TAUTOLOGIES as DEFINITIONS

Discussion:

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Graham Cooper

2012-05-26 00:46:31 UTC

www.tinyurl.com/SetTheory2
So we can make Induction a Tautology in the form A^B->C

TAUTOLOGIES
-----------
A B C TYPE
a a->c c Modus Ponens
d->e e->f d->f Transitivity
!(!d) TRUE d Double Negation
P(0) P(n)->P(S(n)) P(n) Induction Formula

Some TAUTOLOGIES seem to DEFINE the operators (in action)

e.g.
(x+1>x)

~E(R)xeR <-> x~ex
~E(R) R={x | x ~e x}

-----------------------

Could AXIOMS be replaced with TAUTOLOGIES, any formula that always
holds, regardless of the UoD - domain of discourse

What makes an AXIOM that is NOT A TAUTOLOGY?

It must Limit the Domain Of Discourse away from cases where the
formula doesn't hold.

If GODELS PROOF was ignored for now, then FUNCTIONS could prove
theorems.

i.e.
PROOF(THEOREM) = THEOREM v PROOF(A)^PROOF(B)^(A^B->THEOREM)

GODEL'S "IMPOSSIBLE" PROOF PREDICATE!

Then, if you no longer Proved yourselves into an incomplete Logic, an
AXIOMATIC SYSTEM, i.e. formula that hold for that particular theory
would be obsolete.

Herc

m***@gmail.com

2012-05-26 11:14:20 UTC

Permalink

Post by Graham Cooper
www.tinyurl.com/SetTheory2
So we can make Induction a Tautology in the form A^B->C
TAUTOLOGIES
-----------
A B C TYPE
a a->c c Modus Ponens
d->e e->f d->f Transitivity
!(!d) TRUE d Double Negation
P(0) P(n)->P(S(n)) P(n) Induction Formula
Some TAUTOLOGIES seem to DEFINE the operators (in action)
e.g.
(x+1>x)
~E(R)xeR <-> x~ex
~E(R) R={x | x ~e x}
-----------------------
Could AXIOMS be replaced with TAUTOLOGIES, any formula that always
holds, regardless of the UoD - domain of discourse
What makes an AXIOM that is NOT A TAUTOLOGY?
It must Limit the Domain Of Discourse away from cases where the
formula doesn't hold.
If GODELS PROOF was ignored for now, then FUNCTIONS could prove
theorems.
i.e.
PROOF(THEOREM) = THEOREM v PROOF(A)^PROOF(B)^(A^B->THEOREM)
GODEL'S "IMPOSSIBLE" PROOF PREDICATE!
Then, if you no longer Proved yourselves into an incomplete Logic, an
AXIOMATIC SYSTEM, i.e. formula that hold for that particular theory
would be obsolete.
Herc

Observation:The tautology is a surphase structure of depth structure
and this is the categorico-disjunctive infference.For exemple
av-a is the tautology.a/cd is
cdv-cv-d and this is
(-cv-d)c->-d.
Another example;a/cvd
cvdv-c-d and this is
-c-dvcvd.
The mathematical logic research the conditions of categorico
disjunctive infference.

Graham Cooper

2012-05-26 22:03:29 UTC

Permalink

Post by m***@gmail.com

Does Modus Ponens have [disjunctive --> inference] foundation?

a ^ (a->c) -> c

a ^ (-avc) -> c

-(a ^ (-avc)) v c

-(a(-avc))vc

- (a^-a v a^-c) v c

-(a^-c) v c

cv-a v c

c v -a v c

c v -a **DISJUNCTIVE FORM

a->c

OK GIVEN (a->c)
a ^ (a->c) -> c
i.e. a^TRUE->c
therefore a->c

Seems to work.

But then INDUCTION is not considered a Tautology.

-----------------------

What I had in mind was using a self-consistent countable LOGIC using
PROVEN-SET-THEORY

SET AXIOM - The Closure Of Tautologies
LOGIC |- E(Y) Y = {x|f(x)} <-> PROOF( E(Y) Y={x|f(x)} )

SET AXIOM - The Examination of PROVABLE Theories
MATHEMATICS |- E(Y) Y = {x|f(x)} <-> !PROOF( !E(Y) Y={x|f(x)} )

where PROVABLE(THEOREM) <-> !PROOF(!E(THEOREM))

e.g. Goldbach's Conjecture is PROVABLE
as long as there is no counterexample.

This Gives MATHEMATICS a broader scope than TAUTOLOGY DRIVEN LOGIC.

PROOF(THEOREM) <-> THEOREM v PROOF(A)^PROOF(B)^(A^B->THEOREM)

No need to worry about GODEL'S 'impossible' Proof Predicate here!

---------------------------

Here is a simple NUMBER SET THEORY that is closed under POWERSET()

by using the LOGIC SET AXIOM

SET AXIOM - The Closure Of Tautologies
LOGIC |- E(Y) Y = {x|f(x) <-> PROOF( E(Y) Y={x|f(x) )

Russell Set: f(x) = x~ex

but PROOF( !E(RS) ) so f(x) cannot create a set by the axiom.

i.e.
!E(RS)xeRS<->x~ex is provable by tautologies.

This keeps both systems consistent.

Similarly the POWERSET(N) will not stratify, giving a subset of
mathematics (a partition between mathematics and logic) or a closed
LOGIC.

*******COUNTABLE SET THEORY*********

ALPHABET = {0,1,2,3,4,5,6,7,8,9, e, {, }, (, ), ,, =, A, E, ^, v, !,

Post by m***@gmail.com
, <, n, m, o, p ,q, f, g, h, i, j}

DICTIONARY = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...}
Note 1+1 = 2 is irrelevant at this level.

LOGIC SET SPECIFICATION
(E(n) m e n <> f(m)) <> p( (E(n) m e n <> f(m)) )

IMPLIES IS A SINGLE CHAR >
p() is PROOF()

e.g.
E(1) 1 = {1,2,3,4...}
E(2) 2 = {2,4,6}
E(3) 3 = {99,100,101,..999}
E(4) 4 = {4}
E(5) 5 = {4}

The Russell Set cannot exist in this theory
RS = {x|x~ex}
RS = {3,5, ..}

Try and find the PowerSet(N)!

Herc
--
1 X ^ NOT(X)
2 G = NOT(PRV(G))
3 S > INF
4 R = {X | NOT(X e X)}
5 IF HALT() GOTO 5
6 ALL(F) MAX(F)
=
THE 6 DEAD ENDS IN MATHEMATICS