PHIL 2007 - Major Essay

Paradoxes of Confirmation – the Ravens Paradox

Introduction

Hempel’s Ravens paradox is one of the class of the ‘paradoxes of confirmation’ and involves an assertion of confirmation from instances that support a given generalisation

(i.e. that all ravens are black in colour). Furthermore, an equivalent proposition is derived by contraposition (i.e. everything non-black in colour is a non-raven), which is taken as an a priori equivalence, such that any instance of a non-black thing effectively confirms the initial generalisation. The options to respond to this paradox are to:

i) to embrace the paradoxical conclusion;

ii) to deny that the two propositions are a priori equivalent; or

iii) to deny that a generalisation is confirmed by any of its instances.

While Hempel himself takes option (i), Sainsbury goes for option (iii), therein relying on an intuition that some instances that are used to rule in a hypothesis can actually rule it out. While this approach resolves the paradox prima face, this could appear unsatisfactory as there is then a problem with allowing for background information or memory in relation to observations.

Sainsbury also points to a related paradox – the ‘Grue’ paradox, which opens up a characterisation and reference problem. Sainsbury depicts the problem in light of adjusting the response mentioned above in terms of placing further qualification on the instances under which a generalisation is quantified. This is done insofar as the “… hypothesis is confirmed by a body of data containing its instances, and containing no

1 counterinstances, if and only if the data do not say that there is, or even that there is quite likely to be, a property […] such that the examined [observations] are [such] only in virtue of being [otherwise].”1 I shall deal exclusively with the Ravens paradox in this essay. In an attempt to solve the Ravens contradiction, I will intend to show that either

Hempel’s argument is faulty or that the ‘common sense’ view is incorrect.

The Ravens paradox

The Ravens paradox centres on how a generalisation can be confirmed. Before a generalisation can be confirmed its subject must first exist. It is therefore important to understand the necessary conditions for the construction of a simple generalisation such as all A’s are B’s. That is to say, everything has property A also has property B. This can be formulated as: for all x, (Ax→Bx) – that is, for all x, if x has property A then x has property B. There are two necessary conditions for building such a generalisation:

1) There must be at least one instance of some thing that has both property A and property B – for some x, (Ax & Bx); and

2) There must be no instance of anything that has property A and not property B – for no x, (Ax & ~Bx).

These conditions are similar to Nicod’s criterion for the confirmation of generalisations.2

The only difference is that the conditions used above are for the construction of a generalisation whereas Nicod’s criterion is for the confirmation of a generalisation. This is an important point because the construction and confirmation of a generalisation are two different processes.

1 Sainsbury, R. M. (1995) Paradoxes, (2nd ed.) Cambridge University Press. p. 88. 2 Bartha, P. (2008) Lecture Notes - Phil 460. University of British Columbia, http://www.philosophy.ubc.ca/faculty/bartha/p460f08/p460ovh2.pdf

2 The two conditions for generalisation construction are self-evident. For example, if you have observed only one orange and it is (coloured) orange, you can build the generalisation that ‘all oranges are orange’. If you have observed one hundred oranges, all of which are orange then you can also build the same generalisation. If however, you have observed one hundred oranges of which ninety-nine are orange and one is green, you cannot build the generalisation because this conflicts with condition (2).

All generalisations have an ‘identifier property’ – this is the property that denotes what the generalisation is about. For example, the orange above is the identifier in the generalisation ‘all oranges are orange’. The third condition for the formulation of a generalisation is:

3) To form the generalization: for all x, (Ax→Bx) from the observation (Ai & Bi), the identifier property must always be positive, whereas its other property can be either positive or negative.

