Is it necessary to use the induction assumption in an induction proof?
Reason to ask is that I was doing a proof about Markov chains:
Prove that $(X_k, Y_{k+1})_{k \geq 1}$ is Markov chain. $X_0=x_0$ is given and in every round ($k \geq 1$) one first generates $Y_k \sim q(X_{k-1}, \cdot)$ (where $q$ is a probability distribution), then either sets $X_k=Y_k$ or rejects $Y_k$ as invalid and sets $X_k=Y_{k-1}$.
By induction:
$X_0 = x_0$ $Y_1 \sim q(X_0, \cdot)$ $P(Y_1 | X_0)$ OK.
Assume $n=k$ i.e. $P(Y_{k} | Y_0, ..., Y_{k-1})=P(Y_k | Y_{k-1})$ holds.
Prove $n=k+1$ holds:
on the $k$th round $X_k=Y_k$ or reject $Y_k$ and set $X_k=Y_{k-1}$, in either case $X_k$ is the value of the previous r.v.
on the $k+1$th one picks $Y_{k+1} \sim q(X_{k}, \cdot)$ so $Y_{k+1}$ only depends on the previous r.v.
It seems like I didn't need to use the induction assumption at all?
Or perhaps I need to use something to split $P(Y_{k+1} | Y_1, ..., X_k)$ to smaller parts that may include $P(Y_k | Y_{k-1})$, the induction step.
If you can prove it without using the inductive assumption, then that's just fine!
I don't have the background to comment on your specific proof, but here is one that I ran into:
OK, so here I proved that every natural number other than $0$ has a predecessor on the basis of the Peano Axioms. On line $6$ I have proven the base, and on lines $7$ through $11$ I prove the step, with the inductive hypothesis on line $7$. Note that I never end up using this inductive hypothesis. In fact, I don't use any of the Peano axioms except the inductive axiom. And even more strikingly: without induction I could not have proven this, because the statement does not follow from the Peano axioms without the inductive axiom. In other words, here is an example of a non-trivial (necessary!) use of induction, that did not use the inductive hypothesis.