# Paradox of Confirmation Paradoxes Term Paper

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≈ 18 · **File:** .docx · **Level:** College Senior · **Topic:** Black Studies - Philosophy

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[. . .] There are two steps to confirming a generalization: proving the generalization's existence and confirming it. Let's take the simple statement that all As are Bs (all ravens are black). This assumes the existence of two subsets A and B. with the functional definition that if element x is in A, it is in B, that is that if x has the property of subset A, then it has the property of subset B. Mathematically, this can be formulated as (Ax

Bx). In order to prove this, for any element x, one must show that 1) There must be at least one instance of some thing that has both property A and property B.

A x (Ax & Bx) and 2) There must be no instance of anything that has property A and not property B. - x (Ax & -Bx)"

Hence, in order to construct a generalization, at least one instance I must be found so that it fulfils both property A and property B (that is, both property of subset A and B). Following condition 2, the generalization does not stand ground if we find at least one instance that does not fulfill A or B. An example is in order here. If we have observed a green car, we can state that all cars are green (it does not contradict any of the two conditions). If we see five green cars, we can built the same generalization (with an increased probability that it will be true). However, if we see five green cars and a white one, the generalization cannot be made, because it contradicts condition 2. Further more, we can sense that if we see ten cars that are green, the probability that all cars are green is greater than if we had only seen one car green.

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for $19.77 Following these two conditions and referring to the initial three propositions (specifically proposition number 3), we can affirm that:

1. "Any object that has both properties A and B. is a confirming instance of the generalization that everything that has property A also has property B. And 2. "Any observation that is a confirming instance of a generalization is also a confirming instance of any proposition that is logically equivalent to that generalization"

## Term Paper on

Going back to Hempel's paradox, we can assume that in order to prove that all ravens are black, one can either observe ravens (and every black raven observed increases the probability that the statement is true) or observe non- black things (any non-black thing that is not a raven increases the probability that the statement is true). If we limit the area of selection for our observations, for example, to non-black birds, then any non-black bird observed will increase more the probability of our statement.

If we now return to Nicod's criterion and apply it in this case, then the criterion would read as such:

1. Black ravens make the generalization "all ravens are black" more probable

2. Non-black ravens falsify the generalization

3. Non-black non-ravens are irrelevant

Let's take the first conclusion from the criterion and use it in a probabilistic manner. Without any observations made, the statement that "all ravens are black" has a 0 probability. If we assume that the number of ravens in the world is X, than observing 1 raven would increase the probability that the generalization is true to 1/X which is greater than 0. Hence, the more black ravens we observe, the more likely it is that the generalization is true (as the probability increases). Now let's connect Hempel's assumption to Nicod's third statement. If we assume that the number of total objects in the world is immensely larger than X, than the observation of a non-black non- raven (a purple cow), only incrementally increases the probability that our statement is true. Perhaps Nicod's criterion and Hempel's paradox are just two different ways to see one thing: Nicod's criterion states that non-black non-ravens are irrelevant, while Hempel states that such observations do increase, however little, the probability that the statement is true.

I want to summarize all the theoretical implications of the Nicod criterion, of the equivalence condition and Hempel's findings and conclusions in the matter, so as to draw a personal conclusion of the facts.

Nicod's criterion regarding the raven paradox affirms three statements. The first one, black ravens make the generalization "all ravens are black" more probable, is a statement of common sense. It is based on a common sense affirmation that we instinctively sense is true. Indeed, any black raven observed increases the probability that all ravens are black. This is because the set of ravens is to be assumed finite (indeed, there is a certain number of ravens in the world, a determined number, be it 5 or 5,000). Probabilistically speaking, if we assume there is a total number of 5,000 ravens and we have seen one that is black, it is more probable that 4,999 ravens are black than 5,000 ravens. That is why, the more black ravens we observe, the more it is probable that the remaining number of ravens is also black.

The second condition from Nicod's criterion can also be sensed true. If we do see a raven that is not black, than it can no longer be generalized that all ravens are black because, from the set of ravens named A, we have already been able to form two different subsets, B and C, with different properties, one containing white ravens, the other black ravens. Further more, any other color of ravens we may observe will form a different subset so that the statement that all ravens are black, which is equivalent to the statement that there is but one subset of set A (in fact equivalent to set A in that all the elements from this subset are also the elements of set A and vice versa), is true.

The questions that arises is in the third condition of Nicod's criteria, non-black non-ravens are irrelevant, because this comes into a contradiction with the equivalence condition that states that all non-black objects are not ravens. The latter would mean that we can observe non- black objects that are not ravens and, if such objects do exists, it would be a step further in generalizing that all ravens are black. What I have seen and understand from Hempel's solution to the problem is the fact that he denies Nicod's third condition in stating that such an observation of non- black, non- raven objects is a step, even if a very small one, towards proving the generalized statement. As I have shown above, probabilistically speaking, finding a purple cow would increase the probability that the generalization is true by 1/Y, where Y is the total number of non-black, non-raven objects.