Sainsbury claims that the generalisation ‘All places fail to contain my spectacles’ can be made from the not observing spectacles.3 The generalisation can be formulated as: for all x, (Px→~Sx), which would require the observation of (Pi & ~Si). The formulation of this generalisation is now blocked by condition (3) because the spectacles are the identifying property. For all x, (Px→~Sx) can be contraposed to its logically equivalent: for all x,

(Sx→~Px), however, and an observation of (Si & ~Pi) will be sufficient to build the generalisation that ‘All places fail to contain my spectacles’. That is to say the

3 Sainsbury, R. M. (1995) Paradoxes, (2nd ed.) Cambridge University Press. p. 76.

3 observation of my spectacles in no place will be required to build the generalisation. This is an impossible observation and thus the generalisation can never be made. This is so because to create the generalisation one must assume that the spectacles exist; and if they exist they must have the property of being in a place.

It seems evident that the probability of a generalisation is greater if the number of observations is greater. For example the generalisation that ‘all oranges are orange’ built upon the observation of one thousand oranges all of which are orange seems greater than the same generalisation built upon the observation of one orange. It is for this reason that

Nicod’s criterion treats each observation of the form (Ai & Bi) as a confirming instance of: for all x, (Ax→Bx). This is the first part of Hempel’s paradox of the Ravens.

The second part of the paradox comes from logical equivalence. The logical statement: for all x, (Ax→Bx) can be transformed into: for all x, (~Bx→~Ax) by the law of contraposition. These two statements are logically equivalent. It follows from Nicod’s criterion that each observation of the form (~Ai & ~Bi) is treated as a confirming instance of: for all x, (~Bx→~Ax). Since, however, the statement: for all x, (~Bx→~Ax) is logically equivalent to: for all x, (Ax→Bx), each observation of (~Ai & ~Bi) must be a confirming instance of both statements. The same is true of an observation of (Ai & Bi).

This can be summed up as follows:

a) Any object that has both properties A and B is a confirming instance of the generalisation that everything that has property A also has property B.

4 b) Any observation that is a confirming instance of a generalisation is also a confirming instance of any proposition that is logically equivalent to that generalization.

From these two conditions the Ravens paradox can now be formed.4 To confirm the generalisation that ‘all ravens are black’ one can observe ravens. For every black raven observed the probability of the generalisation increases. The statement ‘all non-black things are not ravens’ is logically equivalent to the generalisation that ‘all ravens are black’. Therefore one can just as easily count all non-black things to confirm the generalisation that ‘all ravens are black’. That is to say that it logically follows that the observation of a white shoe is a confirming instance of the generalisation ‘all ravens are black’.

The only problem with this argument is that it hasn't allowed for the presence of background knowledge. Such as knowledge like ‘all ravens are birds’, which is part of the definition of what a raven is. If this knowledge is introduced into the argument then it is not all non-black things that are confirming instances, but only all non-black birds.

Under these circumstances the observation of a white swan will be a confirming instance of ‘all ravens are black’, but the observation of a white shoe will not.

A common counter-example of the Ravens paradox is that the observation of a white shoe can also be used as a confirming instance for a generalisation like ‘all unicorns are black’. But as stated earlier, a generalisation must first exist before it can be confirmed.

As also stated earlier, for the generalization: for all x, (Ax→Bx) to exist there must first

4 Hempel, C. (1945) ‘Studies in the Logic of Confirmation,’ Aspects of Scientific Explanation, New York: Free Press (1965) p. 10.

5 be at least one observation of the form (Ai & Bi). It follows that the observation of a white shoe is a confirming instance of ‘all unicorns are black’, but this generalisation can only exist after at least one black unicorn has been observed. Therefore until a black unicorn has been observed there can be no confirming instance of ‘all unicorns are black’ because there can be no generalisation.

Earlier on, I stated that either Hempel’s argument must be faulty or that the common sense view must be false. I intend to argue that Hempel’s argument is sound and that the common sense view is false. To do this I will have to assert as Hempel does in his argument that Nicod’s criterion for the confirmation of a generalisation contains a contradiction. Nicod’s criterion holds that for the generalization: for all x, (Ax→Bx):

1. Observations of the form (Ai & Bi) confirm the generalization;

2. An observation of the form (Ai & ~Bi) disconfirms the generalization; and

3. Observations of the form (~Ai & Bi) and (~Ai & ~Bi) are neutral.

As shown above: for all x, (~Bx→~Ax) is logically equivalent to: for all x, (Ax→Bx) by the law of contraposition. Therefore, according to 1, observations of the form

(~Ai & ~Bi) must confirm the generalization: for all x, (Ax→Bx). This contradicts part 3 of Nicod’s criterion.

