Discussion:
Simple refutation of the 1931 Incompleteness Theorem
Pete Olcott
2018-02-12 16:42:38 UTC
Raw Message
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<br>
An {a priori fact} is an expression X
of (natural or formal) language L that<br>
has been assigned the semantic property
of True.<br>
<br>
A set T of {a priori facts} of language
L forms the ultimate foundation of the<br>
notion of Truth in language L.<br>
<br>
To verify that an expression X of
language L is True or False only requires<br>
a syntactic logical consequence
inference chain (formal proof) from X or ~X<br>
to one or more elements of T.<br>
<br>
Since an expression X of language L is
not a statement of language L unless<br>
and until it is proven to be True or
False in L, every statement of language<br>
L can be proved or disproved in L.<br>
<br>
<b><span style="background: #ffff00">∀L
∈ Formal_Systems</span></b><b><br>
</b><b>
</b><b><span style="background: #ffff00">∀X
∈ L</span></b><b><br>
</b><b>
</b><b><span style="background: #ffff00">Statement(X)
→ ( Provable(L, X) ∨ Refutable(L, X) )</span></b><b><br>
</b><b>
</b><br>
<b>Stanford Encyclopedia of Philosophy
Gödel’s Incompleteness Theorems</b><b><br>
</b><b>
</b>The first incompleteness theorem states
that in any consistent formal system F<br>
within which a certain amount of
arithmetic can be carried out, <b><span style="background: #ffff00">there
are</span></b><b><br>
</b><b>
</b><b><span style="background: #ffff00">statements
of the language of F which can neither be proved nor disproved
in F.</span></b><br>
<br>
<b>Copyright 2017, 2018 Pete Olcott </b><br>
<br>
<span style="background: #ffff00"></span>
<div class="moz-signature">-- <br>
<meta charset="UTF-8">
<font face="Segoe UI Symbol, sans-serif"><font size="2"><b>
∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) </b></font>
</font></div>
</body>
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Pete Olcott
2018-02-12 18:02:20 UTC
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<div class="moz-cite-prefix">On 2/12/2018 10:42 AM, Pete Olcott
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:UIGdnZNYXKRiXhzHnZ2dnUU7-***@giganews.com">
<meta http-equiv="content-type" content="text/html; charset=utf-8">
<br>
An {a priori fact} is an expression X of (natural or formal)
language L that<br>
has been assigned the semantic property of True.<br>
<br>
A set T of {a priori facts} of language L forms the ultimate
foundation of the<br>
notion of Truth in language L.<br>
<br>
To verify that an expression X of language L is True or False only
requires<br>
a syntactic logical consequence inference chain (formal proof)
from X or ~X<br>
to one or more elements of T.<br>
<br>
Since an expression X of language L is not a statement of language
L unless<br>
and until it is proven to be True or False in L, every statement
of language<br>
L can be proved or disproved in L.<br>
<br>
<b><span style="background: #ffff00">∀L ∈ Formal_Systems</span></b><b><br>
</b><b> </b><b><span style="background: #ffff00">∀X ∈ L</span></b><b><br>
</b><b> </b><b><span style="background: #ffff00">Statement(X) → (
Provable(L, X) ∨ Refutable(L, X) )</span></b><b><br>
</b><b> </b><br>
</blockquote>
<b style="font-weight: bold !important; color: rgb(26, 26, 26);
font-family: Merriweather, Georgia, serif; font-size: 16px;
font-style: normal; font-variant-ligatures: normal;
font-variant-caps: normal; letter-spacing: normal; orphans: 2;
text-align: start; text-indent: 0px; text-transform: none;
white-space: normal; widows: 2; word-spacing: 0px;
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255); text-decoration-style: initial; text-decoration-color:
initial;"><span data-mce-style="background-color: #ffff00;"
style="background-color: rgb(255, 255, 0);"><br>
</span></b><font size="+3"><b><span style="background: #ffff00">Typo
corrected: <br>
</span></b></font><b><span style="background: #ffff00">Statement(L,
X) → ( Provable(L, X) ∨ Refutable(L, X) )</span></b><b><br>
</b><b> </b><br>
<b style="font-weight: bold !important; color: rgb(26, 26, 26);
font-family: Merriweather, Georgia, serif; font-size: 16px;
font-style: normal; font-variant-ligatures: normal;
font-variant-caps: normal; letter-spacing: normal; orphans: 2;
text-align: start; text-indent: 0px; text-transform: none;
white-space: normal; widows: 2; word-spacing: 0px;
-webkit-text-stroke-width: 0px; background-color: rgb(255, 255,
255); text-decoration-style: initial; text-decoration-color:
initial;"><span data-mce-style="background-color: #ffff00;"
style="background-color: rgb(255, 255, 0);"></span></b>
<blockquote type="cite"
cite="mid:UIGdnZNYXKRiXhzHnZ2dnUU7-***@giganews.com"> <b>Stanford
Encyclopedia of Philosophy Gödel’s Incompleteness Theorems</b><b><br>
</b><b> </b>The first incompleteness theorem states that in any
consistent formal system F<br>
within which a certain amount of arithmetic can be carried out, <b><span
style="background: #ffff00">there are</span></b><b><br>
</b><b> </b><b><span style="background: #ffff00">statements of
the language of F which can neither be proved nor disproved in
F.</span></b><br>
<br>
<b>Copyright 2017, 2018 Pete Olcott </b><br>
<br>
<span style="background: #ffff00"></span>
<div class="moz-signature">-- <br>
<meta charset="UTF-8">
<font face="Segoe UI Symbol, sans-serif"><font size="2"><b> ∀X
True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) </b></font> </font></div>
</blockquote>
<p><br>
</p>
<div class="moz-signature">-- <br>
<meta charset="UTF-8">
<font face="Segoe UI Symbol, sans-serif"><font size="2"><b>
∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) </b></font>
</font></div>
</body>
</html>
Richard Tobin
2018-02-12 18:55:31 UTC
Raw Message
Since an expression X of language L is not a statement of language L unless
and until it is proven to be True or False in L, every statement of language
L can be proved or disproved in L.
So your language varies over time?

