MHB What is the negation of this statement and which statement is true, R or ¬R?

[rayne1]

Let P(x) be assertion “x is prime”, E(x) be “x is even”, and D(x, y) be “x divides y” (i.e., y/x is an integer). Consider the following statement:
R = (∀x ∈ Z)(P(x) ⇒ ((∃y ∈ Z)(E(y) ∧ D(x, y))))

Write the negation of R, and determine which statement is true, R or ¬R.

I tried, but I'm not sure if I got the correct answer:
¬R = (∃x ∈ Z)(P(x) ∧ (∀y ∈ Z)((¬E(y)) ∨ (¬D(x, y)))

It seems that ¬R is true.

 

[I like Serena]

Let P(x) be assertion “x is prime”, E(x) be “x is even”, and D(x, y) be “x divides y” (i.e., y/x is an integer). Consider the following statement:
R = (∀x ∈ Z)(P(x) ⇒ ((∃y ∈ Z)(E(y) ∧ D(x, y))))

Write the negation of R, and determine which statement is true, R or ¬R.

I tried, but I'm not sure if I got the correct answer:
¬R = (∃x ∈ Z)(P(x) ∧ (∀y ∈ Z)((¬E(y)) ∨ (¬D(x, y)))

It seems that ¬R is true.

Hi rayne!

Let's define:
Q(x) = (∃y ∈ Z)(E(y) ∧ D(x, y))

Then we have:
R = (∀x ∈ Z)(P(x) ⇒Q(x))

So:
¬R = (∃x ∈ Z)(¬(P(x) ⇒Q(x)))

Did you know that:
(P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x))
(Wondering)

So what does that make ¬(P(x) ⇒Q(x))?

 

[rayne1]

Hi rayne!

Let's define:
Q(x) = (∃y ∈ Z)(E(y) ∧ D(x, y))

Then we have:
R = (∀x ∈ Z)(P(x) ⇒Q(x))

So:
¬R = (∃x ∈ Z)(¬(P(x) ⇒Q(x)))

Did you know that:
(P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x))
(Wondering)

So what does that make ¬(P(x) ⇒Q(x))?

P(x)∧(¬Q(x))

 

[I like Serena]

P(x)∧(¬Q(x))

How did you get that?

 

[rayne1]

How did you get that?

I know that they're logically equivalent

 

[I like Serena]

I know that they're logically equivalent

So you're saying that:
¬(P(x) ⇒Q(x)) = (P(x)∧(¬Q(x)))

Let's take another look.
Substitute (P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x)) to get:
¬(P(x) ⇒Q(x)) = ¬(P(x) ∨ ¬Q(x))

... but wait! Isn't that the negation of what you are saying?

 

[rayne1]

So you're saying that:
¬(P(x) ⇒Q(x)) = (P(x)∧(¬Q(x)))

Let's take another look.
Substitute (P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x)) to get:
¬(P(x) ⇒Q(x)) = ¬(P(x) ∨ ¬Q(x))

... but wait! Isn't that the negation of what you are saying?

I guess so.

 

[rayne1]

I have to go right now, so can't reply in the next 24 hours.

 

[Evgeny.Makarov]

Let P(x) be assertion “x is prime”, E(x) be “x is even”, and D(x, y) be “x divides y” (i.e., y/x is an integer). Consider the following statement:
R = (∀x ∈ Z)(P(x) ⇒ ((∃y ∈ Z)(E(y) ∧ D(x, y))))

Write the negation of R, and determine which statement is true, R or ¬R.

I tried, but I'm not sure if I got the correct answer:
¬R = (∃x ∈ Z)(P(x) ∧ (∀y ∈ Z)((¬E(y)) ∨ (¬D(x, y)))

It seems that ¬R is true.

Rayne, you are right on both counts.

Did you know that:
(P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x))

No, (P(x) ⇒ Q(x)) ⇔ (¬P(x) ∨ Q(x)).

 

[I like Serena]

Rayne, you are right on both counts.

Let's read what R says:
For every x in Z, if x is prime, then there is an y such that y is even and x divides y.
I believe this is true. Pick y=2x.

No, (P(x) ⇒ Q(x)) ⇔ (¬P(x) ∨ Q(x)).

Ah. My bad.

 

[Evgeny.Makarov]

Let's read what R says:
For every x in Z, if x is prime, then there is an y such that y is even and x divides y.
I believe this is true. Pick y=2x.

Yes, of course. I must have read $D(x,y)$ the other way. The negation is still constructed correctly, I believe.

Well, this thread has some unfortunate responses, doesn't it?