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let $ \mathbb{N}:=\{ \emptyset, (\emptyset)^+, ((\emptyset)^+)^+,...,(...((\emptyset)^+)^+...)^+,...\} $, $n \in \mathbb{N} $ with

$ I_n:=\{t \in \mathbb{N}|t \leq n\} $

$ I_n^*:=I_n\backslash \{0\} $

in this case $ 0:=\emptyset $, $ t \leq n $ if $ t \subseteq n $, and $ (A)^+ := A \cup \{A\} $

let $ m,p,o \in \mathbb{N} $, $ o $ is sum of $ m $ and $ p $ if $o \sim (I_m^*\oplus I_p^*) $

in this case $\oplus$ is http://en.wikipedia.org/wiki/Disjoint_union

Is correct? Thanks in advance!

mle
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    Let's see ... $\sim$ not defined. And $\oplus$ is used, not ${}^\ast \oplus$. Why not write out some explanation of what you are trying to do? – GEdgar Aug 08 '13 at 15:24
  • @GEdgar, okok... $\sim$ is http://en.wikipedia.org/wiki/Equipotence.. and $ I_p^* \oplus I_n^* := (I_p^* \times {0}) \cup (I_n^* \times {1})$... with $1:=(\emptyset)^+$ I want to define when $o$ is sum of $m$ and $n$.. – mle Aug 08 '13 at 15:31
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    What does the first line mean? Is $n$ a fixed number, or an arbitrary variable? If the latter, then use appropriate quantifiers. The current version is almost impossible to read. Use words and tell us what it is you are doing, instead of just writing symbols and hoping we can decipher them. – Andrés E. Caicedo Aug 08 '13 at 16:31
  • @AndresCaicedo, sorry..typo!! It is $I_m^* \oplus I_p^* := (I_m^* \times {0}) \cup (I_p^* \times {1})$.. thanks!! – mle Aug 08 '13 at 16:35
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    What is the point of using $I_n^$? Why not redefine $I_n={t\in\mathbb N\mid t<n}$, removing the need to use $I_n^$? (You would look at $I_m\oplus I_p$ in that case.) You really should rewrite the question, so it is clear and human readable. In particular, definitions should come before they are used (for example, line 4 should be before line 1). – Andrés E. Caicedo Aug 08 '13 at 16:42
  • @AndresCaicedo, if I use $I_n:={t \in \mathbb{N}|t <n}$ then, in case $n=0$, $I_0:={}$? – mle Aug 08 '13 at 16:56
  • @AndresCaicedo... of course, is true.. thank you soo much!! – mle Aug 08 '13 at 17:23
  • @AndresCaicedo, but $I_n:={ t \in \mathbb{N}|t < n}$ then $I_n=n$, is correct? Thanks!! – mle Aug 09 '13 at 07:33
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    (Yes. That's correct. So even better, we do not need to use $I_n$ at all, and can directly look at $m\oplus p$.) – Andrés E. Caicedo Aug 09 '13 at 14:54
  • @AndresCaicedo, okok thanks soo much!! ;) – mle Aug 10 '13 at 10:57

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