-- Richard
Pete Olcott
2018-02-13 02:52:28 UTC
Raw Message
Post by Richard Tobin
Since an expression X of language L is not a statement of language L unless
and until it is proven to be True or False in L, every statement of language
L can be proved or disproved in L.
So your language varies over time?
-- Richard
I am not saying that. I am saying that expressions of language
do not count as statements of language until after they have
been proven True or False.

A statement G of language F which can neither
be proved nor disproved in F, is impossible.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Jim Burns
2018-02-13 18:18:56 UTC
Raw Message
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])

This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.

You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).

Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.

I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.

[1]
Define
Q(x) :<-> Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )

[Q(x)] := [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]

Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)

[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]

By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]

Define G_T :<-> Q[Q(x)]

G_T <-> ~Prov_T[G_T]

Corollary:
If theory T is complete, then theory T is inconsistent.
Pete Olcott
2018-02-14 01:04:07 UTC
Raw Message
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.

When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.

There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.

This would be analogous to a guy correctly proving that he never existed.

Post by Jim Burns
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
I don't even need self-reference to refute ~Provable(L, X)
because the evaluation of ~Provable(L, X) inherently requires
a proof or it cannot even be evaluated.
Post by Jim Burns
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
I merely augmented the insufficient specification of the
term of the art: axiom, and provided that actual definition
of [a priori] Truth, this correctly refuting Tarski's claim
that a language can not specify its own Truth predicate.
Post by Jim Burns
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<->  Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] :=  [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T  :<->  Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
Then Truth itself is broken, yet Truth cannot possibly be broken.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Newberry
2018-02-14 02:15:01 UTC
Raw Message
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of inferences
that either prove or disprove G. Why do you keep saying you refuted it?
Post by Pete Olcott
This would be analogous to a guy correctly proving that he never existed.
Post by Jim Burns
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
I don't even need self-reference to refute ~Provable(L, X)
because the evaluation of ~Provable(L, X) inherently requires
a proof or it cannot even be evaluated.
Post by Jim Burns
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
I merely augmented the insufficient specification of the
term of the art: axiom, and provided that actual definition
of [a priori] Truth, this correctly refuting Tarski's claim
that a language can not specify its own Truth predicate.
Post by Jim Burns
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<-> Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] := [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T :<-> Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
Then Truth itself is broken, yet Truth cannot possibly be broken.
Pete Olcott
2018-02-14 04:15:46 UTC
Raw Message
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of inferences that either prove or disprove G. Why do you keep saying you refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Newberry
2018-02-14 05:26:20 UTC
Raw Message
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of
inferences that either prove or disprove G. Why do you keep saying you
refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Gödel's first theorem does not make any such claim. It says that there
are unprovable and undisprovable closed WFFs in the language of PA.
Pete Olcott
2018-02-14 05:57:16 UTC
Raw Message
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of
inferences that either prove or disprove G. Why do you keep saying you
refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Gödel's first theorem does not make any such claim. It says that there are unprovable and undisprovable closed WFFs in the language of PA.
Yet by definition all closed WFF must evaluate to True or False
and as I have shown they only actually evaluate to True or False
on the basis of a formal proof from this WFF to expressions of
language that have been defined to have the semantic property of True.

