Hi I'm reading this great book Foundations of Analysis by Edmund Landau, a really old book that aims to build the foundations of analysis from basic arithmetic.

I'm not that strong on proofing yet but am trying, here is my problem.

Using these 5 axioms for the Natural numbers;

$\displaystyle I) \ 1 \in \mathbb{N}$

$\displaystyle II) \ \forall \ x \ \exists \ x' \ : x' \ = \ x \ + \ 1$

$\displaystyle III) \ x' \ \neq \ 1$

$\displaystyle IV) \ if \ x' \ = \ y' \ then \ x \ = \ y$

$\displaystyle V) \ \exists \ \mathbb{R} \ : \ \mathbb{N} \ \subset \ \mathbb{R} $

with the following properties

$\displaystyle i) \ 1 \ \in \ \mathbb{R}, \ $

$\displaystyle ii) \ if \ x \ \in \ \mathbb{R} \ then \ x' \ \in \ \mathbb{R}$

I want to prove that x' ≠ x. It's the second theorem, the first being;

If x ≠ y then x' ≠ y'

which is proved by assuming x ≠ y and x' = y'

so we find a contradiction with axiomIVabove because x' = y' means x = y.

This is beautiful and understandable but in proving x' ≠ x the proof goes as follows,

By axiomsIandIIIabove,

i)1' ≠ 1 because 1' = 1 + 1.

so 1 belongs in R.

Then it says;

ii)If x ∈ R then x' ≠ x and by theorem 1 (x')' ≠ x' so x' ∈ R

This makes no sense to me, where did it come from

I don't know why you can't use axiom II and just rearrange x' = x + 1 into x = x' - 1 to prove it