Logic

[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