That is simply the way that Truth works and he got it wrong.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Newberry
2018-02-14 07:53:16 UTC
Raw Message
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of
inferences that either prove or disprove G. Why do you keep saying you
refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Gödel's first theorem does not make any such claim. It says that there
are unprovable and undisprovable closed WFFs in the language of PA.
Yet by definition all closed WFF must evaluate to True or False
Well, Gödel proved that there are undecidable closed WFFs in the
language of PA. You are saying that it would imply that such WFFs are
neither true nor false, and that is impossible.

So you are making two claims
1) true = provable
2) Closed WFF that is neither true nor false cannot possibly exist

So there must be an error in Gödel's proof somewhere, but you do not
know where.
Post by Pete Olcott
and as I have shown they only actually evaluate to True or False
on the basis of a formal proof from this WFF to expressions of
language that have been defined to have the semantic property of True.
That is simply the way that Truth works and he got it wrong.
Pete Olcott
2018-02-14 15:08:31 UTC
Raw Message
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of
inferences that either prove or disprove G. Why do you keep saying you
refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Gödel's first theorem does not make any such claim. It says that there
are unprovable and undisprovable closed WFFs in the language of PA.
Yet by definition all closed WFF must evaluate to True or False
Well, Gödel proved that there are undecidable closed WFFs in the language of PA. You are saying that it would imply that such WFFs are neither true nor false, and that is impossible.
So you are making two claims
1) true = provable
2) Closed WFF that is neither true nor false cannot possibly exist
So there must be an error in Gödel's proof somewhere, but you do not know where.
True is a certain kind of provable.
Closed WFF that is neither true nor false is wrong.
Post by Pete Olcott
and as I have shown they only actually evaluate to True or False
on the basis of a formal proof from this WFF to expressions of
language that have been defined to have the semantic property of True.
That is simply the way that Truth works and he got it wrong.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Peter Percival
2018-02-14 15:18:54 UTC
Raw Message
Post by Pete Olcott
True is a certain kind of provable.
Provable in what sense? Provable in some formal calculus?
Post by Pete Olcott
Closed WFF that is neither true nor false is wrong.
True and false in what senses? Do you mean "true in a model", "true in
all models", or what?
Pete Olcott
2018-02-14 15:53:15 UTC
Raw Message
Post by Pete Olcott
True is a certain kind of provable.
Provable in what sense?  Provable in some formal calculus?
Post by Pete Olcott
Closed WFF that is neither true nor false is wrong.
True and false in what senses?  Do you mean "true in a model", "true in all models", or what?
True or False in the sense that a formal proof exists from the
subset T of expressions of language L that have been defined
to have the semantic property of True to the expression X or ~X.

