Re: reductio ad falsum versus reductio ad absurdum


William Elliot wrote:
> On Sun, 28 Aug 2005, futurist wrote:
> To whom are you repling? Removal of immediate references can lead to your
> reply not being noticed by the person to whom you are repling. (I read by
> new arrivals, not by threaded format, as the former is easier for seeing
> what's up.)

sorry about that, didn't mean to delete any name headers...


> Becuase -> combines two formal statements into another.
> P |- Q is a statement about a sequence of formal statements.
> (P |- Q) |- R, is a statment about a sequence of formal statements
> because P |- Q isn't a formal statement.

replace the last is with 'isn't' and i think i understand..



> >
> > Modus ponens
> > P, P->Q |- Q
> >
> > Conditional Proof
> > ???? |- A->Q
> >
> > i don't see a way to fill in the ???? without nesting |-. to be
> > clearer, i have no thesis, i have two questions:
> >
> > 1. why is what i put for the ???? wrong, and what SHOULD go there?
>
> A set of formal statements only.

ok, but which set of formal statements to express the rule of CP?


>
> > 2. (further below) why are two very different inferences both called
> > RAA?
>
> Then ask it further below.
>
> > > What you described can formally be expressed as
> > > Given P,A |- Q, conclude P |- A -> Q
> > > which is called the Deduction Theorem, tho actually it's a metatheorem.
> >
> > in which case it can't be a description of a rule of ND...
> >
> ND has somewhat different syntax than PC, propositional calculus.

i just want a way to express, precisely and formally, the ND rules of
CP, and RAA, and possibly a way to distinguish various forms of RAA.
although at this point i can't see how to do that since |- doesn't
nest...

>
> > when i write:
> >
> > P
> > ----
> > A (assumption)
> > ...
> > Q
> > A->Q (conditional proof)
> >
> > i'm not using any metatheorems--i'm just using some rule (CP) of the
> > object language--but i don't know how to precisely formalize that rule
> > as a sequent... (see above)
> >
> You are mixing ND with PC. Here's ND
>
> assume P
> assume A
> ...
> Q
> A -> Q
> P -> (A -> Q)

i don't know about that--i was taught ND in two separate classes where
we had lists of premises, AND we could introduce assumptions for CP,
which is what i've done here. at any rate, what you gave me was an
example of a proof using a rule, not a formalization of that rule...

>
> > 1. (what i'm inclined to call reductio ad falsum)
> >
> > P
> > ---
> > ...
> > Q
> > A (assumption)
> > ...
> > ~Q
> > ~A
> >
>
> assume P
> ...
> Q
> assume A
> ...
> ~Q
> Q -> F (ND definition ~Q)
> Q
> F
> A -> F
> ~A
> P -> ~A
>
> > (which yields P |- ~A)
> >
> It does not. You assumed P |- Q.


???? P was my premise. A was my assumption A was discharged when i
concluded ~A. thus using only the truth of P i concluded ~A--in other
words i proved that P |- ~A.

P is not atomic here--i'm assuming it has an inner structure that lets
me conclude Q. (i'm not writing out a specific proof, i'm showing a
schematic application of the rule (which is still not really a
*formalization* of the rule))

>
> > 2. (what i'm inclined to call reductio ad absurdum)
> >
> > ---
> > A (assumption)
> > ...
> > Q
> > ...
> > ~Q
> > ~A
> >
> > (which yields |- ~A)
> >
> It does not. You assumed A |- Q.

again, A is not atomic, this is the sort of outline they give in many
books of these sorts of proof.. i.e. imagine that A was "Q&~Q" for the
simplest case. then this would definitely yield a proof of |- ~A .
at any rate these are just examples of these rules and not precise
formulations of them.

>
> > how do i describe the rules applied here in sequents, as all the other
> > rules are described?
> >


> You are hopelessly confusing ND with PC.

my understanding is that PC uses several axioms and one rule (MP),
wheras ND uses many rules and no axioms. i'm not thinking of PC at
all, although i understand that they both begin with a list of
premises.

> Be content with what I think you've stated
> P -> Q, Q -> (A -> ~Q) |- ~A
> P -> Q, Q -> ~Q |- ~P
>
> As for a distinction
> P -> ~P |- ~P

this one doesn't seem sufficiently general-- couldn't you have P->Q&~Q
withought P->~P? or at the very least, you could get the former in
fewer steps... so we can say:

P -> Q&~Q |- ~P


> P -> Q, P -> ~Q |- ~P

yes i know this one as 'impossible antecent'

so, the formulations immediately above are examples of rules that could
work *in conjunction* with CP to yield the proofs in my example--this
places the burden of describing the rule on CP--how do we precisely
describe the rule of Conditional Proof?

.

Relevant Pages

  • reductio ad falsum versus reductio ad absurdum
    ... It is a statement of the metalanguage ... >> A set of formal statements only. ... same theorem or rule schemata. ... > describe the rule of Conditional Proof? ...
    (sci.logic)
  • Re: reductio ad falsum versus reductio ad absurdum
    ... reply not being noticed by the person to whom you are repling. ... Becuase -> combines two formal statements into another. ... P |- Q is a statement about a sequence of formal statements. ... (what i'm inclined to call reductio ad absurdum) ...
    (sci.logic)