Prove ##1 + a=s(a)=a+1## for ##a \in \mathbb{N}##

[issacnewton]

Homework Statement

Prove $1 + a=s(a)=a+1$ for $a \in \mathbb{N}$ where $s$ is a successor function from Peano postulates

Relevant Equations

Peano postulates

I have to prove that $1 + a = s(a) = a + 1$ using Peano postulates if $a \in \mathbb{N}$. The book I am using ("The real numbers and real analysis" by Ethan Bloch) defines Peano postulates little differently.
Following is a set of Peano postulates I am using. (Axiom 1.2.1 in Bloch's book)

There exists a set $\mathbb{N}$ with an element $1 \in \mathbb{N}$ and a function $s: \mathbb{N} \rightarrow \mathbb{N}$ that satisfy the following three properties.

1) There is no $n \in \mathbb{N}$ such that $s(n) = 1$

2) The function $s$ is injective.

3) Let $G \subseteq \mathbb{N}$ be a set. Suppose that $1 \in G$, and that if $g \in G$ then $s(g) \in G$. Then $G = \mathbb{N}$

And addition operation is given in Theorem 1.2.5 as follows

There is a unique binary operation $+: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ that satisfies the following two properties for all $n, m \in \mathbb{N}$

a. $n + 1 = s(n)$
b. $n + s(m) = s(n + m)$

With this background, we now proceed to present my proof. Let us define a set

$$ G = \{ z \in \mathbb{N} | 1 + z = s(z) = z + 1 \} $$

We want to prove that $G = \mathbb{N}$.
Obviously, $G \subseteq \mathbb{N}$. Using definition of addition, $1 + 1 = s(1) = 1 + 1$. Since $1 \in \mathbb{N}$, it follows that $1 \in G$.

Now, suppose $r \in G$. This means that $r \in \mathbb{N}$ and

$$ 1 + r = s(r) = r + 1 \cdots\cdots (1) $$

Since $s: \mathbb{N} \rightarrow \mathbb{N}$ is a function, we have,

$$ 1 + r = s(r) $$
$$ s(1+r) = s(s(r)) $$
Using addition definition, part (b),

$$ 1 + s(r) = s(s(r)) $$

And from addition definition, part (a), we have $s(s(r)) = s(r) + 1$. We got

$$ 1 + s(r) = s(s(r)) = s(r) + 1 $$

Since $s(r) \in \mathbb{N}$, we have $s(r) \in G$. So, using Peano postulates, part 3), we have $G = \mathbb{N}$. Now if $a \in \mathbb{N}$ is arbitrary, $a \in G$. It follows that $1 + a = s(a) = a + 1$. And since $a$ is arbitrary,

$$ \forall a \in \mathbb{N}\quad 1 + a = s(a) = a + 1 $$

Is the proof good enough ?

Thanks

 

[PeroK]

Looks good to me.