All expressions X of language L that are statements of L:
Statement(L, X) are Provable(T, X) or Provable(T, ~X).
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Newberry
2018-02-14 16:54:51 UTC
Raw Message
Post by Pete Olcott
Post by Peter Percival
Post by Pete Olcott
True is a certain kind of provable.
Provable in what sense? Provable in some formal calculus?
Post by Pete Olcott
Closed WFF that is neither true nor false is wrong.
True and false in what senses? Do you mean "true in a model", "true
in all models", or what?
True or False in the sense that a formal proof exists from the
subset T of expressions of language L that have been defined
to have the semantic property of True to the expression X or ~X.
Not generally accepted, but we do not need to go into that right now.
Post by Pete Olcott
Statement(L, X) are Provable(T, X) or Provable(T, ~X).
How do you know that? Are you suggesting that only provable or
disprovable formulae count as statements? That's fine, you can do that.
But Gödel did not use the word "statement" in this sense [if he used the
word at all.] He referred to a string of symbols with certain syntactic
properties.
Pete Olcott
2018-02-14 17:24:55 UTC
Raw Message
Post by Newberry
Post by Pete Olcott
Post by Pete Olcott
True is a certain kind of provable.
Provable in what sense?  Provable in some formal calculus?
Post by Pete Olcott
Closed WFF that is neither true nor false is wrong.
True and false in what senses?  Do you mean "true in a model", "true
in all models", or what?
True or False in the sense that a formal proof exists from the
subset T of expressions of language L that have been defined
to have the semantic property of True to the expression X or ~X.
Not generally accepted, but we do not need to go into that right now.
Until it is generally accepted everyone in the world remains incorrect.
Post by Newberry
Post by Pete Olcott
Statement(L, X) are Provable(T, X) or Provable(T, ~X).
How do you know that?
I just proved that in the part you chose to ignore.
Post by Newberry
Are you suggesting that only provable or disprovable formulae count as statements?
No more than that, unless a WFF is provably True or Provable False
it does not count as a statement.

That's fine, you can do that. But Gödel did not use the word "statement" in this sense [if he used the word at all.] He referred to a string of symbols with certain syntactic properties.
If the Stanford University summation of what he said is a correct paraphrase
Statement(F, G) ∧ ~Provable(F, G) ∧ ~Refutable(F, G)
Then it makes no difference at all what words he used, he is still wrong.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Newberry
2018-02-14 16:48:07 UTC
Raw Message
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of
inferences that either prove or disprove G. Why do you keep saying you
refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Gödel's first theorem does not make any such claim. It says that there
are unprovable and undisprovable closed WFFs in the language of PA.
Yet by definition all closed WFF must evaluate to True or False
Well, Gödel proved that there are undecidable closed WFFs in the
language of PA. You are saying that it would imply that such WFFs are
neither true nor false, and that is impossible.
So you are making two claims
1) true = provable
2) Closed WFF that is neither true nor false cannot possibly exist
So there must be an error in Gödel's proof somewhere, but you do not know where.
True is a certain kind of provable.
Closed WFF that is neither true nor false is wrong.
It may well be wrong. But why do you assert it does not exist?
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
and as I have shown they only actually evaluate to True or False
on the basis of a formal proof from this WFF to expressions of
language that have been defined to have the semantic property of True.
That is simply the way that Truth works and he got it wrong.
Pete Olcott
2018-02-14 16:56:33 UTC
Raw Message
Post by Newberry
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Newberry
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of
inferences that either prove or disprove G. Why do you keep saying you
refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Gödel's first theorem does not make any such claim. It says that there
are unprovable and undisprovable closed WFFs in the language of PA.
Yet by definition all closed WFF must evaluate to True or False
Well, Gödel proved that there are undecidable closed WFFs in the
language of PA. You are saying that it would imply that such WFFs are
neither true nor false, and that is impossible.
So you are making two claims
1) true = provable
2) Closed WFF that is neither true nor false cannot possibly exist
So there must be an error in Gödel's proof somewhere, but you do not know where.
True is a certain kind of provable.
Closed WFF that is neither true nor false is wrong.
It may well be wrong. But why do you assert it does not exist?
I never asserted that it does not exist. The intersection of
the set of closed WFF of L that do not evaluate to True or False
and the set of WFF that are statements of L is the empty set.

