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John D. Norton

2. Symmetry Theorem
for deductively definable inductive logics

The fact that an inductive logic is deductively definable places a strong restriction on the logic. Since the inductive logic is only allowed to "know" about deductive relations between propositions, we will see that its inductive strengths must conform to the simple formula (S) presented below. The essential idea expressed in (S) is that any symmetry of the deductive structure must also be a symmetry of the inductive logic. For a symmetry transformation of the deductive structure is one that makes no change in the deductive structure. Therefore it can make no change in the inductive logic as well, if that logic is deductively defined.

Symmetries of deductive relations...

To get a sense of how this works, imagine a case in which there are ten mutually exclusive options and that these ten are the logically strongest options available. It will be helpful to picture these as ten tiles.

1 2 3 4 5 6 7 8 9 10

This picture is just intended to evoke the idea that these ten options are exclusive: you cannot have both 3 and 5. They are also the most specific; the tiles do not have logical parts (yet). That is, they are "atoms."

The deductive relations between these options are of the form


(D1) (1 or 2) deductively entails (1 or 2 or 3 or 4).


One can see immediately that this deductive relation among 1, 2, 3 and 4 is mirrored exactly by the corresponding relation among 7, 8, 9 and 10. For, analogously, we have

(D2) (7 or 8) deductively entails (7 or 8 or 9 or 10).

In fact the deduction (D1) is just the same as the deduction (D2) in all aspects other than the naming of the tiles. To visualize this, imagine we peel off the labels "1" and "7" from tiles 1 and 7 and switch them, where ever they occur; and so on for the the tiles 2and 8, 3 and 9, 4 and 10. If we do that, the first deduction (D1) turns into the second from (D2) vice versa.


It is easy to see that all that matters in affirming the sameness of the deductions (D1) and (D2) is that, in each case, we are inferring from a disjunction ("or'ing") of a pair of tiles to a disjunction of 4 tiles that contains them. Once we have those tile counts, we can produce many variant manifestations of the one deductive relation ("2 goes to 4") merely by sticking different labels on the tiles. They are all the same "2 goes to 4" deduction, just labeled differently.

This analysis is easily extended. Let us say that we have two propositions, A and B, and some deductive relations between them: A may entail B; or B may entail A; or neither entails the other; or A may entail B with the possibility of a distinct C intervening, etc. How much do we need to know to fix precisely which deductive relation obtains between A and B?

To answer, let us use the symbol # to represent the count of tiles (or, equivalently, atoms) so that

#A = number of tiles in proposition A
= number of atoms in proposition A

All we need to know are the counts of tiles in four sets, #A&B, #A&~B, #~A&B and #~A&~B.

In the case of (D1) above, we set

A = 1 and B = ( 1 or 2 or 3 or 4)

and read #A&B=1, #A&~B=0, #~A&B=3, #~A&~B=6.

Now consider another pair of propositions, A' and B'. The deductive relations between A and B are the same as those obtaining between A' and B' just when the tile/atom counts in these four sets match up. That is,

#A&B = #A'&B'
#A&~B = #A'&~B'
#~A&B = #~A'&B'
#~A&~B = #~A'&~B'

The inference (D2) is the same as (D1) because the tile/atom counts in these four sets match up.

This is a significant result that deserves special display:

The deductive
relations between
propositions A and B
are fully fixed
by the
atom counts
#A&B
#A&~B
#~A&B
#~A&~B
In more abstract terms, this captures an important symmetry of a Boolean algebra of propositions. Any permutation of its atoms is a symmetry of the deductive structure. That just means that the deductive structure is unaffected by a permutation ("switching around") of the atoms. The symmetry is expressed here metaphorically by saying that any rearrangement of the labels on the tiles leaves the deductive relations unchanged (as long as we are consistent in our relabeling).

...constrain the inductive logic

This last fact about deductive relations is decisive when it comes to formulating any deductively definable inductive logic. In brief, if an inductive logic is defined in terms of the deductive relations alone, then any symmetry of the deductive relations must also be a symmetry of the inductive logic.

We can see how this works in a few simple cases. Continuing the case of the ten options above, we can immediately see that the three strengths

[ 1 | 1 or 2 or 3 ] and [ 8 | 8 or 9 or 10 ] and [ 3 | 1 or 2 or 3 ]

must be the same. For each is of the form [A|B] where the deductive relation between A and B is same. That means that we can convert one into the other merely by switching labels on the tiles. We convert the first into the second merely by switching the labels 1 and 8; 2 and 9; and 3 and 10, where ever they occur. We convert the first into the third by switching the labels 1 and 3, where ever they occur.

Generalizing this case, we can see that the strengths [A|B] and [A'|B'] in a deductively definable inductive logic will be the same just in case the tile/atom counts match up as before. That is,

#A&B = #A'&B'
#A&~B = #A'&~B'
#~A&B = #~A'&B'
#~A&~B = #~A'&~B'

Another way to say that is to say that the inductive strength [A|B] is fixed once we know the tile/atom counts in the four sets mentioned. That is, the strength [A|B] is a function of these four counts.

For what follows, it will be convenient to present these counts a little differently. Since, we know these four counts must sum to 10, the total number of tiles/atoms. We need only fix the first three counts #A&B, #A&~B, #~A&B, since the value of the fourth, #~A&~B, is recoverable by subtraction.

Let us now generalize from the case of an algebra of propositions with 10 atoms to one with N atoms. The basic theorem is

If an inductive logic is deductively definable in an algebra with N atoms, there exists a function fN such that

(S)[A|B] = fN(#A&B, #A&~B, #~A&B)

This is the most useful form of the theorem. It is stated slightly more strongly than it need be. We do not need to require that the inductive logic is deductively definable. A weaker condition is sufficient for (S) to hold; it is that the symmetries of the deductive structure are also symmetries of the inductive logic.


0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 What Logics of Induction are There?
John D. Norton