The Paradox of the Ravens |
Back |
Originator: |
Carl Hempel |
||||||||||||||||||||||
Publication: |
Theoria (1937) - A Swedish periodical |
||||||||||||||||||||||
The Paradox: |
Carl Hempel's 'Ravens paradox' was first published in 1937 in the Swedish periodical Theoria1. The paradox results from Hempel's conclusion that a simple generalisation such as 'all Ravens are black' can be confirmed by the observation of a white handkerchief2. This seems to contradict the common sense view that the observation of a white handkerchief has no bearing upon the generalisation that 'all Ravens are black'. To solve the contradiction and hence the Ravens paradox, I will have to show that either Hempel's argument is faulty or that the common sense view is incorrect. The Ravens paradox centres on how a generalisation can be confirmed. Before a generalisation can be confirmed it 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 mathematically formulated as "x (Ax®Bx) - for all x, if x has property A then x has property B. There are two necessary conditions for building such a generalisation:
These conditions are similar to Nicod's criterion3 for the confirmation of generalisations. The only difference is that the conditions I have used are for the construction of a generalisation whereas Nicod's criterion is for the confirmation of a generalisation. This is an important point because as I intend to show later in the essay, the construction and confirmation of a generalisation are two different processes. Condition (1) states that at least one observation of the form (Ai & Bi) - where i is an individual instance - must be made before a generalisation can be constructed. That is to say that a memory of past observation is required before a generalisation can be made. If there has been no observation then there can be no generalisation. This is also an important point that which I will touch on again later in the essay. The two conditions for generalisation construction are self-evident. For example say you have observed only one apple and it is red. You can build the generalisation that 'all apples are red'. If you have observed ten apples, all of which are red then you can also build the same generalisation. If however, you have observed ten apples of which nine are red and one is green, you cannot build the generalisation that 'all apples are red' because this conflicts with condition (2) - the green apple does not have the property of being red. The generalisation 'there are no cars in my room' can be mathematically formulated as "x (Cx®-Rx) - for all x if x is a car then x is not in my room. The conditions for making this generalisation are the observation of at least one car not in my room and of none in my room. I can see the need to make an addition to the rules for the formation of a generalisation. All generalisations have what I will call an identifier property. This is the property that denotes what the generalisation is about. For example the car above is the identifier in the generalisation 'there are no cars in my room' and the apple above is the identifier in the generalisation 'all apples are red'. The third condition for the formulation of a generalisation is:
That is to say that the generalisation above "x (Cx®-Rx) can only be formed from an observation of (Ci & -Ri). Sainsbury claims that the generalisation 'All places fail to contain my spectacles' can be made from the not observing spectacles4. The generalisation can be mathematically formulated as "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. "x (Px®-Sx) can be contraposed to its logically equivalent "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 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 self-evident 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 apples are red' built upon the observation of one hundred apples all of which are red seems greater than the same generalisation built upon the observation of one red apple. It is for this reason that Nicod's criterion treats each observation of the form (Ai & Bi) as a confirming instance of "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 "x (Ax®Bx) can be transformed into "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 "x (-Bx®-Ax). Since however, the statement "x (-Bx®-Ax) is logically equivalent to "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:
From these two conditions the paradox of the Ravens can now be formed. Hempel assumes the generalisation that 'all ravens are black'. To confirm the generalisation 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 handkerchief 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 handkerchief will not. A common counter example5 of the Ravens paradox is that the observation of a white handkerchief can also be used as a confirming instance for a generalisation like 'all unicorns are black'. But as stated earlier in the essay, a generalisation must first exist before it can be confirmed. And as also stated earlier, for the generalisation "x (Ax®Bx) to exist there must first be at least one observation of the form (Ai & Bi). It follows that the observation of a white handkerchief 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. At the beginning of the essay 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 generalisation "x (Ax®Bx):
As I have shown above "x (-Bx®-Ax) is logically equivalent to "x (Ax®Bx) by the law of contraposition. Therefore according to 1. observations of the form (-Ai & -Bi) must confirm the generalisation "x (Ax®Bx). This contradicts part 3. of Nicod's criterion. I now intend to argue why part 3. of Nicod's criterion - and in particular the assertion that observations of the form (-Ai & -Bi) are neutral - is incorrect. So far I have discussed what observations are confirming instances of generalisations. Now I want 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. This is a founding rule of all human thought. In the case of the generalisation "x (Ax®Bx), any instance of the form (Ai & -Bi) is in direct contradiction. Therefore the only way to establish the truth of "x (Ax®Bx) is to show that (Ai & -Bi) cannot exist. In the case of the Ravens paradox that 'all ravens are black', one must show that a non-black raven does not exist. There are three ways that this can be achieved:
Rational action states that the best action is the one that is most efficient towards achieving a given goal. In this case the goal is to find wether 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 (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 cars in my room' is equivalent to the generalisation that 'all cars are not in my room' and can be mathematically formulated as "x (Cx®-Rx). Likewise this can be confirmed by counting cars or by counting things in my room. It is obvious that in this case the rational action is to count all the things in my room. Surely once I have counted all the things in my room and found none of them to be cars I can state that the generalisation 'there are no cars in my room' is confirmed. That is to say that each individual instance of (-Ci & Ri) is a confirming instance. What I have in effect tried to show in this essay is that the common sense view is incorrect and not Hemple's argument. Of course a white handkerchief 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. Once one has a clear understanding of this point the many objections made towards Hempel's reasoning are falsified. What is left of the paradox is a common sense view against Hemple's argument with no valid reasoning to back it up. And thus there is no paradox, and Hempel's reasoning can be held to be true.
1 Mathematical Puzzles and Diversions - page 54
Cohen, L. Jonathan (1989) The Philosophy of Induction and Probability, Clarendon Press - Oxford. |