Semantically incorrect finite strings do exist, I never said that they do not.
Pete Olcott
2018-02-15 02:06:58 UTC
Raw Message
<html>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<body text="#000000" bgcolor="#FFFFFF">
<div class="moz-cite-prefix">On 2/14/2018 7:19 PM, Newberry wrote:<br>
</div>
<blockquote type="cite" cite="mid:p62n8a\$r58\$***@dont-email.me">Pete
Olcott wrote:
<br>
<blockquote type="cite">On 2/14/2018 11:27 AM, Newberry wrote:
<br>
<blockquote type="cite">On Wednesday, February 14, 2018 at
8:56:41 AM UTC-8, Pete Olcott wrote:
<br>
<blockquote type="cite">On 2/14/2018 10:48 AM, Newberry wrote:
<br>
<blockquote type="cite">Pete Olcott wrote:
<br>
<blockquote type="cite">On 2/14/2018 1:53 AM, Newberry
wrote:
<br>
<blockquote type="cite">Pete Olcott wrote:
<br>
<blockquote type="cite">On 2/13/2018 11:26 PM,
Newberry wrote:
<br>
<blockquote type="cite">Pete Olcott wrote:
<br>
<blockquote type="cite">On 2/13/2018 8:15 PM,
Newberry wrote:
<br>
<blockquote type="cite">Pete Olcott wrote:
<br>
<blockquote type="cite">On 2/13/2018 12:18 PM,
Jim Burns wrote:
<br>
<blockquote type="cite">On 2/12/2018 9:52
PM, Pete Olcott wrote:
<br>
<blockquote type="cite">On 2/12/2018 12:55
PM, Richard Tobin wrote: <br> <blockquote type="cite">In article <a class="moz-txt-link-rfc2396E" href="mailto:UIGdnZNYXKRiXhzHnZ2dnUU7-***@giganews.com">&lt;UIGdnZNYXKRiXhzHnZ2dnUU7-***@giganews.com&gt;</a>,
<br>
<br>
</blockquote>
</blockquote>
<br>
<blockquote type="cite">
<blockquote type="cite">
<blockquote type="cite">Since an
expression X of language L is not a
statement of
<br>
language L unless and until it is
proven to be True or
<br>
False in L, every statement of
language L can be proved
<br>
or disproved in L.
<br>
</blockquote>
<br>
So your language varies over time?
<br>
</blockquote>
</blockquote>
<br>
<blockquote type="cite">I am not saying
that.
<br>
</blockquote>
<br>
When you say
<br>
"expressions ... do not count ...
until after ... ",
<br>
you are saying that.
<br>
<br>
<blockquote type="cite">I am saying that
expressions of language do not count as
<br>
statements of language until after they
have been proven
<br>
True or False.
<br>
</blockquote>
<br>
For a wide range of theories T, there is a
closed formula
<br>
G_T for which it has been proved
<br>
T  |-  G_T &lt;-&gt;
~Ex:Proof_T(x,[G_T])
<br>
</blockquote>
<br>
All of those proofs are necessarily
semantically incorrect.
<br>
<br>
When-so-ever any closed WFF X of any
language L is evaluated
<br>
for True or False this semantically requires
a formal proof
<br>
from X to expressions T of language L that
have been defined
<br>
to have the semantic property of True.
<br>
<br>
There cannot possibly be a complete
inference chain from an
<br>
expression X of language L that makes
~Provable(L, X) true
<br>
because both True and False are only defined
as provable
<br>
from T.
<br>
</blockquote>
<br>
That is what Gödel's theorem says, that there
is no chain of
<br>
inferences that either prove or disprove G.
Why do you keep
<br>
saying
<br>
you
<br>
refuted it?
<br>
</blockquote>
<br>
To claim that an expression of language is True
or False
<br>
without being
<br>
provably True or False is religion not logic.
<br>
</blockquote>
<br>
Gödel's first theorem does not make any such
claim. It says that
<br>
there
<br>
are unprovable and undisprovable closed WFFs in
the language of PA.
<br>
<br>
</blockquote>
<br>
Yet by definition all closed WFF must evaluate to
True or False
<br>
</blockquote>
<br>
Well, Gödel proved that there are undecidable closed
WFFs in the
<br>
language of PA. You are saying that it would imply
that such WFFs are
<br>
neither true nor false, and that is impossible.
<br>
<br>
So you are making two claims
<br>
1) true = provable
<br>
2) Closed WFF that is neither true nor false cannot
possibly exist
<br>
<br>
So there must be an error in Gödel's proof somewhere,
but you do not
<br>
know where.
<br>
<br>
<br>
</blockquote>
<br>
True is a certain kind of provable.
<br>
Closed WFF that is neither true nor false is wrong.
<br>
</blockquote>
<br>
It may well be wrong. But why do you assert it does not
exist?
<br>
<br>
</blockquote>
<br>
I never asserted that it does not exist. The intersection of
<br>
the set of closed WFF of L that do not evaluate to True or
False
<br>
and the set of WFF that are statements of L is the empty
set.
<br>
<br>
Semantically incorrect finite strings do exist, I never said
that
<br>
they do not.
<br>
</blockquote>
<br>
According to you T/F = provable/refutable. And you are saying
that
<br>
there are closed WFFs that are neither provable nor refutable.
Is that
<br>
correct?
<br>
<br>
</blockquote>
<br>
You said it incorrectly.
<br>
T/F → provable/refutable
<br>
</blockquote>
<br>
T cannot be a proper subset of provable because the system would
be unsound.
<br>
<br>
</blockquote>
<b></b><br>
<b>Provable(L, X) means:</b><b><br>
</b>For expression X of language L, that there exists a formal proof
from <br>
expression X of language L to expression Y of language L. <br>
<br>
<b>True(L, X) means: </b><b><br>
</b>For expression X of language L, that there exists a formal proof
from <br>
expression X of language L to expression Y of language L <font
size="+1"><b><br>
and expression Y of language L is a member of T. </b></font><br>
<br>
<br>
<blockquote type="cite" cite="mid:p62n8a\$r58\$***@dont-email.me">
<blockquote type="cite">Here is a closed WFF that is not provable
or refutable:
<br>
semantics
<br>
</blockquote>
<br>
Gödel's sentence is also a closed WFF that is not provable or
refutable. Why do you have a problem with that?
<br>
<br>
</blockquote>
<br>
To see exactly why Gödel is incorrect only requires understanding <br>
exactly why the Liar Paradox is incorrect and exactly how the Liar <br>
Paradox is analogous to the 1931 GIT. <br>
<br>
Copyright 2018 (and other years) Pete Olcott <br>
<br>
<div class="moz-signature">-- <br>
<meta charset="UTF-8">
<font face="Segoe UI Symbol, sans-serif"><font size="2"><b>
∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) </b></font>
</font></div>
</body>
</html>
Pete Olcott
2018-02-14 23:27:31 UTC
Raw Message
Post by Pete Olcott
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
That is what Gödel's theorem says, that there is no chain of inferences that either prove or disprove G. Why do you keep saying you refuted it?
To claim that an expression of language is True or False without being
provably True or False is religion not logic.
Consider the Goldbach Conjecture. At the moment is neither
provable nor disprovable. Hence, by your proclamation it
is religion. We don't know whether it will ever be proven or
disproven. For rather obvious reasons it is suspected that
it might be true in real actual numbers but not provable.
I have already carefully thought of this. It is not the same
thing to not know a proof procedure as it is to know that a
proof procedure cannot possibly exist. The GC is the former
and not the latter.

