Suppose $R$ is a relation on $A$ and $S$ is the transitive closure of $R$. If $(a, b) \in S$, then there is some positive integer $n$ such that $(a, b) \in R^n$, and therefore by the well-ordering principle, there must be a smallest such $n$. We define the distance from $a$ to $b$ to be the smallest positive integer $n$ such that $(a, b) \in R^n$, and we write $d(a, b)$ to denote this distance.
(a) Suppose that $(a, b)$ and $(b, c)$ (and therefore $(a, c) \in S$ since $S$ is transitive). Prove that $d(a, b) \leq d(a, b) + d(b, c)$.
(b) Suppose $(a, c) \in S$ and $0 < m < d(a, c)$. Prove that there is some $b \in A$ such that $d(a, b) = m$ and $d(b, c) = d(a, c) - m$.
Here are my attempts at proving the above. Are they correct?
Proof of (a):
Let $d(a, b) = m$ and $d(b, c) = n$. Then $(a, b) \in R^m$ and $(b, c) \in R^n$. Therefore $(a, c) \in R^n \circ R^m = R^{n + m}.$ Let,
$$ T = \{ q : (a, c) \in R^q \}. $$
Then it follows that $m + n \in T$. Since the smallest element of $T$ is $d(a, c)$ we must have that,
$$ d(a, c) \leq m + n = d(a, b) + d(b, c). \square $$
Proof of (b):
Let $n = d(a, c)$. Then $(a, c) \in R^n = R^{n - m} \circ R^m$. This means that there exists a $b \in A$ such that $(a, b) \in R^m$ and $(b, c) \in R^{n - m}$. Hence $d(a, b) \leq m$ and $d(b, c) \leq n - m$. We also have from part (a) that,
$$ d(a, c) \leq d(a, b) + d(b, c). $$
Now suppose that $d(a, b) < m$. Then we would have,
$$ d(a, c) \leq d(a, b) + d(b, c) < m + n - m = n $$
which is impossible since $d(a, c) = n$. So $d(a, b) = m$. Similarly, if we suppose that $d(b, c) < n - m$ we are led to the contradiction that $d(a, c) < n$. Hence, $d(b, c) = n - m$.
So there is some $b \in A$ such that $d(a, b) = m$ and $d(b, c) = d(a, c) - m$. $\square$
I think the proof of part (a) is correct but I'm not so sure about part (b). The reason I think part (b) might not be correct is that this exercise is from the chapter on induction and I haven't used induction in the proof.