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| trancaholic |
I'm looking for an example, and it is so hard that I have seriously considered whether there really is a real life example that adheres to the following constraints:
Proposition C normally implies D, unless propositions A and B are known to be the case. Semi-formally:
C -> D
and
(A & B) => not(C -> D)
Neither A, B, nor A & B implies the negation of C or D (|| means or)
not((A || B) => not(C))
not((A || B) => not(D))
And neither A nor B breaks the connection between C and D alone
not(A => not(C -> D))
not(B => not(C -> D))
The simpler case, where only one proposition (A) breaks the deduction of D from C, is quite easy to exemplify:
C -> D:
§X of the Yth amendment states that Z, therefore Z
A:
The Yth amendment has been repealed in a later amendment
Clearly A neither entails that "§X of the Yth amendment doesn't state Z" nor that one should conclude "not Z", but does break the inference step from the one to the other. However, finding such a "real" life example where two (or more) aspects are needed to break the connection is much harder. Any ideas? |
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| pkcRAISTLIN |
| ahhhh, principles of argument. the reason i never studied philosophy after first year :D its like mathematics for humanities students. yuk! good luck ;) |
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| occrider |
| So lemme get this straight, you're looking for a real life example whereby P(A & B) alone negates the conditional probability of (C | D) (I always thought "|" indicated conditional probability :conf: ) such that C and D become completely independant events? |
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| trancaholic |
| quote: | Originally posted by occrider
So lemme get this straight, you're looking for a real life example whereby P(A & B) alone negates the conditional probability of (C | D) (I always thought "|" indicated conditional probability :conf: ) such that C and D become completely independant events? |
I'll try to be more clear. I'm not thinking in probabilities, but rather in abstract argumentation terms. An argument (C->D) would be something which given the truth of C normally would allow one to conclude D (say, "I will post this serious question in the COR" would normally allow me to conclude that "the most helpful answer to my question will involve a humming bee, two ounces of sugar, and vast quantities of human feces"). However, sometimes an extra known fact (A) will make the deduction from C to D unreasonable (like, "Recently several PFRs have started posting in the COR").
I stress that A does not in itself entail the negation of C or D, but just invalidates the justified deduction of D from C. So A does not allow me to conclude that C or D is not the case - it just prevents me from accepting D as a valid truth solely on the basis of the truth of C.
The problem arises when I start looking for two - essentially different - facts, A and B, to play the role (which above was played only by A) of the exception to the otherwise reasonable argument. The best example that I've come up with so far is:
C: It says in the Bible that God is all good
D: God is all good
A: There's evil in the world
B: God is omnipotent
Taken together A and B invalidates the otherwise reasonable argument "It says in the Bible that God is all good, so God is all good".
The reason why I'm not very fond of this example, is because A and B doesn't really attack the inference step from C to D, but rather provides a case for the negation of D, viz. "God is not all good". |
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| DrUg_Tit0 |
Hmm..how about this..
Fiat makes crappy cars
Model 101 is a Fiat and therefore sucks
Zastava also makes model 101
Zastava cars rock
So in conclusion, not all models 101 suck. |
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| trancaholic |
| ^^^ Appreciate the effort, but isn't it the case that already when we're told that "Zastava also makes model 101", then the argument is broken. Whether Zastava makes good or bad cars is irrelevant for the "model 101->bad car" argument. |
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| Moongoose |
Well Zastava made cars on license from fiat (500, 128, Punto) so they are basicly the same car with different badges, but thats not the point here really.
However if i did the math correctly, it is not only hart to find an example, but under the conditions you have given us it is in fact impossible. Heres why.
Condition 1:
Neither A, B, nor A & B implies the negation of C or D (|| means or)
In order for not((A || B) => not(C)) to be true C needs to be true
In order for not((A || B) => not(D)) to be true D needs to be true
Condition 2:
And neither A nor B breaks the connection between C and D alone
In order for not(A => not(C -> D)) to be true A needs to be true
In order for not(B => not(C -> D)) to be true B needs to be true
So with this limitations all the propositions need to be true. now to the problem
(A & B) => not(C -> D)
Since A and B are both true, (A&B) is true.
Since C and D are both true, (C->D) is true.
So now we have "True -> not True" whic is false
At least i think its like that, its been a year now since ive been to a discrete structures lesson and ive yet to succesfully complete the exam so i may be completely wrong here :crazy: |
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| DrUg_Tit0 |
| quote: | Originally posted by trancaholic
^^^ Appreciate the effort, but isn't it the case that already when we're told that "Zastava also makes model 101", then the argument is broken. Whether Zastava makes good or bad cars is irrelevant for the "model 101->bad car" argument. |
Hm, but if Zastava would also make crappy cars, then all the models 101 would still suck. |
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| Moongoose |
| Well Yugos are quite crappy :D |
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| trancaholic |
| quote: | Originally posted by DrUg_Tit0
Hm, but if Zastava would also make crappy cars, then all the models 101 would still suck. |
I might have misunderstood the C->D argument then. I thought it was "something is a Fiat -> it's crap".
| quote: | Originally posted by Moongoose
However if i did the math correctly, it is not only hart to find an example, but under the conditions you have given us it is in fact impossible. Heres why. |
Your analysis is correct, except that "->" is a defeasible relation, and as such does not have a boolean value defined by its elements, and that "=>" is not meant to be the material implication, but simply a warrant for using modus ponens for concluding the consequent from the antecedent. That's why I said the "math" was only semi-formal. Sorry for the confusion. |
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| DrUg_Tit0 |
| quote: | Originally posted by trancaholic
I might have misunderstood the C->D argument then. I thought it was "something is a Fiat -> it's crap". |
Hm, actually when I look at it, I used a bit different form because my second statement is actually composed of 2 statements. So instead of having C->D, I said that (C&E)->D, but if (A&B), then the correlation is broken.
Fiat makes crappy cars (C)
Fiat makes model 101 (E)
Model 101 sucks (D)
Zastava makes model 101 (A)
Zastava cars rock (B)
So by knowing that Fiat makes models 101 and that Fiats suck, we would immediately conclude that every model 101 we see on the street sucks. But then we learned that Zastava came along and started producing the licensed copy of model 101 which was of much higher quality (well, it isn't but let's say that it is for the sake of the argument) and our first conclusion that all models 101 suck became incorrect.
I guess you could combine C and E into saying that Fiat makes crappy models 101. |
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| DrUg_Tit0 |
| quote: | Originally posted by Moongoose
Well Yugos are quite crappy :D |
No they aren't. It's just that their qualities are a bit more difficult to recognize than those of other cars :)
Say a 1.1 Yugo is faster than a 1.2 Opel Corsa..And when you crash it it's 10 times cheaper to repair :) |
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