We do not know how to do X is not at all the same
thing as saying we know that doing X is impossible.

The GC is not undecidable, it is merely undecided.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Jim Burns
2018-02-14 02:24:45 UTC
Raw Message
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
It seems that what you mean by "necessarily semantically
incorrect" is that those theorems disagree with you.

I think most people would find it a very strange world
if what is _actually the case_ was always what we thought
was the case, if we were always correct.

Anyway, those proofs are _formally_ correct. That has been
the claim beginning with the title of Godel's 1931 paper.
"On Formally Undecidable Propositions of
Principia Mathematica And Related Systems"

Your POsemantics are not interesting. You can't even tell if
a typical closed formula is a sentence or not. What do you
do for Act Two, after you've declared yourself infallible
in Act One?

Also, to the extent that you've ever understood what our
semantics are, you've always found them acceptable -- even
_too_ obvious, I suspect. I could almost hear your brain
going "WHERE'S THE TRICK?" But _the whole point_ is to be
as dead-obvious as imaginable. This is to be a _proof_ ,
not a magic trick.
Post by Pete Olcott
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
Let
G_T :<-> Q[Q(x)]
defined as before.

Then it's a _theorem_ that
G_T <-> ~Prov_T[G_T]
and
~G_T <-> Prov_T[G_T]

Adding more axioms doesn't get rid of theorems.
Adding axioms doesn't _get rid of_ any complete inference
chains. If you _add_ the requirements
G_T <-> Prov_T[G_T]
and
~G_T <-> Prov_T[~G_T]
you still have those two other things.