Thus, from my point-of-view, Hempel's solution to the problem resides in the fact that he manages to fit in both Nicod's criterion and the equivalence condition by introducing probabilities. He denies 3 from Nicod's criterion, adopting the equivalence condition, and proving that the observation of a non-black, non-raven object increases the probability that the statement is true, even if infinitesimally.

Bibliography

1. Friedemann, L. Sara. Why Hempel's Paradox Isn't a Paradox (And What We Could Do If It Was). May 2003. On the Internet at http://www.ellipsis.cx/~liana/phi/920spr03.pdf

2. Carnap, Rudolf. Logical Foundations of Probability. 1950. Can be found on the Internet at http://www.lawrence.edu/fac/boardmaw/Carnap_Explica2.html

3. http://users.skynet.be/bs661306/peter/doc/hpv00r03-476.htm

4. Mackie J.L. The paradox of Confirmation. The British Journal for the Philosophy of Science, 13, 265-277. 1963.

5. Andrew MacMillan at http://www.paradoxes.info / http://home.sprynet.com/~owl1/confirma.htm

http://www.uni-konstanz.de/ppm/Research.htm

http://www.geocities.com/CapitolHill/Lobby/3022/hempel.html

http://monami.us/joshua/chid/spring2002/phil460/philessay2.php

10. Hempel, Carl G, "Studies in the Logic of Confirmation" in P. Achinstein, "The concept of Evidence"

From the internet at http://users.skynet.be/bs661306/peter/doc/hpv00r03-476.htm

Friedemann, L. Sara. Why Hempel's Paradox Isn't a Paradox (And What We Could Do If It Was). May 2003. On the Internet at http://www.ellipsis.cx/~liana/phi/920spr03.pdf

Carnap, Rudolf. Logical Foundations of Probability. 1950. Can be found on the Internet at http://www.lawrence.edu/fac/boardmaw/Carnap_Explica2.html

As shown, among others, by Professor J.L. Mackie in his The paradox of Confirmation, published in The British Journal for the Philosophy of Science, 13, 265-277, in 1963.

That is why Hempel's conclusion implies an infinitesimal proof of the statement by the observation of a purple cow. Intuitively, if we observe one object from a non-set, it would not generalize the set.

Professor Mackie refers to this as the Alternative Outcomes Principle

These conclusions seem ambiguous, but professor Watkins follows the idea of alternative principles and their outcome in deciding that both alternatives cannot qualify.

The discussion revolves around an excellent article by Andrew MacMillan that can be found on the Internet at http://www.paradoxes.info/

From MacMillan [END OF PREVIEW] . . . READ MORE

[. . .] There are two steps to confirming a generalization: proving the generalization's existence and confirming it. Let's take the simple statement that all As are Bs (all ravens are black). This assumes the existence of two subsets A and B. with the functional definition that if element x is in A, it is in B, that is that if x has the property of subset A, then it has the property of subset B. Mathematically, this can be formulated as (Ax

Bx). In order to prove this, for any element x, one must show that 1) There must be at least one instance of some thing that has both property A and property B.

A x (Ax & Bx) and 2) There must be no instance of anything that has property A and not property B. - x (Ax & -Bx)"

Hence, in order to construct a generalization, at least one instance I must be found so that it fulfils both property A and property B (that is, both property of subset A and B). Following condition 2, the generalization does not stand ground if we find at least one instance that does not fulfill A or B. An example is in order here. If we have observed a green car, we can state that all cars are green (it does not contradict any of the two conditions). If we see five green cars, we can built the same generalization (with an increased probability that it will be true). However, if we see five green cars and a white one, the generalization cannot be made, because it contradicts condition 2. Further more, we can sense that if we see ten cars that are green, the probability that all cars are green is greater than if we had only seen one car green.

Buy full paper

for $19.77 Following these two conditions and referring to the initial three propositions (specifically proposition number 3), we can affirm that:

1. "Any object that has both properties A and B. is a confirming instance of the generalization that everything that has property A also has property B. And 2. "Any observation that is a confirming instance of a generalization is also a confirming instance of any proposition that is logically equivalent to that generalization"

## Term Paper on *Paradox of Confirmation Paradoxes Seem* Assignment

Going back to Hempel's paradox, we can assume that in order to prove that all ravens are black, one can either observe ravens (and every black raven observed increases the probability that the statement is true) or observe non- black things (any non-black thing that is not a raven increases the probability that the statement is true). If we limit the area of selection for our observations, for example, to non-black birds, then any non-black bird observed will increase more the probability of our statement.If we now return to Nicod's criterion and apply it in this case, then the criterion would read as such:

1. Black ravens make the generalization "all ravens are black" more probable

2. Non-black ravens falsify the generalization

3. Non-black non-ravens are irrelevant

Let's take the first conclusion from the criterion and use it in a probabilistic manner. Without any observations made, the statement that "all ravens are black" has a 0 probability. If we assume that the number of ravens in the world is X, than observing 1 raven would increase the probability that the generalization is true to 1/X which is greater than 0. Hence, the more black ravens we observe, the more likely it is that the generalization is true (as the probability increases). Now let's connect Hempel's assumption to Nicod's third statement. If we assume that the number of total objects in the world is immensely larger than X, than the observation of a non-black non- raven (a purple cow), only incrementally increases the probability that our statement is true. Perhaps Nicod's criterion and Hempel's paradox are just two different ways to see one thing: Nicod's criterion states that non-black non-ravens are irrelevant, while Hempel states that such observations do increase, however little, the probability that the statement is true.

