Prove $x' \neq x$ using Peano axioms I am looking at Edmund Landau, Foundation of analysis and do not agree with is proof of Theorem 2 part 2. I put the pages here for easy reference (http://pbrd.co/1y89p7b and  http://pbrd.co/1y89A2s). 
Problem I have with Landau, is theorem 2,  part 2  does not prove the theorem (I think). 
Theorem is \begin{equation}x’ \neq x\end{equation} 
1. He assumes the theorem and goes on with the result.


*Another issue, if we assume successor function is injective does not mean that x and suc(x) can be one and the same in value. Lastly, injection is a big assumption. Axiom 2 “exactly 1” but does not say mutually exclusive.

*I have a separate proof via contradiction. Basically I assume x’ = x and using axiom 4, disprove it as below:
Assume
\begin{equation}x’ = x\end{equation} 
Also,from Axiom 4
\begin{equation} 
x' = y' => x = y
\end{equation} 
This leads to the following contradiction from our assumption and 1st part of axiom 4 that:
\begin{equation}
x = y'
\end{equation} 
This cannot be true because of part 2 of axiom 4 which says:
\begin{equation}
x = y
\end{equation} 
Since the axiom must hold true making our assumption as incorrect. Hence the theorem is proved i.e.
\begin{equation} 
x' \neq x
\end{equation} 
Would appreciate any insights. Is it because this book is a translation of the original german book and someone messed it up ?

Base Case:
$$0' \ne 0$$
That is an axiom: nothing precedes $0$.
Inductive case:
$$\begin{align}
x' \ne x & \Rightarrow x'' \ne x' &\text{Inductive Statement} \\
x'' = x' &\Rightarrow x' = x  & \text{Contrapositive} \\
y' = x' &\Rightarrow y = x  & \text{by letting y =  x'} \\
\end{align}$$
That is also an axiom:  Successor is injective (invertible).
And induction is itself an axiom.
