Is the Negation of 'Some Box Contains 11 or More Balls' Correctly Translated?

[adambk]

"Some box contains 11 or more balls."

My answer:

Not (some box contains 11 or more balls)

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

\equiv no box contains 11 and more balls

Is this correct?

 

[Office_Shredder]

If a box contains 11 AND more balls... I think we can all see here how what you just posted is true by necessity, not the negation of a statement. 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.

 

[CRGreathouse]

"Some box contains 11 or more balls."

My answer:

Not (some box contains 11 or more balls)

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

\equiv no box contains 11 and more balls

Is this correct?

\nexists B|c(B)\ge11

\nexists B|c(B)=11\wedge\nexists B|c(B)>11

\nexists B|\left(c(B)=11\vee c(B)>11\right)

 

[NickJ]

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.

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."

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."

 

[Office_Shredder]

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."

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 said

"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

 

[honestrosewater]

Right, the answer isn't necessarily a single formula. The answer is a set of (equivalent) formulas for each interpretation. And we get to choose our interpretation(s) since we weren't given any. (adambk?) Assuming a reasonable first-order language interpretation and the usual relations on N, you can say more simply All boxes contain less than 11 balls. Letting < be an abbreviation for the negation of > is simpler than splitting up > into its strict order and equality. Your language or structure might not even have equality.

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

The logical symbols are as usual.
x is a variable.
B is a 1-ary relation symbol. (Bx = x is a box.)
C is a 1-ary function symbol. (Think of it as a function from the set of boxes to N. Cx = the number of balls that x contains.)
> is a 2-ary relation symbol. (Think of it as the usual greater-than-or-equal-to relation.)
< is an abbrevation for the negation of >, i.e., t < u = ~(t > u) for any terms t and u. (Think of it as the usual less-than relation.)
11 is a constant symbol. (Think of it as the number 11.)

1) \neg (\exists x \ [Bx \ \wedge \ (Cx \geq 11)]) (hypothesis)
2) \forall x \ [\neg(Bx \ \wedge \ (Cx \geq 11))] (1: quantifier negation equivalency)
3) \forall x \ [\neg Bx \ \vee \ \neg (Cx \geq 11)] (2: De Morgan's)
4) \forall x \ [\neg Bx \ \vee \ (Cx &lt; 11)] (3: abbreviation or Trichotomy)
5) \forall x \ [Bx \ \rightarrow \ (Cx &lt; 11)] (4: Implication)

Note that if you don't restrict your variable's range (as B does above), the standard interpretation would have it range over your entire domain, making your formula actually say that nothing contains more than 11 balls. In order to be equivalent to the original negation, every individual in your domain would have to be a box, which restricts what you can do with C, makes 11 a box, and is also a problem if you want your domain to have balls.

I think the above is probably the intended interpretation. But there's no reason that, for example, C needs to be a function symbol or, under further interpretation, a function. The assumptions that you make in the interpretation of a string of symbols are just assumptions, and adambk's original answer is only wrong with respect to the standard or intended interpretation. Maybe a box can contain both 11 and, say, 17 balls. Maybe balls can contain other balls. Maybe C's codomain isn't N. Maybe you can cast C as a relation on your domain. Maybe Some box contains 11 or more balls means that some box contains the number 11 or more, i.e., other, balls. Maybe balls are boxes. Maybe 11 is a box. Maybe more is a type of ball. And so on.


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

 

[HallsofIvy]

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