showing $\nexists\;\beta\in\mathbb N:\alpha<\beta<\alpha+1$ I want to prove that $\nexists\; \beta\in\mathbb N$ such that $\alpha<\beta<\alpha+1$ for all $\alpha\in\mathbb N$. I just want to use the Peano axioms and $+$ and $\cdot$
If $\alpha<\beta$ then there is a $\gamma\in\mathbb N$ such that $\beta=\alpha+\gamma$.
If $\beta<\alpha+1$ then there is a $\delta\in\mathbb N$ such that $\alpha+1=\beta+\delta$.
Now I tried to equalize the two equations and I got $\gamma\le0$ which is  contradictory to $\gamma\in\mathbb N$. But I used $\alpha+1-\delta=\beta$ in which the $-$ is problematic because I am not allowed to use it.
Anbody knows a better solution? Thanks a lot!
 A: Fix $\alpha\in\Bbb N$; the result will follow almost immediately if you can prove that if $\alpha<\beta$, then $S\alpha\le\beta$. 
Let $A=\{\gamma\in\Bbb N:\gamma=0\text{ or }S\alpha\le\alpha+\gamma\}$; clearly $0\in A$. Suppose that $\gamma\in A$. If $\gamma=0$, then $S\gamma=1$, and $S\alpha=\alpha+1=\alpha+S\gamma$, so $S\gamma\in A$. Otherwise, $$S\alpha\le\alpha+\gamma<\alpha+\gamma+1=\alpha+S\gamma\;,$$ and again we conclude that $S\gamma\in A$. The induction axiom now implies that $A=\Bbb N$.
Now suppose that $\alpha<\beta$. Then $\beta=\alpha+\gamma$ for some $\gamma\in\Bbb N\setminus\{0\}$, and it follows from the previous paragraph that $S\alpha\le\alpha+\gamma=\beta$.
A: I'm not sure if the following is allowed but:
Inserting the first equation in the second we get:
$\alpha+1=\alpha+\gamma+\delta$.
Now we can substract $\alpha$ at both sides to get $1=\gamma+\delta$, which is a contradiction.
A: I assume that $0\notin \mathbb N$ and $x<y:\Leftrightarrow \exists z\colon x+z=y$.
You may already have proved 
$$\tag1\forall x,y\colon \exists z\colon x+y=S(z)$$
and
$$\tag2 \forall x,y,z\colon (x+y=x+z\rightarrow y=z).$$
Now the assumption $\alpha<\beta \land \beta<\alpha+1$ implies that there are $x,y$ such that $\alpha+x=\beta$ and $\beta+y=S(\alpha)$, hence $\alpha+x+y=S(\alpha)$.
Using $(1)$, there exists $z$ such that 
$\alpha+1=\alpha+S(z)=$ and by $(2)$, we conclude $1=S(z)$ contrary to $\neg\exists x\colon S(x)=1$.

In case you don't have $(1)$ or $(2)$ yet, they are readily proved by induction:


*

*If $x=1$, then $x+y=S(y)$. For the induction step, $S(x)+y=S(x+y)$ gives us $z=x+y$ as candidate immediately.

*If $x=1$, then $S(y)=x+y=x+z=S(z)$, i.e. $S(y)=S(z)$, implies $y=z$. Otherwise $S(x+y)=S(x)+y=S(x)+z=S(x+z)$ implies $x+y=x+z$ and thus $x=z$ by induction hypothesis.

