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!