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My answer:

Not (some box contains 11 or more balls)

[tex]\equiv[/tex] (not some box contains 11) and (more balls) {by de Morgan}

[tex]\equiv[/tex] no box contains 11 and more balls

Is this correct?

- Thread starter adambk
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- #1

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My answer:

Not (some box contains 11 or more balls)

[tex]\equiv[/tex] (not some box contains 11) and (more balls) {by de Morgan}

[tex]\equiv[/tex] no box contains 11 and more balls

Is this correct?

- #2

Office_Shredder

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so you get No box contains 11 or more balls.

- #3

CRGreathouse

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[tex]\nexists B|c(B)\ge11[/tex]adambk said:

My answer:

Not (some box contains 11 or more balls)

[tex]\equiv[/tex] (not some box contains 11) and (more balls) {by de Morgan}

[tex]\equiv[/tex] no box contains 11 and more balls

Is this correct?

[tex]\nexists B|c(B)=11\wedge\nexists B|c(B)>11[/tex]

[tex]\nexists B|\left(c(B)=11\vee c(B)>11\right)[/tex]

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I disagree. "Some box contains 11 or more balls" should be translated as "There exists a box x such that either x contains 11 balls or x contains more than 11 balls."Office_Shredder said:I'm pretty sure it's just supposed to be 11 or more balls as a single point, not as a logical OR statement.

so you get No box contains 11 or more balls.

Then, when this is negated, we get: It is not the case that there is a box such that either the box contains 11 balls or the box contains more than 11 balls.

Let's put this in symbols, with E = ...contains 11 balls; M = ...contains more than 11 balls.

Then: It is not the case that there exists an x such that (Ex or Mx).

Pushing the negation through, we get: For all x, not-(Ex or Mx).

Then: For all x, not-Ex and not-Mx.

So: No box contains 11 balls and no box contains more than 11 balls.

That's the negation of "Some box contains 11 or more balls."

- #5

Office_Shredder

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That's logically equivalent to what I said actually.... that no single box contains (11 or more) balls. This is vs. what the OP had saidNickJ said:So: No box contains 11 balls and no box contains more than 11 balls.

That's the negation of "Some box contains 11 or more balls."

"no box contains 11 and more balls"

In which is appears to be reading No box contains 11 AND more balls, i.e. no single box contains 11 balls and more balls at the same time. Maybe I'm reading it wrong, but that's what it looks like to me

- #6

honestrosewater

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Anywho, here's that derivation, with a little closer attention to detail.

The logical symbols are as usual.

1) [tex]\neg (\exists x \ [Bx \ \wedge \ (Cx \geq 11)])[/tex] (hypothesis)

2) [tex]\forall x \ [\neg(Bx \ \wedge \ (Cx \geq 11))][/tex] (1: quantifier negation equivalency)

3) [tex]\forall x \ [\neg Bx \ \vee \ \neg (Cx \geq 11)][/tex] (2: De Morgan's)

4) [tex]\forall x \ [\neg Bx \ \vee \ (Cx < 11)][/tex] (3: abbreviation or Trichotomy)

5) [tex]\forall x \ [Bx \ \rightarrow \ (Cx < 11)][/tex] (4: Implication)

Note that if you don't restrict your variable's range (as

I think the above is probably the intended interpretation. But there's no reason that, for example,

By the bye, if you know Aristotelian (a.k.a. categorical and syllogistic) logic, you could just recall that

- #7

HallsofIvy

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More simply, "every box contains less than 11 balls".

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