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I have a question in Logic:

If I am asked to construct a formula, using the '=' predicate, that shows that there are exactly n objects, I need to show that there are no n+1 objects, right?

For example, to show that there are exactly 3 objects, I will show that there are no 4 objects.

Also, I need to show that there exist n objects.

My question is, do I need to show that there is no case where only n-1 objects exist?

For example, to show that there are exactly 3 objects, do I need to show that there is no case in which there are only 2 objects?

If so, is that the correct form to do so? :


1) Showing that there is no case of 4 elements:

$ {\lnot}{\exists}x{\exists}y{\exists}z{\exists}w(x{\neq}y{\land}x{\neq}z{\land}x{\neq}w{\land}y{\neq}z{\land}y{\neq}w{\land}z{\neq}w) $


2) Showing that there are 3 elements:

$ {\exists}x{\exists}y{\exists}z(x{\neq}y{\land}x{\neq}z{\land}y{\neq}z) $


3) Showing that there is no case of only 2 elements:

$ {\lnot}{\exists}x{\exists}y{\lnot}{\exists}z(x{\neq}y{\land}x{\neq}z{\land}y{\neq}z) $


And finally, combining the three:

$ (1){\land}(2){\land}(3) $


I am really not sure.

Thanks in advance

EDIT:

Actually, assuming that I do need to show that there is no case of less than 3 elements (n elements), I would probably need to show that there is no case where there is just one element as well (from 1 until n-1), am I correct?

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  • $\begingroup$ Where's the predicate? $\endgroup$ – Luke Mathieson Jan 17 '16 at 5:45
  • $\begingroup$ Work through some examples. Can you find an example where (2) is true but (3) is false? $\endgroup$ – D.W. Jan 17 '16 at 6:53
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    $\begingroup$ What is the question now? Should we just check your answer and confirm? That does not make for a good SE question. $\endgroup$ – Raphael Jan 17 '16 at 10:37
  • $\begingroup$ Well, that is what I need to know. I searched online and in my course materials. Our course materials are fairly poor.. and online I found this question math.stackexchange.com/questions/139378/… From which I understood (once again, please correct me if I'm wrong) that I can just write the (1) and (2) parts, meaning I do not need to do the part where I check if there are less than 3 objects $\endgroup$ – eevee25 Jan 17 '16 at 10:50
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    $\begingroup$ If you found the answer below solved your problem, you can mark it as the accepted answer by ticking the "checkmark" on the left of the answer. $\endgroup$ – D.W. Jan 18 '16 at 21:58
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You need to say that (a) there are at least $n$ elements, and (b) there are at most $n$ elements. To express (a), $$ L_n := \exists x_1\dotsc \exists x_n\, \left( \bigwedge_{1\le i < j \le n} x_i\ne x_j \right). $$ To express (b), $$ M_n := \forall x_1\dotsc \forall x_{n+1}\, \left( \bigvee_{1\le i < j \le n+1}x_i = x_j\right). $$ So the sentence $L_n \land M_n$ holds iff there are exactly $n$ elements.

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