge Q(y))$

$\exists x P(x) \wedge \forall y \neg (M(x,y)\wedge Q(y))$

$\exists x P(x) \wedge \forall y (\neg M(x,y) \vee \neg Q(y))$

To translate this statement back to English would be

Some person is majoring in math and everyone is either not a
friend of this person or doesn't need help in homework.

b) let Q(x,y)=x and y are roommates
M(x,y)=x likes y

The statement would be

$\forall x \left[\exists y (Q(x,y) \wedge \forall z(\neg M(y,z)))\right]$

so the nagated statement would be

$\neg \forall x \left[\exists y (Q(x,y)\wedge \forall z(\neg M(y,z)))\right]$

$\exists x \forall y \left[\neg Q(x,y) \vee \exists z(M(y,z)) \right]$

Translation:

Either there is some person who is not roommate with anybody or there is
someone who is liked by all.

c)let P(x)= x is in freshman class.
M(x)=x has a roommate.

The statement would be

$\exists x \left[ P(x)\wedge \neg M(x) \right]$

So the negated statement is

$\neg \exists x \left[ P(x)\wedge \neg M(x) \right]$

$\forall x \left[ \neg P(x) \vee M(x) \right]$

Translation:

Everyone either is not in freshman class or has a roommate.

d)let M(x,y)= x likes y

The statement would be

$\forall x \left[ \exists y (M(x,y))\wedge \exists z \neg M(x,z) \right]$

The negated statement would be

$\neg \forall x \left[ \exists y (M(x,y))\wedge \exists z \neg M(x,z) \right]$

$\neg \left[\forall x \exists y (M(x,y))\wedge \forall x \exists z \neg M(x,z) \right]$

$(\neg \forall x \exists y M(x,y)) \vee (\neg \forall x \exists z \neg M(x,z) )$

$\left[\exists x \forall y \neg M(x,y)\right] \vee \left[\exists x \forall z M(x,z)\right]$

Translation:

Either there is someone who likes everyone or there is someone who doesn't like
everyone.

Please comment

Thanks

**Plato** · July 10th 2011, 10:48 AM

Originally Posted by issacnewton

I have to negate the following statements and then express again
in English. I need to know if I am making any mistakes.

a)Everyone who is majoring in math has a friend who needs
help with his homework.

b)Everyone has a roommate who dislikes everyone.

c)There is someone in the freshman class who doesn't
have a roommate.

d)Everyone likes someone,but no one likes everyone.

Please comment

You are making too much out of this for my tastes, but maybe that is required.

Examples:
a) Someone is majoring in mathematics and none of his friends needs help with homework.

c) Everyone in the freshman class has a roommate.

**issacnewton** · July 10th 2011, 10:56 AM

Thanks Plato for reply.

Where am I making mistakes then ? Thats what I want to know. I am using standard procedure as given in Daniel Velleman's book

**Plato** · July 10th 2011, 11:06 AM

Originally Posted by issacnewton

Where am I making mistakes then ? Thats what I want to know. I am using standard procedure as given in Daniel Velleman's book

As I did say, you may be required to do all of that symbolic workings. Frankly, you loose me in your process. I don't understand why for instance you introduce a non-friend in part (a)?
You have to admit that my negation is a natural way of saying that.

**issacnewton** · July 10th 2011, 11:16 AM

So Plato, in part a), are my symbolic equations correct ? Or is it just the problem with the translation back in English ? In part a) , which equation do you
think I took wrong step ?

**Plato** · July 10th 2011, 11:36 AM

Originally Posted by issacnewton

So Plato, in part a), are my symbolic equations correct ?

I your OP you said:

Originally Posted by issacnewton

I have to negate the following statements and then express again in English. I need to know if I am making any mistakes.

There is nothing in the directions about putting these into symbolic form. So I ask you if you are required to do so? If not why do it?
It seems to me the directions “to negate the following statements and then express again in English” are very straightforward. So once again, I do not follow your symbolic translations. So about the symbols I cannot say. Sorry.

**issacnewton** · July 10th 2011, 11:49 AM

Oh I am sorry Plato. I probably didn't say it correctly. Yes , I am required to put this in the form of symbolic equations, complete the negation using
various rules about the quantifiers and then translate back into the English. The book "How to Prove it" is teaching in chapter 2 , the use of
quantifiers. Author did give some examples and I am doing the exercises....

**Plato** · July 10th 2011, 12:00 PM

Originally Posted by issacnewton

I am required to put this in the form of symbolic equations, complete the negation using various rules about the quantifiers and then translate back into the English.

Originally Posted by issacnewton

a) Let P(x)= x is majoring in math
Q(x)=x needs help with homework
M(x,y)=x and y are friends

Well the symbolic forms complicate things.
Using your notation the a) statement becomes:
$\left( {\forall x} \right)\left[ {P(x) \to \left( {\exists y} \right)\left[ {M(x,y) \wedge Q(y)} \right]} \right]$

The negation of which is $\left( {\exists x} \right)\left[ {P(x) \wedge \left( {\forall y} \right)\left[ {\neg M(x,y) \vee \neg Q(y)} \right]} \right]$ .

As you no doubt can see translating that back to ‘natural’ English is complicated.

