Re: looking for a predicate hierarchy

"Dmitry A. Kazakov" <mailbox@xxxxxxxxxxxxxxxxx> wrote in
news:1hqbtgcvxtpln.kwy2q4wiyosc.dlg@xxxxxxxxxx:

On Sat, 23 Dec 2006 19:38:48 +0100 (CET), V.J. Kumar wrote:

"Dmitry A. Kazakov" <mailbox@xxxxxxxxxxxxxxxxx> wrote in
news:o68o7slu8pj7$.1cpgy64c622lt$.dlg@xxxxxxxxxx:

Formulas with your implication potentially cannot handle
contradiction, that's what. The whole point of having 'T' as a
designated truth value is to allow models for expressions like (F
/\ ^F).

You forgot that everything is in the inference rules.

Not necessarily.

Necessarily. The properties of a logical system are determined by the
inference rules.

Well, no. According to the completeness theorem validity and provability
in first-order logic, and of course in propositional logic as well,
coincide. For example whatever you can prove with truth tables in
propositional logic, you can do the same with axioms and inference rules.
It is known that the four value logic is sound and complete (would be
strange if it were otherwise !). Whatever way one chooses for a proof
depends usually on efficiency of one vs. the other.


Your argumentation seems to be based on an assumption that

(A /\ not A) => B

were somehow equivalent to

{ A, not A } |= B

but that's wrong.

It is not wrong thanks to the deduction theorem, but let's not dwell on
it. Instead, let's forget about models, trivialization and stuff.
Here's a simple proof for simple minds courtesy of Lewis and ? circa 1930
that any formula in classic logic can be proved given a contradiction:

1. A and not A
2. A -- from step 1
3. A or F -- where F an arbitrary formula
4. not A -- from step 1
5. F -- because of 3 and 4

Step 5 is known by a weird name of disjunctive syllogism, given the
knowledge that (A or F) hold *and* that A does not, surely F must hold.
The proof shows that admitting contradiction trivializes classic locgic.

In a logic that handles contradiction step 5 is invalidated by allowing A
and ^A to hold at the same time thus making step 5 incorrect -- F may or
may not hold. By introducing your '~' connective, you revive step 5, now
~A and A cannot hold at the same time as can easily be seen from the '~'
truth table. So the formula (A and ~A) trivializes your logic.


Firstly the former by no means implies the latter.

Secondly the former is not trivially true:

A (A /\ not A) => B
------------------------------
0 1
1 1
_|_ 1 if B=1 or B=_|_, otherwise T
T 1 if B=1 or B=T, otherwise _|_

The consequence relation { |- ) can be defined either
semantically (e.g. with truth tables for the connectives) or
syntactically (axioms and inference rules).

For a logic to be able to handle contradictions like (F /\ ~ F) for
example,

This is not considered as a contradiction. Contradiction out of
certain evidences {0,1} is not constructed with either V or /\.

It does not matter what you call the formula (A and ~A), what's important
is that it makes the logic trivial: any arbitrary formula is a theorem.


Yes, a
contradiction cannot be constructed from 0 and 1 using /\ (AND), or
any conventional logic operators. This was a *desired* property,
that 1 /\ 0 = 0, 1 V 0 = 1, after all. That alone does not make it
trivial, because the contradiction and uncertainty can still be
produced. For example by operations like consensus(+) and
gullibility(*):

1 + 0 = _|_ 1 * 0 = T

??? What has it got to do with the price of fish ? What make the
4-valued logic trivial is your "~" because as was said before,

Note, that ~ is not used in place of negation. This is again a jumpy
assumption.

Again, the name is immaterial, the connective usage causes
trivialization.



.

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