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Addition(+) is defined using the successor function(++) in Peano arithmetic as:

  1. 0 + m = m
  2. (n++) + m = (n + m)++

While these are intuitive axioms that are consistent with my previous, elementary, understanding of addition, I don't understand how it follows from these axioms that n + m is the same thing as incrementing m n times.

Although I can see that it is true for specific cases:

  1. 1 + m = (0++) + m = (0 + m)++ = m++
  2. 2 + m = (1++) + m = (1 + m)++ = (m++)++
  3. etc.

Thanks very much.

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    $\begingroup$ You continue the pattern and prove with induction. $\endgroup$ – Gareth Ma Apr 3 at 15:53
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    $\begingroup$ I think I would have defined it as $m+0=m$ and $m+(n++)=(m+n)++$. $\endgroup$ – Angina Seng Apr 3 at 16:02
  • $\begingroup$ @AnginaSeng Isn't that equivalent to "0 + m = 0" and "(n++) + m = (n + m)++"? i.e. You could just do induction to get the other. $\endgroup$ – Haziq Muhammad Apr 3 at 16:44
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We can prove it by induction on $m$. If $m=0$, $n+m=n+0=n$ is the result of not incrementing $n$ at all, i.e. doing it $0$ times. Suppose $n+k$ is the result of $k$ increments starting from $n$. For the inductive step, associativity of $+$ gives $n+(k+1)=(n+k)+1$, i.e. we increment $k$ times, then once more.

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  • $\begingroup$ Thanks, that makes a lot of sense. But I was a bit unsure about doing induction like that because I didn't know how I would define "k increments." $\endgroup$ – Haziq Muhammad Apr 3 at 16:30
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    $\begingroup$ @HaziqMuhammad You would define it recursively: $0$ increments leaves $n$ unchanged, $1$ changes it to $n$'s successor $n+1$, & $k+1$ increments applies $k$ then $1$. $\endgroup$ – J.G. Apr 3 at 16:32
  • $\begingroup$ Thanks again. It's nice to know that I won't have to go through Peano's axioms every time I want to add two numbers. $\endgroup$ – Haziq Muhammad Apr 3 at 16:46
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    $\begingroup$ @HaziqMuhammad Ditto with multiplication, for which Presburger arithmetic is at a distinct disadvantage. $\endgroup$ – J.G. Apr 3 at 16:58

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