Proof by somewhat convoluted "finite" induction I am doing a somewhat strange proof by induction. The situation is as follows.
I have 4 statements $A,B,C,D$ which I have to prove for $t = 1,2,\dots,k$.
So I am trying to do an induction from $1$ to $k$ (finite induction?!).
It goes like this.

*

*For $t=1$ I have the statements proved, no problem here.


*I assume all $4$ statements are true for $t=s \le k-1$.


*Then I am trying to prove the 4 statements for $t=s+1$. While doing that I am proving $A(s+1),B(s+1),C(s+1)$ in this order using only $A(j),B(j),C(j)$ for $j \le s$ (which I already have from the induction hypothesis). But then to prove $D(s+1)$, I need to use $A(s+1)$ in my arguments. I kind of wonder if I am allowed to do that.
I think I am allowed because I already proved $A(s+1)$, right? And its proof has nothing to do with $D(s+1)$. So I think I am not going in a logical loop here.
Also I wonder in general if there's such induction at all which is not done for all $n \in \mathbb{N}$ but only for $n \le k$.
 A: I'm writing this as an answer as it would be a little long for a comment: In a nutshell, it's perfectly valid to use $D(s)$ to prove $A(s+1)$. Here is an example that uses two different assumptions (from a proof I gave in this question) although it isn't nearly as complicated as your problem. Define
$$b_n=\frac{1}{3}+\frac{1}{b_{n-1}}$$
with $b_1=\frac{5}{4}$. We will show that $1<b_n<\frac{3}{2}$. Let $P(n)$ be the proposition that $1<b_n$ and let $Q(n)$ be the proposition that $b_n<\frac{3}{2}$.
Base Cases: Obviously, the propositions both hold for $n=1$.
Induction Step: Assume that $P(n)$ and $Q(n)$ are true for some $n$. This implies both
$$1<b_n$$
$$b_n<\frac{3}{2}$$
This gives us
$$b_{n+1}=\frac{1}{3}+\frac{1}{b_n}>\frac{1}{3}+\frac{1}{3/2}=1$$
$$b_{n+1}=\frac{1}{3}+\frac{1}{b_n}<\frac{1}{3}+\frac{1}{1}=\frac{4}{3}<\frac{3}{2}$$
and we are done. As you can see in this example, we used $P(n)\Rightarrow Q(n+1)$ and $Q(n)\Rightarrow P(n+1)$. What's important for multiple proposition induction proofs is that
$$P_1(n)\Rightarrow P_2(n)\Rightarrow ...\Rightarrow P_k(n)\Rightarrow P_1(n+1)\Rightarrow P_2(n+1)\Rightarrow ...\Rightarrow P_k(n+1)$$
A: Induction will hold on $(X,f,x_0)$ if we have $f:X\to X$ and $x_0\in X$ and
$~~~~X=\{x_0, f(x_0), f(f(x_0)),~ \cdots \}$
regardless of whether or not $X$ is infinite.
Finite Example
Consider $X =\{0, 1 \}$, $~f(0)=1$, and $f(1)=0$.
We have $f: X \to X~$ and $X=\{ 0, f(0)\}$.
It is trivial to prove that induction holds, that is:
$~~~~\forall P\subset X:[0\in P ~\land ~\forall x\in P: f(x)\in P \implies P=X]$
Hint: There are only 4 subsets of $X$ to consider.
