Weak Induction implies Strong Induction Proof Inquiry

Question

I am trying to improve my ability to think formally and write public posts informally. This requires a sort of "toggling" between one and the other. In the following question, I would like to confirm that I am correctly producing this conversion.

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I read a post (What exactly is the difference between weak and strong induction?) about the distinctions (really, semantic equivalence) between weak induction and strong induction.

The accepted and heavily thumbed-up answer shows that weak induction implies strong induction and that strong induction implies weak induction.

The following definitions were provided for weak induction and strong induction:

Weak Induction:

"Let $S(n)$ denote a statement regarding an integer $n$, and let $k\in\mathbb{Z}$ be fixed. If

• (i) $S(k)$ holds, and
• (ii) for every $m\geq k, S(m)\to S(m+1)$,

then for every $n\geq k$, the statement $S(n)$ holds."

Strong Induction: "Let $S(n)$ denote a statement regarding an integer $n$. If

• (i) $S(k)$ is true and
• (ii) for every $m\geq k, [S(k)\land S(k+1)\land\cdots\land S(m)]\to S(m+1)$,

then for every $n\geq k$, the statement $S(n)$ is true. "

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Here is the proof offered for Weak Induction Implies Strong Induction:

Assume that for some $k$, the statement $S(k)$ is true and for every $m\geq k, [S(k)\land S(k+1)\land\cdot\land S(m)]\to S(m+1)$. Let $B$ be the set of all $n>m$ for which $S(n)$ is false. If $B\neq\varnothing, B\subset\mathbb{N}$ and so by well-ordering, $B$ has a least element, say $\ell$. By the definition of $B$, for every $k\leq t<\ell, S(t)$ is true. The premise of the inductive hypothesis is true, and so $S(\ell)$ is true, contradicting that $\ell\in B$. Hence $B=\varnothing$. $\blacksquare$

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Here is my conversion to a formal FOL expression (step by step).

The claim to prove:

$\forall k \Big ( \Big [ S(k) \land \forall m \geq k \big (S(m) \rightarrow S(m+1) \big) \Big] \rightarrow \Big[S(k) \land \forall m \geq k \big(S(k) \land S(k+1) \land ... \land S(m) \rightarrow S(m+1) \Big] \Big ) $

The author proceeds with a proof by contradiction which amounts to negating the previous statement and arriving at a contradiction. The negation of $P \rightarrow Q$ is $P \land \neg Q$. And the negation of $R \land S$ is $\neg R \lor \neg S$. Using this, the negated statement is:

$\exists k \Big ( \Big [ S(k) \land \forall m \geq k \big (S(m) \rightarrow S(m+1) \big) \Big] \land\Big[\neg S(k) \lor \neg [\forall m \geq k \big(S(k) \land S(k+1) \land ... \land S(m) \rightarrow S(m+1)] \Big] \Big ) $

In order for this statement to be true, it is clear that $S(k)$ is true, which means that $\neg S(k)$ cannot also so be true. Therefore, in the disjunction, it must be the formula $\neg [\forall m \geq k \big(S(k) \land S(k+1) \land ... \land S(m) \rightarrow S(m+1)]$ that is true.

This is equivalent to $\exists m \geq k \Big ( S(k) \land S(k+1) \land ... \land S (m) \land \neg S(m+1) \Big )$.

It seems like we could really just stop right here...because $\neg S(m+1)$ would contradict the fact that the weak induction hypothesis gives us $\forall n \geq k \ S(n)$. If someone could explain why the author continues with his set $B$ construction, I would greatly appreciate it. Even if we were to construct a set $B:=\{n \in \mathbb N| n \gt m\ \land \neg S(n)\}$, it is still the case that $\neg S(m+1)$ is a contradictory statement. The construction of $B$ seems unnecessary.

Thank you!

Answer 1

The author was simply giving an ordinary mathematical proof. He didn’t continue with his set $B$ construction: he started with it. The FOL manipulations with which you started probably never even entered his mind; they certainly would not have entered mine, and they strike me as completely unnecessary. Given the equivalence between weak induction and well-ordering of $\Bbb N$, this ‘least counterexample’ approach is a standard, natural way to use induction.

One can, however, also prove the result directly from the statement of weak induction. Assume the hypothesis of strong induction. For $m\ge k$ let $P(m)$ be the statement $\bigwedge_{i=k}^mS(i)$. Then $P(k)=S(k)$ is true, and $P(m)\to P(m+1)$ for each $m\ge k$, so by weak induction $P(n)$ is true for each $n\ge k$, and therefore $S(n)$ is true for each $n\ge k$.