Second Doubt

[marcos]

Hello William Elliot, and everybody else. I am reading "Introduction
to mathematical logic", by Elliot Mendelson, and this is my doubt:
First of all I must explain: In any proof, one wf depends upon other
if the first one needs the presence of the second to achieve the
proof.
Then it comes deduction theorem: " Assume that, in some deduction
showing that there is a deduction of C from a set of wfs T and the wf
B, no application of Generalization to a wf that depends upon B has as
its quantified variable a free variable of B. Then we can proof B->C
from the set of wfs T."
I've got no doubt yet. But then the book states that when we wish to
apply the deduction theorem several times in a row to a given
deduction (for example, to obtain T |-- D->(B->C) from
T, D, B |-- C), the following additional conclusion can be drawn from
the deduction theorem:
" The new proof of T |-- B->C involves an application of
Generalization to a wf depending upon a wf E of T only if there is an
application of Generalization in the given proof of T, B |-- C that
involves the same quantified variable and is applied to a wf that
depends upon E."
This last paragraph contains my doubt: How does it help me applying
the deduction theorem several times this conclusion?.
Thanks.

[William Elliot]

On Thu, 19 Nov 2009, marcos wrote:

> Then it comes deduction theorem: " Assume that, in some deduction
> showing that there is a deduction of C from a set of wfs T and the wf
> B, no application of Generalization to a wf that depends upon B has as
> its quantified variable a free variable of B. Then we can proof B->C
> from the set of wfs T."

The deduction theorem is:
. . T, B |- C iff T |- B->C

Rule G is: if for all wwf F in T, a does not appear free in F, then
. . T |- C, implies T |- (a)C
Alternatively, let (a)T = { (a)F | F in T }. Then
. . T |- C implies (a)T |- (a)C

Elliot has a strange mix of the two in his deduction theorem.

> I've got no doubt yet. But then the book states that when we wish to
> apply the deduction theorem several times in a row to a given
> deduction (for example, to obtain T |-- D->(B->C) from
> T, D, B |-- C), the following additional conclusion can be drawn from
> the deduction theorem:

> " The new proof of T |-- B->C involves an application of
> Generalization to a wf depending upon a wf E of T only if there is an
> application of Generalization in the given proof of T, B |-- C that
> involves the same quantified variable and is applied to a wf that
> depends upon E."

> This last paragraph contains my doubt: How does it help me applying
> the deduction theorem several times this conclusion?.

Is he making it much more complicated than need be?
T, A, B |- C iff T |- A -> (B -> C) is obvious.

As for the quantification, I suppose it could be like
when a is free in A and not F in T with free a in F,
. . T, A |- C implies T |- A->C implies T |- (a)(A -> C)

Dang if can make out what he's thinking.
I suggest you keep the deduction theorem and rule G
separate in your thinking and mix them only as needed.

If you want to clarify the setting he's using, I'll consider a second
look. Have you considered a different text for a second opinion? When
was Elliot's book published?

[marcos]

Elliot Mendelson's book was first published in 1964, and I am reading
fourth edition (1997). I have two books: "Introduction to mathematical
logic", by Elliot Mendelson, and "Logic for mathematicians", by A.G.
Hamilton. Do you suggest me any other?.

[William Elliot]

I learned from "Logic for Mathematicians" by Rosseur, 1960's.
Quine's "Mathematical Logic" is considered a classic. Do the
books you have also cover set theory?

[marcos]

Yes, they do. I am interested in logic because I am interested in set
theory.
I've solved the question I had this last time. The deduction theorem
mentioned in my book is a strange mix of the deduction theorem and
rule G, as you said, just because in the deduction there might be free
variables; and the doubt I had was stupid somehow. It is obvious once
readen two times.