Note: This question has already been answered here Proving mathematical induction with arbitrary base using (weak) induction.

I was trying to 'reconstruct' at least one proof given in this question or to grasp the idea. I believe my main struggle is the setup and find something to even work with. I will summarize it here again in my own words.

Proposition: Let $n_0 \in \mathbb{N}=\lbrace0,1,2, \dots \rbrace$ be arbitrary and $A(n)$ be a property for $n \in \mathbb{N}$.
If $A(n_0)$ is true and $\forall n \geq n_0: A(n) \implies A(n+1)$ then $\forall n \geq n_0:A(n)$

This is something that can be shown by induction. However I am having great trouble with the setup of it. I could copy-paste what's already written in the linked question but I have no understanding about how they compose their statements there.

The Proposition above can be summarized as: Induction works for any base case $n_0 \in \mathbb{N}$, prove that.

  • So one setup they use in the linked question is the following. Let $Q$ be the statement $Q(n): n \geq n_0 \implies A(n)$. This makes somehow sense to me, because I read it as if $n \geq n_0$ is true, then $A(n)$ better be true as well.

But I don't see how it is linked to the proposition above, because the proposition uses much more than just that, it also makes use of that $\forall n \geq n_0: A(n) \implies A(n+1)$ so I can only guess that they make use of the transitive property of the implication (law of syllogism)

  • The second setup as suggested by Professor Brian M. Scott is to define $Q(n):A(n+n_0)$ which seems very counter intuitive to me. It might be a good trick but I have no idea how to work with it.

Let me try to work with the hint given by Brain M. Scott. Define $Q(n): A(n+n_0)$

$Q(0): A(n_0)$ and now I am already stuck. Does this expression direct me to go back up to the proposition and continue reading from there? If so, how is that supposed to complete the base case?


1 Answer 1


The "standard" expression of Induction principle is :

$(A(0) \land \forall n (A(n) \rightarrow A(n+1))) \rightarrow \forall n A(n)$.

In words :

if we can prove that "property" $A$ holds for $0$ and we can prove that : if it holds for a number $n$ whatever, then it holds also for its successor (i.e. $n+1$), then "property" $A$ holds for every natural numbers.

The "trick" with

Proposition : Let $n_0 \in \mathbb N = \{ 0,1,2,\ldots \}$ be arbitrary and $A(n)$ be a property for $n \in \mathbb N$. If $A(n_0)$ is true and $∀n \ge n_0 : A(n) \rightarrow A(n+1)$ then $∀n \ge n_0 : A(n)$

is quite simple; define a new property $A'(n) := A(n+n_0)$ and apply "normal" Induction.

For $n=0$, we prove $A'(0)$, which is $A(0+n_0)=A(n_0)$.

Then we prove ; $\forall n(A'(n) \rightarrow A'(n+1))$, which amount to : $\forall n \ge n_0(A(n) \rightarrow A(n+1))$.

Thus, by Induction we conclude with :

$\forall n A'(n)$,

which is : $\forall n \ge n_0 : A(n)$.

  • $\begingroup$ Thanks for your answer, I believe to understand this already better. The base case with the new defined property would be true by definition, because $A(n_0)$ is true. Do I read that right? Next I assume the Proposition to be true for our newly defined property $A'(n)=(An+n_0)$ and continue from there? $\endgroup$
    – Spaced
    Oct 5, 2014 at 14:15
  • $\begingroup$ @Spaced - in general, we have to prove the base case also; but usually, it is a quite simple calculation. Having proved or verified the base case, YES, you have to prove the induction step for $A'$. Having done this, you can conclude that the property $A$ holds for all numbers "starting from" the base one. $\endgroup$ Oct 5, 2014 at 14:20
  • $\begingroup$ I think what confuses me in your description is the phrase $A'(n) \to A'(n+1)$ which amount to $A(n) \to A(n+1)$ you mean that they have the same significance or that they are really the same? $\endgroup$
    – Spaced
    Oct 5, 2014 at 14:28
  • $\begingroup$ @Spaced - I've written : "...which amount to : $∀n≥n_0(A(n)→A(n+1))$" and not $\forall n(A(n)→A(n+1))$ ... $\endgroup$ Oct 5, 2014 at 14:34
  • $\begingroup$ I think I get it, I have to show more empathy to the quantors in order to understand this step :) $\endgroup$
    – Spaced
    Oct 5, 2014 at 14:34

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