Re: sentential logic - curious problem


On Oct 2, 12:15 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
|The axioms and inference rule I mentioned are complete for a system
|with primitive set {~ ->}.
|
|But { | } is interdefinable with {~ ->}.
|
|So if we translate the axioms and inference rule into just { | },
then
|they should be complete for a system with primitive set { | }.
|
|Is my conclusion a non sequitur?

You overlooked a detail. Suppose you have a derivation D in
the system that's complete for {~, ->} (and although I
haven't checked, I suspect your system actually is complete
for those). You provide a translation into the language
where | is the only operator, which I think is also correct,
in the sense that D translates into a derivation D'
involving just | using only the new axioms and rules.

The conclusion you can draw is only that the new rules are
complete for formulas that have been translated from the
old language. The problem is that you have additional
formulas, P|Q for an obvious example, that aren't.

We ran into essentially the same issue once on the
phil-logic mailing list. The key additional thing you need
to ensure that | is ~(P&Q) is that it respects substitution
of equivalents. Naturally this is the point where your
attempt broke down. Herbzet's example evaluates 2<->3 to
1, but (2|3)<->(3|2) doesn't evaluate to 1.

With more axioms or rules you can get the system to respect
substitution of equivalents. Respecting equivalence and
being equivalent itself to ~(P&Q) then jointly would ensure
that the system is complete, by ensuring that each valid
formula is derivable from the formula you would get by
first translating into the language without a |, producing
a derivation there, then translating back into the language
with only |.

Keith Ramsay

.

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