Re: If (P & ~P) -> Q is not derivable then Goedel's formula is not derivable


lugita15@xxxxxxxxx wrote:
Newberry wrote:
lugita15@xxxxxxxxx wrote:
Newberry wrote:
lugita15@xxxxxxxxx wrote:
Newberry wrote:
R. Srinivasan wrote:
On Nov 21, 8:27 pm, "Newberry" <newbe...@xxxxxxxxxx> wrote:
R. Srinivasan wrote:
On Nov 18, 10:00 pm, "Newberry" <newbe...@xxxxxxxxxx> wrote:
[...]

More info on Occurrence-relevance Logic:
M. Richard Diaz, Topics in the Logic of Relevance, Philosophia Verlag,
1981

I also suggest my paper:http://xnewberry.tripod.com/IBL_2006_11_11.html
It treats the topic from a sligthly diferent persepctive.

Let us start with your formula (1) in the above link, in which it is
asserted that:

~(P & ~P & Q) is not derivable for arbitrary P and Q. (1)

Since P and Q are arbitrary in (1), they can be interchanged. Hence

~(Q & ~Q & P) is not derivabe for aribitrary P and Q (2)

From (1) and (2), it follows that

(~(P & ~P & Q) & ~(Q & ~Q & P)) is not derivable (3)

~((P & ~P & Q) v (Q & ~Q & P)) is not derivable.

~(P & Q & (~P v ~Q)) is not derivable

~((P & Q) & ~(P & Q)) is not derivable (4).

Since P and Q are arbitrary, we may replace (P&Q) by P in (4), to
obtain:

~(P & ~P) is not derivable.
(5)

Unless you can find some objection to my argument, I conclude that your
formula (1) entails (5). In particular, do you object to (3) above? Or
do you object to going to (5) from (4)? Note that one can also start
with (5) and work backwards to conclude (1).

Regards, RS

I object to (4) to (5). You certainly cannot go backwards because
o-relevance is not preserved under uniform substitution.


So you accept my derivation up to (4), which means that you accept
that:

~((P & Q) & ~(P & Q)) is not derivable (4)

I can make the passage to (5) less restrictive that I had stated above.
All we need for ~(P&~P) to be not derivable is for at least one
instance of it to be not derivable. Take "P" to be "The sun exists" and
"Q" to be "The sun rises in the east". Since you accept (4), you accept
that the following proposition is underivable:

It is not the case that ((The sun exists and the sun rises in the east)
and it is not the case that (the sun exists and the sun rises in the
east)) (5a)

But I could take "P" as "The sun exists and the sun rises in the east"
in (5a) and conclude that this instance of ~(P&~P) is not derivable in
o-relvance logic. From which the general conclusion (5) follows:

~(P & ~P) is not derivable. (5)

It seems to me that you would have to make further restrictions in your
logic such as, making the following illegal:

(A&B) -> A

and

(A&B) -> B

These certainly are illegal in o-relevant logic.

That makes o-relevant logic extremely counterintuitive. Is it not
obviously and manifestly true that if A and B is true, then both A must
be true and B must be true? Is that not obvious by the very meaning of
the word "and"? A and B is true when and only when both A and B are
true!

It is not manifestly true that (A & B) implies A

(A&B) -> A

as it is by no means obvious what B has to do with implying A. In fact
B is irrelevant for deriving A and therefore the formula above is not
valid.
Why does it matter whether B is relevant or not? I don't think it
does. Part of logical reasoning is sifting relevant information from
irrelevant information.
For example, let us say that we have the following information about
the murder of President Lincoln:
1. Lincoln was shot by Booth
2. Lincoln wore was shot in the 1800's
Let us say that a detective knew that both statement 1 and statement 2
are true. Then couldn't he deduce from this information that Lincoln
was shot by Booth?

If the detective knew that 1 is true then he does not need to deduce
it.

This is expressed by (1&2)->1.

In o-relevant logic it expresses that both 1 and 2 are required to
conclude 1, which obviously is not true.

Depending on what you mean by "required," the propositional calculus
can translate the word in different ways:

If in order to conclude B you require A, then this is expressed as
(B->A). If this is the interpretation you are looking for, then you
must accept the "implication" 1->(1&2). This means that the truth of
1 is required to conclude the truth of 1 and 2.

If it is both required that A be true and sufficient that A be true in
order to conclude B, then this is expressed by A<->B. This seems to
better match the idea of "required to conclude," which you are denoting
with the implication sign.

As an aside, I would be interested to see how o-relevant logic fares
when it is expanded into first-order logic.

if |- (x)Ax or |- (x)Bx then |/- (x)(Ax v Bx)

if |- ~(Ex)Ax or |- ~(Ex)Bx then |/- ~(Ex)(Ax & Bx)

if |- ~(Ex)Pxg then |/- ~(Ex)(Ey)(Pxy & Qy)

where Pxy means x is the proof of y, Q has been constructed such that
only one y satisfies it - namely g, the Goedel number of
~(Ex)(Ey)(Pxy & Qy).


Where is the flaw in my argument?




in order to invalidate my derivation. Or you may have to invalidate
propostions like "The sun exists and the sun rises in the east", which
is to say, that you may have to place restrictions on taking
conjunctions. The bottom line is that you can't get away with keeping
all the classical rules for implication, negation, conjunction,
disjunction, constructing wffs, etc. and invalidate a fundamental
classical derivation. Any restrictions that you place on these should
not seem to be ad-hoc (i.e., arbitrary) in order for you to convince
people to accept o-relevance logic.

There is nothing ad hoc here. A variable is irrelevant when a formula
evaluated according to these tables

v | F x T & | F x T ->| F x T
---------- ---------- ----------
F | F x T F | F F F F | T T T
x | x x T x | F x x x | x x T
T | T T T T | F x T T | F x T

is a tautology even though said varible was assigned only x (i.e. not T
or F.)


Examples:

P | Q | P v (~P v Q)
---+---+--------------
T | x | ..T..F..x
F | x | ..T..T..T

We observe that the formula evaluates as a tautology (the column under
the main connective has only Ts) without assigning T or F to Q.

P | Q | P v (~P v Q)
---+---+--------------
x | T | ..T..x..T
x | F | ..x..x..x

There is one x in the column under the main connective. We cannot
determine if the formula is a tautology without assigning the truth
values to P.

.

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