Logic

[The Stranger]

Can someone please help me with this problem :P

Hempel's Criteria: Jus(E, H), evidence E justifies (Jus) hypothesis H.
1. Implication-condition: if E → H , then Jus(E, H) .
2. Consistency-condition : if Jus(E, H) and Jus(E, F), then H and F are consistent.
3. Special consequence-condition : if Jus(E, H) and H → F , then Jus(E, F).
4. Reversed Consequence-condition: if Jus(E, H) and F → H , then Jus(E, F).

an explication of Jus(E, H) is formed thus: Jus(E, H) if and only if …. E ….. H …… ;
Both E and H have to appear in this sentence atleast once without being quantified.

Prove that no explication can exist that meets all the 4 Hempel conditions simultaniously.

[Husar]

I cannot help you with that logic stuff but "than" is used for comparisons (bigger than, wider than etc.), what you were looking for is "then", since it appears in almost every line of yours, I can only assume it's not a typo but a logical flaw.

What surprises me about the actual logic stuff, to me it looks like 3 and 4 only work if 2 is true, that would mean to me that 3 and 4 always apply both, because if H -> F then H and F are consistent and then F -> H, or not? (I only ever did logic in german and it's been a while )

1 is a precondition for 2, 3 and 4 and as such it should be implied that they can be met simultaneously, which leaves me to believe that there is something I'm misunderstanding, could be the term "consistent" in 2.

I'm assuming it has something to do with 3 and 4 and Hypothesis H and F cannot circle-prove/justify eachother or so.

[Tellos Athenaios]

Can someone please help me with this problem :P

Hempel's Criteria: Jus(E, H), evidence E justifies (Jus) hypothesis H.
1. Implication-condition: if E → H , then Jus(E, H) .
2. Consistency-condition : if Jus(E, H) and Jus(E, F), then H and F are consistent.
3. Special consequence-condition : if Jus(E, H) and H → F , then Jus(E, F).
4. Reversed Consequence-condition: if Jus(E, H) and F → H , then Jus(E, F).

As it stands here, for (3) and (4) to be true at the same time it must follow that H and F are logically equivalent. (H -> F && F -> H). Except that from F ->H we may not derive H -> F, and from H -> F we may not derive F ->H.

I.e.:

E
Jus(E,H)
F -> H
--------
E -> H
E
F -> H
--------
H
F -> H
--------

How do you get from there to “H -> F” ?

[The Stranger]

ye, i just did this with my dad and we came to the same conclusion, but you explain it better than he did :P so thanx!

so the contradiction would be that H → F and F → H can't be true at the same time while the conditions demand them to be?

[Tellos Athenaios]

ye, i just did this with my dad and we came to the same conclusion, but you explain it better than he did :P so thanx!

so the contradiction would be that H → F and F → H can't be true at the same time while the conditions demand them to be?

No the problem is that rule (3) and rule (4) require logical equivalence of F and H and it is not possible to prove this in the general case (i.e. using only the conditions/propositions given for any F, E, or H).

But it is perfectly possible there exist two propositions F, and H such that F -> H && H -> F:

Consider many of the basic examples of boolean algebra:
F: !A || (B && A)
H: !A || B
H <-> F

[The Stranger]

hmm but then that would mean that there is an explication that meets all 4 hempel criteria, right? or does explication only apply to general cases?

[Tellos Athenaios]

hmm but then that would mean that there is an explication that meets all 4 hempel criteria, right? or does explication only apply to general cases?

But what do you actually know about F or H? Nothing. Of course if your assignment is “refute that there exists any explication for any E, F, H such that it yields these 4 conditions as truths” then we're barking up the wrong tree. But so long as they are left completely arbitrary you can observe that the explication (therefore) has to be able to work on arbitrary (universal qualifiers) E, F and H; and that the task, then, is to show that there is no explication which can do that.

This is made easy when you realise that F<-> H must be logically equivalent for the explication to work at all. Since this equivalence would be a property of F, and H, it cannot be assumed in the general case. So all you need to do is to show how this reliance breaks down when you produce a pair of F, H such that F -> H but not H -> F and call it job done.