I want to summarize all the theoretical implications of the Nicod criterion, of the equivalence condition and Hempel's findings and conclusions in the matter, so as to draw a personal conclusion of the facts.

Nicod's criterion regarding the raven paradox affirms three statements. The first one, black ravens make the generalization "all ravens are black" more probable, is a statement of common sense. It is based on a common sense affirmation that we instinctively sense is true. Indeed, any black raven observed increases the probability that all ravens are black. This is because the set of ravens is to be assumed finite (indeed, there is a certain number of ravens in the world, a determined number, be it 5 or 5,000). Probabilistically speaking, if we assume there is a total number of 5,000 ravens and we have seen one that is black, it is more probable that 4,999 ravens are black than 5,000 ravens. That is why, the more black ravens we observe, the more it is probable that the remaining number of ravens is also black.

The second condition from Nicod's criterion can also be sensed true. If we do see a raven that is not black, than it can no longer be generalized that all ravens are black because, from the set of ravens named A, we have already been able to form two different subsets, B and C, with different properties, one containing white ravens, the other black ravens. Further more, any other color of ravens we may observe will form a different subset so that the statement that all ravens are black, which is equivalent to the statement that there is but one subset of set A (in fact equivalent to set A in that all the elements from this subset are also the elements of set A and vice versa), is true.

The questions that arises is in the third condition of Nicod's criteria, non-black non-ravens are irrelevant, because this comes into a contradiction with the equivalence condition that states that all non-black objects are not ravens. The latter would mean that we can observe non- black objects that are not ravens and, if such objects do exists, it would be a step further in generalizing that all ravens are black. What I have seen and understand from Hempel's solution to the problem is the fact that he denies Nicod's third condition in stating that such an observation of non- black, non- raven objects is a step, even if a very small one, towards proving the generalized statement. As I have shown above, probabilistically speaking, finding a purple cow would increase the probability that the generalization is true by 1/Y, where Y is the total number of non-black, non-raven objects.

Thus, from my point-of-view, Hempel's solution to the problem resides in the fact that he manages to fit in both Nicod's criterion and the equivalence condition by introducing probabilities. He denies 3 from Nicod's criterion, adopting the equivalence condition, and proving that the observation of a non-black, non-raven object increases the probability that the statement is true, even if infinitesimally.

Bibliography

1. Friedemann, L. Sara. Why Hempel's Paradox Isn't a Paradox (And What We Could Do If It Was). May 2003. On the Internet at http://www.ellipsis.cx/~liana/phi/920spr03.pdf

2. Carnap, Rudolf. Logical Foundations of Probability. 1950. Can be found on the Internet at http://www.lawrence.edu/fac/boardmaw/Carnap_Explica2.html

3. http://users.skynet.be/bs661306/peter/doc/hpv00r03-476.htm

4. Mackie J.L. The paradox of Confirmation. The British Journal for the Philosophy of Science, 13, 265-277. 1963.

5. Andrew MacMillan at http://www.paradoxes.info / http://home.sprynet.com/~owl1/confirma.htm

http://www.uni-konstanz.de/ppm/Research.htm

http://www.geocities.com/CapitolHill/Lobby/3022/hempel.html

http://monami.us/joshua/chid/spring2002/phil460/philessay2.php

10. Hempel, Carl G, "Studies in the Logic of Confirmation" in P. Achinstein, "The concept of Evidence"

From the internet at http://users.skynet.be/bs661306/peter/doc/hpv00r03-476.htm

Friedemann, L. Sara. Why Hempel's Paradox Isn't a Paradox (And What We Could Do If It Was). May 2003. On the Internet at http://www.ellipsis.cx/~liana/phi/920spr03.pdf

Carnap, Rudolf. Logical Foundations of Probability. 1950. Can be found on the Internet at http://www.lawrence.edu/fac/boardmaw/Carnap_Explica2.html

As shown, among others, by Professor J.L. Mackie in his The paradox of Confirmation, published in The British Journal for the Philosophy of Science, 13, 265-277, in 1963.

That is why Hempel's conclusion implies an infinitesimal proof of the statement by the observation of a purple cow. Intuitively, if we observe one object from a non-set, it would not generalize the set.

Professor Mackie refers to this as the Alternative Outcomes Principle

These conclusions seem ambiguous, but professor Watkins follows the idea of alternative principles and their outcome in deciding that both alternatives cannot qualify.

The discussion revolves around an excellent article by Andrew MacMillan that can be found on the Internet at http://www.paradoxes.info/

From MacMillan [END OF PREVIEW] . . . READ MORE

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