**issacnewton** · July 10th 2011, 12:14 PM

Yes Plato, I see the point. I think author just wants the students to get the hang of it I guess. But the way you have written the affirmative part of part a) looks different from the way I have done. Or I am doing the same thing ?

Probably the mathematical statements , when negated , won't be so hard to interpret to us mortals......

**Plato** · July 10th 2011, 12:18 PM

Originally Posted by issacnewton

But the way you have written the affirmative part of part a) looks different from the way I have done. Or I am doing the same thing ?

Of course it is different. It is correct. I told you that I could not follow what you did much less why you did it.

**issacnewton** · July 10th 2011, 12:29 PM

Thanks. So I am correct but did differently. These quantifiers are really brain twisters I guess. But understanding it is of great importance to appreciate
the logical structure in mathematics. One of the books which I got is Copis's Introduction to Logic. Apart from symbolic logic, he goes in great detail about
discussing general logic , which is applied in various fields. So is mathematical logic a subset of philosophical/general logic ? Because, Copi discusses lot of
things ( usually having some Latin/Greek names) without any symbols.

**Plato** · July 10th 2011, 12:34 PM

Originally Posted by issacnewton

So I am correct but did differently.

No. But now I give up.

**issacnewton** · July 10th 2011, 12:45 PM

Originally Posted by Plato

Of course it is different. It is correct.

I thought that you meant I was correct. some misunderstanding ?

**emakarov** · July 11th 2011, 02:53 AM

First, I pointed out in another thread that it's easier to check your work if you choose mnemonic names for predicates.

Originally Posted by issacnewton

a)Everyone who is majoring in math has a friend who needs
help with his homework

The statement would be

$\exists x (P(x))\rightarrow \exists y (M(x,y)\wedge Q(y))$

The closing parenthesis after P(x) should be moved to the end of the formula.

$\exists x P(x) \wedge \forall y (\neg M(x,y) \vee \neg Q(y))$

To translate this statement back to English would be

Some person is majoring in math and everyone is either not a
friend of this person or doesn't need help in homework.

Correct. I prefer to turn the disjunction $\neg M(x,y) \vee \neg Q(y)$ into an implication $M(x,y)\to\neg Q(y)$ , so the English version is "Someone is majoring in mathematics and none of his friends needs help with homework," as Plato said. It is more natural to say "All people are mortal" (using an implication) than "Everyone is either not a person or is mortal" (using a disjunction).

b)Everyone has a roommate who dislikes everyone.

b) let Q(x,y)=x and y are roommates
M(x,y)=x likes y

The statement would be

$\forall x \left[\exists y (Q(x,y) \wedge \forall z(\neg M(y,z)))\right]$

so the nagated statement would be

$\neg \forall x \left[\exists y (Q(x,y)\wedge \forall z(\neg M(y,z)))\right]$

$\exists x \forall y \left[\neg Q(x,y) \vee \exists z(M(y,z)) \right]$

Translation:

Either there is some person who is not roommate with anybody or there is someone who is liked by all.

The negated formula is correct; the English translation is not. The formula corresponding to the English phrase is

$(\exists x \forall y \neg Q(x,y)) \vee (\exists x \forall y\,M(y,x))$

Again, I prefer to turn $\neg Q(x,y) \vee \exists z(M(y,z))$ into $Q(x,y) \to \exists z(M(y,z))$ . Then the English translation is "There is a person such that all of his/her roommates love somebody."

c)There is someone in the freshman class who doesn't
have a roommate...

c)let P(x)= x is in freshman class.
M(x)=x has a roommate.

The statement would be

$\exists x \left[ P(x)\wedge \neg M(x) \right]$

So the negated statement is

$\neg \exists x \left[ P(x)\wedge \neg M(x) \right]$

$\forall x \left[ \neg P(x) \vee M(x) \right]$

Translation:

Everyone either is not in freshman class or has a roommate.

Or: Everyone in the freshman class has a roommate.

d)Everyone likes someone,but no one likes everyone.

d)let M(x,y)= x likes y

The statement would be

$\forall x \left[ \exists y (M(x,y))\wedge \exists z \neg M(x,z) \right]$

I would say the direct formalization is

$(\forall x \exists y M(x,y))\wedge (\forall x\neg\forall z M(x,z))$

This is equivalent to your formula, but it is closer to English because the main connective is a conjunction, which corresponds to "but."

The negated statement would be

$\neg \forall x \left[ \exists y (M(x,y))\wedge \exists z \neg M(x,z) \right]$

$\neg \left[\forall x \exists y (M(x,y))\wedge \forall x \exists z \neg M(x,z) \right]$

$(\neg \forall x \exists y M(x,y)) \vee (\neg \forall x \exists z \neg M(x,z) )$

$\left[\exists x \forall y \neg M(x,y)\right] \vee \left[\exists x \forall z M(x,z)\right]$

Translation:

Either there is someone who likes everyone or there is someone who doesn't like
everyone.

The negated formula is correct. Concerning English, I think "there is someone who doesn't like everyone" means $\exists x \neg\forall y M(x,y)$ . To express $\exists x \forall y \neg M(x,y)$ , I would say, "there is someone who doesn't like anybody."

**issacnewton** · July 12th 2011, 01:46 AM

Priviet Makarov

Thanks for the detailed response. Which introductory book on logic did you use as an undergrad ?

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