Challenging the Assertion: Everyone Has a Disliked Relative

[bonfire09]

Negate the statement is "Everyone has a relative he doesn't like"

Let x-person
Let y- relative
R(x,y)- x is related to y
L(x,y)-x likes y

∀x ∃y(R(x,y)→~L(x,y))
∃x~∃y(R(x,y)→~L(x,y))
∃x ∀y~(R(x,y)→~L(x,y))
∃x ∀y(R(x,y)^L(x,y))
is this correct? It reads for someone x who is related to all y and x likes all y.

 

[bonfire09]

do they both sound opposite from each other?

 

[MLP]

Without actually doing the translation, here is an English rendering:

For every x, there is a y, such that y is related to x and x does not like y.

 

[SW VandeCarr]

I don't know if it's proper, but negating the "forall" operator \sim\forall (x) would convey your intention. However this is equivalent to \exists (x) or \sim \exists (x). To remove the ambiguity you might be able to use the negation of the "forall" operator with the affirmative existential operator if that's what you want to say in the negation. It's clear that negating the forall operator is redundant if you negate the existential operator.

 

[Bacle2]

In plane English, I would believe that the negation is that someone likes all their

relatives. Maybe rewriting it : "Everyone dislikes at least one of their relatives",

would be negated as "Someone likes all their relatives", or, saying that a counter

to "Everyone dislikes at least one of their relatives" is : not true, there is someone

who actually likes all their relatives .

 

[SW VandeCarr]

Taking the initial English statement "Everyone has a relative they don't like". The negation would be "Some people (at least one), but not everyone, has a relative they don't like."

 

[Bacle2]

Taking the initial English statement "Everyone has a relative they don't like". The negation would be "Some people (at least one), but not everyone, has a relative they don't like."

Wouldn't it be: " Not everyone has a relative they don't like" , aka:

" Someone does not have a relative they don't like " ?

Problem is English, as every-day languages are imprecise; there is a tradeoff of

flexibility for precision, so maybe other interpretations are possible.

 

[SW VandeCarr]

Wouldn't it be: " Not everyone has a relative they don't like" , aka:

" Someone does not have a relative they don't like " ?

Problem is English, as every-day languages are imprecise; there is a tradeoff of

flexibility for precision, so maybe other interpretations are possible.

The negation of the forall operator is ambiguous by itself because it can mean some (at least one) or none. If you use it with the affirmative existential operator, I think the ambiguity is removed. The point I think the OP wants to convey with the negation is that not everyone has a relative they don't like. In logic, you have to specify whether "not everyone" means "some" or "none" although I suppose you could construct a statement with the "or" conjunction.

 

[bonfire09]

. The original statement says everyone has a relative he does not like I think it can be negated to form Not (everyone has a relative he does not like) which means Someone has relatives he likes.

 

[SW VandeCarr]

. The original statement says everyone has a relative he does not like I think it can be negated to form Not (everyone has a relative he does not like) which means Someone has relatives he likes.

Yes. "Not everyone has a relative he does not like" is the negation. That's what I said. "Not everyone" meaning at least one can be logically expressed as a negation by negating "forall" and affirming \exists (x). If you just assert "someone", that's not a logical negation. Logic is concerned with the formal structure of statements, not with their linguistic meaning.

 

[bonfire09]

I'm not sure why but I have practiced this quite a bit but still I'm having trouble forming predicate formula's out of informal sentences.

 

[Bacle2]

Well, a thought: my prof.used to say that when one goes from , say, English ( or any

language used in daily exchanges ), to a formal language, one is not translating, but

transcribing, because the two --"street language" and formal languages-- are

intrinsically different: daily language is very fluid but ambiguous, and the opposite

is the case for formal languages. This makes the back-and-forth much harder than

going between formal languages or between street languages: a good chunk of the

intended meaning will disappear or be distorted.

 

[MLP]

I gave an English rendering of the non-negated sentence earlier. This translates to:

(\forallx)(\existsy)(Rxy \wedge \negLxy)

Negating this we get

\neg(\forallx)(\existsy)(Rxy \wedge \negLxy)

Driving the negation successively inward we get:

(\existsx)\neg(\existsy)(Rxy \wedge \negLxy)

(\existsx)(\forally)\neg(Rxy \wedge \negLxy)

(\existsx)(\forally)(\negRxy \vee\neg\negLxy)

Getting rid of the double-negation:

(\existsx)(\forally)(\negRxy \vee Lxy)

Appealing to the definition of the material conditional:

(\existsx)(\forally)(Rxy \rightarrow Lxy)

This comes pretty close to what was said earlier: "Someone does not have a relative they don't like " or, put positively, "Someone likes all of their relatives".

 

[SteveL27]

Negate the statement is "Everyone has a relative he doesn't like"

Let x-person
Let y- relative
R(x,y)- x is related to y
L(x,y)-x likes y

∀x ∃y(R(x,y)→~L(x,y))

I haven't read the other responses so apologies if someone already mentioned this.

Your last line there says that:

For all x, there exists y such that: (if x and y are related, then x does not like y).

That is a little different than the original statement, which says that everyone has a relative they don't like.

Example: Your formulation is true for someone who has no relatives. But the original formulation is false.

To see this, let x be someone with no relatives. And let y be anyone else. Then it's true that IF x and y are related, then x dislikes y. It's true because x and y aren't related!

However, it's NOT true that x has a relative he doesn't like. So you have not accurately captured the original proposition.

 

[bonfire09]

I noticed that afterwards haha that the statement could still be true even if x and y were not related. Thanks for the help everyone