Talk:Generalization (logic)
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It is not explained why the deduction theorem does not apply in this case. ---- NoizHed (talk) 18:13, 16 November 2007 (UTC)
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(I am not a logician)
Let 1) If |-P(x) then |- AllxP(x)
2) P(x)|-AllxP(x)
3)|-[P(x) --> AllxP(x)] False
I think (1) does not give (2).
I) Enderton has (1) as a Metatheorem [p117, use Gamma as empty] and has Unrestricted Deduction Metatheorem [p118] If (1) gives (2) then these combine to give the false (3).
II) Mendelson and Detlovs/Podniak have (2) [in the form of a Rule of Inference]. However, they have a Restricted Deduction Theorem which does not apply to (2). Please correct me if I am misreading something.
(It is clear that (2) gives (1). Starting with a deduction of P(x) one attaches the deduction (2) to get a deduction of AllxP(x).)
128.241.40.80 (talk) 22:35, 22 November 2007 (UTC)A. Reader
Supems(17:03, 16 April 2008 (UTC)
I am not a expert on this area but I need knowledge in this area to pursue my Mathematical studies.
I feel that it is not a good idea to consider the generalization as a |-P(x) -> |-for all x P(x) and talk about the using of deduction theorem because when we prove P(x) for arbitrary x we get (for all x P(x)) which is the generalization so there is no point of talking about using deduction theorem. So it is better to use the gen abbrv and use it as a mere rule of inference. And if there is a restriction of using deduction theorem the restriction is not quite clear in the article including the example proof it is bit confusing when reading the proof and the restriction said above in the article. so it will be better if someone who knows about this clarify this to me. And please let me know if there are any restrictions for deduction thm (not only in this case)
Thanks Supems (talk) 17:09, 16 April 2008 (UTC)