And all four of them together proves that T is inconsistent,
whether G_T or Prov_T[G_T] individually is true or false.
Post by Pete Olcott
This would be analogous to a guy correctly proving that
he never existed.
No, this is analogous to a guy correctly proving
Either there is an alien civilization on the moon Titan
or there isn't such a civilzation there.
without even trying to check either way.

There are certain statements that are true _no matter what_
their terms are interpreted as. P \/ ~P is true _for any P_

You keep talking about true and false. This misses the point.

Theorems are _valid_ . They are true for _any semantics_ that
conform to the assumptions.
(Semantics that don't conform are irrelevant.
The theorem makes no claims about those.)
Post by Pete Olcott
Post by Jim Burns
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
I don't even need self-reference to refute ~Provable(L, X)
because the evaluation of ~Provable(L, X) inherently requires
a proof or it cannot even be evaluated.
There is a proof of the claim actually made, which is
G_T <-> ~Prov_T[G_T]
Post by Pete Olcott
Post by Jim Burns
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references
used by the definitions used by G_T).
I merely augmented the insufficient specification of the
term of the art: axiom, and provided that actual definition
of [a priori] Truth, this correctly refuting Tarski's claim
that a language can not specify its own Truth predicate.
So, you changed Godel's proof and (allegedly) showed
the new proof is not valid -- and this is Godel's fault
somehow.
Post by Pete Olcott
Post by Jim Burns
If theory T is complete, then theory T is inconsistent.
Then Truth itself is broken, yet Truth cannot possibly
be broken.
_Your_ capital-T Truth is your own invention. Yes, your
invention is broken (makes theory T inconsistent), but
your invention is not truth -- you just call it that.
Pete Olcott
2018-02-14 04:31:27 UTC
Raw Message
Post by Jim Burns
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
It seems that what you mean by "necessarily semantically
incorrect" is that those theorems disagree with you.
No I means all the expressions of language that are analogous to
the Liar Paradox, yet taken to be true.

There must be an unbroken chain of inference from an expression
of language to one or more elements of the set of expressions
of language that have been defined to have the semantic property
of True for any expression of language to be true.

Every closed WFF of any language that anyone would ever write is
necessarily evaluated this same way.
Post by Jim Burns
I think most people would find it a very strange world
if what is _actually the case_ was always what we thought
was the case, if we were always correct.
It is all a matter of semantic logical entailment from known truths
to other expressions of language.
Post by Jim Burns
Anyway, those proofs are _formally_ correct. That has been
the claim beginning with the title of Godel's 1931 paper.
"On Formally Undecidable Propositions of
Principia Mathematica And Related Systems"
The formalism is necessarily insufficiently expressive of the
underlying semantics.

The reason that I have known that I must be correct all of these
years is that if there really is undecidable decision problems
then Truth itself is necessarily broken, yet it is not possible
for Truth itself to be broken.

Until you fully understand this part you will remain ignorant
John Gabriel
2018-02-14 17:25:50 UTC
Raw Message
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
"This would be analogous to a guy correctly proving that he never existed."

It wouldn't be 'analogous', it would be *exactly* like he correctly proving that he never existed.

