Comments on: When can we do induction?

By: Remember me

Remember me — Wed, 05 Aug 2015 12:34:16 +0000

Great Blog!!

By: Mario Carneiro

Mario Carneiro — Fri, 13 Mar 2015 05:11:28 +0000

To be more precise about the reason I suggest an "exists" interpretation, the way I originally read it (with my "formalist" hat on) was to "repair" the free variable m by universally quantifying the whole formula to get "for all m, ((m < n -> m in A) -> n in A)", which is propositionally equivalent to "(exists m:(m < n -> m in A)) -> n in A" and is not what you intended. The first boxed equation is stated correctly, and "strong" induction is called that because you are allowed to make a stronger assumption in your induction proof, hence the presence of an additional quantifier by comparison to the first theorem. I would write the boxed equation as:

Theorem Let $A\subseteq\Bbb N$ such that for all $n\in\Bbb N$ we have $(\forall m<n,\ m\in A)\implies n\in A$. Then $A=\Bbb N$.

By: Bristol

Bristol — Wed, 11 Mar 2015 16:04:05 +0000

From a computer science perspective, induction on the natural numbers is just a very special case of structural induction - if we define a type Nat = Zero | Succ of Nat (borrowing syntax from Standard ML) to mimic the set-theoretic construction then what we're really saying that for a predicate p on Nat, if p(Zero) holds and for any n in Nat we hage p(n) => p(Succ(n)) then p must be true over all of Nat. Of course this misses the mathematically interesting point that all such "recursively defined structures" can be ordered nicely, by definition. But it gives a host of examples on which one can do induction: lists, binary trees, in fact pretty much any useful data structure.

By: Tobias Kildetoft

Tobias Kildetoft — Wed, 11 Mar 2015 09:34:41 +0000

Thank you for the comment. I agree that your version is technically more correct, but I disagree that it could be interpreted differently. Implications will generally be interpreted in the way I intend here. This does mean that I have been a bit inconsistent, as I could have left out a "for all" in the box preceding it. I felt that adding the extra "for all" in the second box (and those others with similar statements) would make the formulation overly clunky to read, but I will consider if it might be possible without disturbing the readability too much.

By: Mario Carneiro

Mario Carneiro — Wed, 11 Mar 2015 08:09:25 +0000

The second boxed theorem should make it clear that the $m$ is universally quantified within the antecedent, i.e. $(\forall m (m < n \implies m\in A\right))\implies n\in A$. Otherwise it could be interpreted as a $\exists m$ and this leads to something definitely false.