I now wish to discuss what is needed to establish a generalisation as being true. By definition, a proposition is true if it is the only possibility. To show that a proposition is the only possibility one has to show that no contradictory proposition can exist. In the

6 case of the generalization: for all x, (Ax→Bx), any instance of the form (Ai & ~Bi) is in direct contradiction. Therefore the only way to establish the truth of: for all x, (Ax→Bx) is to show that (Ai & ~Bi) cannot exist. In the case of the Ravens paradox, one must show that a non-black raven does not exist. There are three ways that this can be achieved:

a) By only looking for things with both the properties of being a raven and not being black (Ri & ~Bi). In this case every observation will be a disconfirming instance.

b) By only looking for things with the property of being a raven (Ri). In this case, of all the ravens observed only the ones with the property of being not black will be disconfirming instances. All the observed ravens that have the property of being black are counted as being confirming instances of the generalisation.

c) By only looking for things with the property of not being black (~Bi). In this case, of all the non-black things observed only the ones with the property of being a raven (Ri) will be disconfirming instances. All the observed non-black things that have the property of being ravens are counted as being confirming instances.

Rational action states that the best choice is the one that is most efficient towards achieving a given goal. In this case the goal is to find whether any instance of (Ri & ~Bi) exists. Option (a) requires the least amount of counting so it would seem that this is the rational course of action to take. There is one drawback to this approach however, because there is no feedback on the probability of the generalisation. The other two approaches give a degree of probability, whereas option (a) can only deny. Of the two options (b) and (c) it is rational to choose the option that will involve the least amount of observations to establish its truth. That is the option that has the least total of individuals.

Every observation from this option will have a greater confirming degree than the other option because the confirming degree is one divided by the total to be counted. In the case of the generalisation that ‘all ravens are black’ the most efficient method is option

7 (b), to count everything with property of being a raven. This decision lies on the assumption that there are more non-black things than ravens.

The generalisation that ‘there are no elephants in my kitchen’ is equivalent to the generalisation that ‘all elephants are not in my kitchen’ and can be formulated as: for all x, (Ex→~Kx). Likewise this can be confirmed by counting elephants or by counting things in my kitchen. It is obvious that in this case the rational action is to count all the things in my kitchen. Surely once I have counted all the things in my kitchen and found none of them to be elephants I can state that the generalisation ‘there are no elephants in my kitchen’ as confirmed. That is to say that each individual instance of (~Ei & Ki) is a confirming instance.

Conclusion

What I have, in effect, attempted to show is that the common sense view is incorrect and not Hempel’s argument. Of course a white shoe is only a confirming instance of the generalisation that ‘all ravens are black’ if it is not known that all ravens are birds. I have also tried to show that there is a clear distinction between the construction of a generalisation and the confirming of that generalisation. What is left of the paradox is a common sense view against Hemple's argument with no valid reasoning to back it up.

Thus there is no paradox and Hempel's reasoning can be held to be true.

8 References

Bartha, P. (2008) Lecture Notes - Phil 460. University of British Columbia, http://www.philosophy.ubc.ca/faculty/bartha/p460f08/p460ovh2.pdf

Hempel, C. (1945) ‘Studies in the Logic of Confirmation,’ pp. 3-46 of C. Hempel Aspects of Scientific Explanation, New York: Free Press (1965).

Nicod, J. (1923) ‘The Logical Problem of Induction,’ translated by Woods in J. Nicod Geometry and Induction, Berkeley: University of California (1970).

Sainsbury, R. M. (1995) Paradoxes, (2nd ed.) New York: Cambridge University Press.