I've proven that every positive natural number has a unique predecessor using Peano's axioms. But now, I was wondering how I could prove that every natural number has a unique successor using the same axiomatic system. Of course, by Peano's axioms, if $n\in\mathbb{N}$ then $\sigma(n)\in\mathbb{N}$. But how do I show that this natural number is unique?
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2$\begingroup$ $\sigma$ is supposed to be a function. $\endgroup$– amegCommented May 26 at 14:43
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$\begingroup$ @ameg so is the unique successor fact implicitly assumed since $\sigma$ is a function? $\endgroup$– AryaanCommented May 26 at 14:45
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$\begingroup$ Right. It's an unstated axiom. $\endgroup$– MJDCommented May 26 at 15:16
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$\begingroup$ Another unstated axiom is the existence if $0$. It's quite consistent with the other axioms for there to exist no natural numbers at all. $\endgroup$– MJDCommented May 26 at 15:38
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$\begingroup$ MJD, in what sense are these axioms "unstated"? Normally, the existance of 0 (or 1) is the first axiom. Then there is an set of axioms about a successor function that should exist and have such-and-such properties. $\endgroup$– Nadav Har'ElCommented May 26 at 16:08
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2 Answers
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The successor function, $\sigma(n)$, defines the successor of a number $n$. You don't need to prove that it's unique - it's a function - i.e., for any $n$ this function gives just one value (usually called $n+1$) which will be defined as the successor. The only successor.
After understanding that, you might wander now whether is is possible that two different numbers $m$ and $n$ have the same successor $k$. This is not possible because of the predecessor uniqueness that you said you proved: both $m$ and $n$ are predecessors of $k$ so if the predecessor is unique, they must be equal.
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As others have pointed out, this trivially follows from the fact that the successor function is just that: a function.
But if it helps, here is a formal proof:
