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In a step of a proof about Peano arithmetic the following calculation is made: $N(ss^k0)=N(s)N(s^k0)$ where $N$ is the interpretation which identifies $sx$ with $S(x)$ where $S$ is the successor function, and identifies $+xy$ with $x+y$ (the addition of natural numbers), and so on...

A general fact about interpretations, $I$,is that to every singular term $f^nt_1t_2\ldots t_n$ where $f^n$ is an $n$-place function symbol and $t_i$ are singular terms, $I$ assigns the $F(\alpha_1,\alpha_2,\ldots, \alpha_n)$ where $F = I(f^n)$ and $\alpha_k = I(t_k)$.

$$I(f^nt_1\ldots t_n) = I(f^n)(I(t_1), I(t_2),\ldots, I(t_n))=F(\alpha_1,\ldots, \alpha_n) $$

I'm struggling to see how this makes $N(ss^k0) = N(s)N(s^k0)$

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I think this is the idea..... but hoping for someone to confirm my intuitions:

$$N(s^{k+1}0) = N(s^1t_1) = N(s^1)N(t_1) = N(s)N(s^k0)$$

where $t_1$ is the singular term $s^k0$

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  • $\begingroup$ Yes, you are right. Note that $N(s) N(s^k0)$ has to be intended as $N(s) \big( N(s^k0) \big)$, i.e. the successor function $S$ applied to interpretation of $s^k0$ (which is, in turn, the $k$-iterated succession function $S$ applied to $0$). $\endgroup$ – Taroccoesbrocco Sep 19 '18 at 14:52

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