It is however instructive to remind yourself that while in the general case these 4 cannot be true at the same time, that does not mean that no two propositions F and H can ever be logically equivalent!

[tibilicus]

Maths is completely alien to me. I understand foreign languages better than maths.

[The Stranger]

same... thats why im struggling, even tho tellos prolly explained it as clearly as possible im still finding it difficult to understand.

[The Stranger]

Sigh... its only getting more complicated, i dont even know why i should learn this -_-

if anyone knows anything about Bayesian probability please HELP!

prove the following axiomas, H and E are not contradictions (value of 0) or logical constructions such as the bachelor is unmarried (value of 1) (Pr means probablity)

a. Pr(H ∧ E) ≤ Pr(H) , and then ofcourse also Pr(H ∧ E) ≤ Pr(E)

b. If H → ¬E, then Pr(H | E) = 0

c. Pr(H | E) + Pr(¬H | E) = 1

[Kival]

Did you have some degree of set theory? Do you now that Pr(H ∧ E)=Pr(H ∩ E)? a and b are pretty simple then, c might be easy to, I did not check.

If not I need to know your definitions, because I've only made probability theory based on set theory.

[The Stranger]

no i dont have a degree tho i did know Pr(H ∧ E)=Pr(H ∩ E). i do understand the implication of all these things but i dont know how to write down the proof.

i guess the definitions are the same.

[Kival]

Oh, sorry, I did not want to ask if you have a degree, only if you know enough set theory. Okay, ignore my mistakes in english, I hope you can understand the ideas:

Okay let's work with set theory then. a is actually quite trivial but I'll try to give you an idea how to prove it:


Pr(E | H) = Pr(H ∩ E)/Pr(H) (Definition of baysian probability)
=> Pr(H ∩ E) = Pr(E | H)*Pr(H) (just simple transformation)

As Pr(E | H) has a value range in the interval [0;1] (it's a probability function), obviously it follows:

Pr(H ∩ E) <= Pr(E | H)*Pr(H)

Because of Pr(H | E)=Pr(H ∩ E)/Pr(E), the second part of a) follows trivially.

for b) you just need the equations and knew two things:

H → ¬E -> (H ∩ E) = {} and Pr ( {} )=0 ({} being the empty set)

c) Obviously (H ∩ E) and (H^(c) ∩ E) are disjoint sets, so you can use Sigma additivity (H^(c) being the complement of H) . Do you know where to go from there?

[The Stranger]

i get B and C.

I still dont get A however, why is the probability of (H&E) smaller or the same as the probability of (H)?

[Kival]

Pr(E | H) = Pr(H ∩ E)/Pr(H) (Definition of baysian probability)
=> Pr(H ∩ E) = Pr(E | H)*Pr(H) (just simple transformation)

As Pr(E | H) has a value range in the interval [0;1] (it's a probability function), obviously it follows:

Pr(H ∩ E) <= Pr(E | H)*Pr(H)

Because of Pr(H | E)=Pr(H ∩ E)/Pr(E), the second part of a) follows trivially.

From the above you can agree on:

Pr(H ∩ E) = Pr(E | H)*Pr(H) ?

Now just remember, what you know about the Pr-function. It's a function which gives you the probabilty of an event. The event is described as a set or as a logical combination of events. What do you know in general about the probability function? It has only values between 0 and 1. So 0<=Pr(A)<=1 for all possible events A. In mathematical detail, Pr(. | B), the baysian probability function is an own function but it's still a probability function and so it still has only values from 0 to 1. If you multiply Pr(H) with a value between 0 and 1, it can only be Pr(H) or smaller.

[The Stranger]

ah... if u put it like that it seems simple XD

[Reenk Roink]

First hard homework in your philosophy major career TS?

[The Stranger]

omg yes... first 2 years i was cruising, now im in some rough waters, the courses are really challenging and this tops it till now :P. I wish i had more time, but it seems like i am trying to do too much stuff :P

tho really thank you Kival, you actually made me understand it haha, appreciated.

[Kival]

tho really thank you Kivak, you actually made me understand it haha, appreciated.

It's Kival, but you're welcome. ;-)

[The Stranger]

typo :P