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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.

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

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

  3. 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 $n \le k$.

peter.petrov
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    Yes, its perfectly valid to prove $D(s+1)$ using $A(s+1)$. As to your second point, if you are trying to prove some proposition $P(n)$ for a $1\leq n\leq k$, you are really just doing regular induction on the proposition $Q(n)$ defined as: $P(n)$ for $1\leq n\leq k$ and true for $n>k$. Its just for $n>k$ this is self-evident. – QC_QAOA May 06 '22 at 21:52
  • @QC_QAOA Thanks, I also kind of felt it should be valid, but I wasn't fully convinced. What if the proof of $A(s+1)$ was using the induction hypothesis $D(s)$. Would my approach still be valid then? Or would that cause a loop? – peter.petrov May 06 '22 at 21:54
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    If it makes you more comfortable, you might try a proof by well-ordering: assume a minimal counterexample, then prove there is an even smaller one. These methods are equivalent, of course, but have different flavors. – While I Am May 06 '22 at 22:56

2 Answers2

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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

$$P_1(n)\Rightarrow P_2(n)\Rightarrow ...\Rightarrow P_k(n)\Rightarrow P_1(n+1)\Rightarrow P_2(n+1)\Rightarrow ...\Rightarrow P_k(n+1)$$

QC_QAOA
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  • Thanks. Yes, I see. Your example is a bit more symmetric. I've done such proofs many times, I know these are valid. In my case I was asking this: if I use $D(s)$ to prove $A(s+1)$ i.e. to make the inductive step for $A$, and then if use $A(s+1)$ to prove $D(s+1)$ i.e. to make the inductive step for $D$, would that still be valid? – peter.petrov May 06 '22 at 22:14
  • Because in your example if you denote $C(n) = P(n)\ and\ Q(n)$ then there's no problem, you're using part $P$ of $C(n)$ to prove part $Q$ of $C(n+1)$ and vice versa. But it's all clear and it feels valid even intuitively. – peter.petrov May 06 '22 at 22:16
  • Yes, that's fine. What you are doing is basically $A(s)\Rightarrow D(s)\Rightarrow A(s+1)\Rightarrow D(s+1)$ – QC_QAOA May 06 '22 at 22:59
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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.

Dan Christensen
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