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I am quite new to $\lambda$-calculus.

In order to proof any statement $P(n)$ in $\lambda$-calculus for all Church numerals $n$, is it sufficient to proof $P(0)$ and $P(n) \rightarrow P(succ \; n)$?

(Using $0:= \lambda fx.x$ and $succ := \lambda nfx.f(nfx)$.)

Wouldn't it be necessary to proof first that $0$ is the successor of no numeral and that $succ$ is injective?

If this is the case, how can you proof the injectiveness of $succ$ without recurring to induction? All proofs I have seen so far, rely on the internal structure of Church numerals ($\lambda fx.f^n x$) but how can we assume that all Church numerals have this form without induction? (What is this exponent $n$ even? Some concept of natural number we are taking from outside of $\lambda$-calculus?)

This question is motivated by the fact that e.g. in ZF you first proof that 0 has no predecessor and that the successor function is injective and until that moment you take the jump to Peano arithmetics and gain access to induction. Maybe my error is to seek similarities between ZF and $\lambda$-calculus where there are none.

Chirmol Studio
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  • You don't need to know that $0$ is not a successor, and that $succ$ is injective, to use induction. Consider (for example) the set of integers. In this set induction is still true, in the sense that $P(0)\land(\forall n. P(n)\implies P(n+1))$ implies $\forall n\in\Bbb N. P(n)$. But in the integers, zero is the successor of $-1$. – MJD May 01 '23 at 01:57
  • Also note that $P(a)$ and $\forall x. P(x)\implies P(f(x))$ implies (by ordinary induction of the natural numbers) that $P(f^n(a))$ for all natural numbers $n$, and this is regardless of whether $f$ is injective. Your succ function is of this type. – MJD May 01 '23 at 02:00
  • @MJD But where does this "n" come from? What does "for all natural numbers" mean? Do we need an outside idea/notion/definition of natural numbers (maybe from ZF/Peano) in order to use induction in lambda calculus? – Chirmol Studio May 01 '23 at 19:37
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    Why not? Unless you are trying to create all of mathematics from nothing but lambda-calculus, you already have an idea of how integers work, and several models for them. The point of the Church numerals $\lambda f x. f(\dots f x)$ is to show that lambda-terms can also model the integers. – MJD May 01 '23 at 19:52
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    Compare with Why can we use induction when studying metamathematics?. – MJD May 01 '23 at 19:55
  • @MJD Thank you very much. I think this was my error. As in ZF you define the natural numbers from "within", for some weird reason I thought the same would be true for lambda-calculus. I think I have been comparing pears with apples. – Chirmol Studio May 01 '23 at 19:55
  • In ZF the situation is the same: you come to ZF with an idea of the natural numbers, and then you demonstrate that the Von Neumann ordinals are a model of the Peano axioms. – MJD May 01 '23 at 19:58
  • Are Church numerals a model for the Peano/Robinson axioms? – Chirmol Studio May 01 '23 at 19:59
  • Let us continue this discussion in chat. – Chirmol Studio May 01 '23 at 20:06

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