Prove: If each of m and n is a counting number, then m+n is a counting number.
A counting number is defined as a number in the minimal induction set.
The minimal induction set is the set of all counting numbers and is given by C.
Below is the sketch of the proof I've tried,
Lemma 1. For a transitive set t, $\displaystyle \bigcup(t^+) = t $.
Proof.
$\displaystyle \bigcup t^+ = \bigcup (t \cup \{t\})
=\bigcup t \cup \bigcup \{t\}
=\bigcup t \cup t
=t
$
The last step is followed from $\displaystyle \bigcup t \subseteq t$ for a transitive set t.
1. Show that the minimal induction set (a.k.a "a set of natural numbers") is transitive using the above lemma.
For a set theoretic construction of natural numbers, the followings hold (since it is transitive)
$\displaystyle 0 \in 1 \in 2 \in 3 ....$
$\displaystyle 0 \subseteq 1 \subseteq 2 \subseteq 3 \subseteq...$.
2. Using a recursion theorem, show that for each $\displaystyle m \in \omega$, there exists a unique function
$\displaystyle A_m:\omega \rightarrow \omega$ (guarantees that it is a function from a set of natural numbers to a set of natural numbers) such that
$\displaystyle A_m(0) = m, A_m(n) = m + n $.
Now, combine 1 & 2 show that if each of m and n is a counting number, then m+n is a counting number.