Thursday, March 01, 2012

Ockham on induction

I have made a start on book III-3 of Ockham's Summa Logicae by translating some of the chapters on induction. The part 3 of the third part is received little attention from philosophers, and as far as I know has never been translated. The chapters I have looked at come across as terribly weak, unlike the powerful and insightful work of the first two parts.

Ockham says that induction "is a progression from singulars to the universal", which is pretty much the modern understanding of the term. He gives the example "this [man] runs, and that one, and so on, therefore every man runs". What does the 'and so on' mean? If it means 'every other man apart from this one and that one', then the argument amounts to 'Man a and man b and every other man apart from a and b run, therefore every man runs', which is trivial, though admittedly valid. Or does he mean that the 'and so on' is a placeholder for a longer proposition which enumerates every man there is, and so does not include the word 'every', but concludes with the word 'every'? But that would not work, because the enumeration would have to conclude 'and there are no more men'. Mathematical induction, of course, is different from any of these. As I understand it, it is an inference from the properties of the successor of every number, to the properties of all numbers of a particular type (the naturals). Thus the antecedent and the consequent of mathematical induction both contain the word 'every'. I may be wrong, I'm sure my mathematically minded commenters will leap in to correct me if so.

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11 Comments:

Blogger David Brightly said...

Hi Ed, Mathematical induction (which is of course deductive rather than inductive) is an argument of the form



1.  for all n,  P(n) --> P(n+1)                 (the inductive case)

2.  P(1)                                               (the base case)

-----------------------------------

3.  for all n, P(n)



with P(n) some formula with n free.  The classic example takes P(n) to be the formula the sum of the first n integers = n(n+1)/2.  The hardest bit is usually proving the inductive case by showing that P(n)-->P(n+1) for arbitrary n (and thus for all n).  The base case is usually trivial.  I first met this aged 16 and thought it was wonderful---I could prove an infinity of theorems!  The hardest part was a notational hurdle---one had to stop seeing P(n) as a function application and start seeing it as a logical proposition.

9:31 am  
Blogger David Brightly said...

This comment has been removed by the author.

9:31 am  
Blogger Tony Lloyd said...

Is it really all that weak?

He was writing a few years before Hume and even post-Hume there are plenty of people who agree with Ockham.

11:25 pm  
Blogger Edward Ockham said...

>>Is it really all that weak?

Reading on to the next few chapters, I now see where Ockham is coming from. Not Hume's direction, but more later.

8:14 am  
Blogger Edward Ockham said...

David, thanks for that reminder. But proves my point, no? Your (1) has an 'every' in it.

By the way, isn't there a missing assumption in there? Doesn't there have to be an induction axiom or similar? Long time since I looked at this.

8:15 am  
Blogger Edward Ockham said...

In fact, don't we need an axiom that allows precisely the jump from your 1 & 2 to your 3. That's what the Wikipedia article suggests, but of course I caution everyone to be wary of Wikipedia. (That said, I think the maths articles are generally better than many).

9:12 am  
Blogger Anthony said...

What is your point which you say has been proven?

The only assumptino I see missing from David's proof is that n is restricted to positive integers.

2:55 pm  
Blogger Edward Ockham said...

I think his proof should read

1. P(1)

2. for all n in N (P(n) -> P(n+1))

2a [P(1) & for all n in N (P(n) -> P(n+1)) ] -> for all n P(n)

------------------------------
3. for all n P(n)

where 2a is the axiom of induction. If 3 followed simply from 1 and 2 alone, we wouldn't need the axiom.

4:07 pm  
Blogger David Brightly said...

Yes, the inference from 1 and 2 to 3 is an instance of the 'axiom schema of induction' of Peano arithmetic, your 2a.  This being a logic and not a maths blog we should make it explicit.  The for every n in (1) follows from the logical principle that a proof of phi(n) for arbitrary n is a proof of for all n, phi(n).  The inference to (3) is valid for every n reachable from 1 by finitely many applications of successor.  There are models of first order PA containing non-standard naturals which are not so reachable, though each of these models contains a copy of the standard naturals.  In effect the axiom schema of induction is saying that these are the only naturals.   Agree about WP maths articles, though they are at wildly varying levels.

5:08 pm  
Blogger Anthony said...

Interesting. What about 2b) (A & B & (A & B -> C)) -> C?

Rhetorical question. ;)

1:21 pm  
Blogger Edward Ockham said...

>>Interesting. What about 2b) (A & B & (A & B -> C)) -> C?

Tortoise

6:40 pm  

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