Sadly, you won't get far with this because it is too simple a refutation of the bogus completeness theorem.
Post by Pete Olcott
Post by Jim Burns
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
I don't even need self-reference to refute ~Provable(L, X)
because the evaluation of ~Provable(L, X) inherently requires
a proof or it cannot even be evaluated.
Post by Jim Burns
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
I merely augmented the insufficient specification of the
term of the art: axiom, and provided that actual definition
of [a priori] Truth, this correctly refuting Tarski's claim
that a language can not specify its own Truth predicate.
Post by Jim Burns
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<->  Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] :=  [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T  :<->  Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
Then Truth itself is broken, yet Truth cannot possibly be broken.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
John Gabriel
2018-02-14 17:31:09 UTC
Raw Message
Post by John Gabriel
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
"This would be analogous to a guy correctly proving that he never existed."
It wouldn't be 'analogous', it would be *exactly* like he correctly proving that he never existed.
Sadly, you won't get far with this because it is too simple a refutation of the bogus completeness theorem.
or incompleteness theorem. Both are nonsense.
Post by John Gabriel
Post by Pete Olcott
Post by Jim Burns
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
I don't even need self-reference to refute ~Provable(L, X)
because the evaluation of ~Provable(L, X) inherently requires
a proof or it cannot even be evaluated.
Post by Jim Burns
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
I merely augmented the insufficient specification of the
term of the art: axiom, and provided that actual definition
of [a priori] Truth, this correctly refuting Tarski's claim
that a language can not specify its own Truth predicate.
Post by Jim Burns
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<->  Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] :=  [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T  :<->  Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
Then Truth itself is broken, yet Truth cannot possibly be broken.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
p***@gmail.com
2018-02-15 23:58:19 UTC
Raw Message
Post by John Gabriel
Post by Pete Olcott
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of  language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T  |-  G_T <-> ~Ex:Proof_T(x,[G_T])
All of those proofs are necessarily semantically incorrect.
When-so-ever any closed WFF X of any language L is evaluated
for True or False this semantically requires a formal proof
from X to expressions T of language L that have been defined
to have the semantic property of True.
There cannot possibly be a complete inference chain from an
expression X of language L that makes ~Provable(L, X) true
because both True and False are only defined as provable
from T.
"This would be analogous to a guy correctly proving that he never existed."
It wouldn't be 'analogous', it would be *exactly* like he correctly proving that he never existed.
That is what analogous means.
Post by John Gabriel
Sadly, you won't get far with this because it is too simple a refutation of the bogus completeness theorem.
Post by Pete Olcott
Post by Jim Burns
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
I don't even need self-reference to refute ~Provable(L, X)
because the evaluation of ~Provable(L, X) inherently requires
a proof or it cannot even be evaluated.
Post by Jim Burns
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
I merely augmented the insufficient specification of the
term of the art: axiom, and provided that actual definition
of [a priori] Truth, this correctly refuting Tarski's claim
that a language can not specify its own Truth predicate.
Post by Jim Burns
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<->  Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] :=  [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T  :<->  Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
Then Truth itself is broken, yet Truth cannot possibly be broken.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
bassam king karzeddin
2018-02-14 06:53:48 UTC
Raw Message
Strange that a parrot proven case (Jim Burs) with too elementary matters repeats blindly so much nonsense, trying tirelessly to protect the global ignorance everywhere he goes! wonder!

BKK
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<-> Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] := [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T :<-> Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
John Gabriel
2018-02-14 17:04:55 UTC
Raw Message
Post by bassam king karzeddin
Strange that a parrot proven case (Jim Burs) with too elementary matters repeats blindly so much nonsense, trying tirelessly to protect the global ignorance everywhere he goes! wonder!
Post by bassam king karzeddin
BKK
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<-> Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] := [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T :<-> Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
p***@gmail.com
2018-02-15 23:56:14 UTC
Raw Message
Post by bassam king karzeddin
Strange that a parrot proven case (Jim Burs) with too elementary matters repeats blindly so much nonsense, trying tirelessly to protect the global ignorance everywhere he goes! wonder!
BKK
Post by Jim Burns
Post by Pete Olcott
Post by Richard Tobin
Post by Pete Olcott
Since an expression X of language L is not a statement of
language L unless and until it is proven to be True or
False in L, every statement of language L can be proved
or disproved in L.
So your language varies over time?
I am not saying that.
When you say
"expressions ... do not count ... until after ... ",
you are saying that.
Post by Pete Olcott
I am saying that expressions of language do not count as
statements of language until after they have been proven
True or False.
For a wide range of theories T, there is a closed formula
G_T for which it has been proved
T |- G_T <-> ~Ex:Proof_T(x,[G_T])
This closed formula G_T is not self-referential. Substituting
something else for G_T which _is_ self-referential does not
make G_T self-referential, either.
You could verify the non-self-referentiality of G_T by
looking at the definition[1] of G_T, if you were the sort of
person who cared about things like whether what you claim
is true.
(Note that this is NOT the definition of G_T
G_T <-> ~Ex:Proof_T(x,[G_T])
)
Of course, you POsimplified away the definition of G_T,
in order to be able to claim without objections that the
definition of G_T is not semantically grounded
(ie, to claim that there are loops in the references used
by the definitions used by G_T).
Your claim is false, but you seem to think that, if you
don't look, we can't see it either. That is about par for
the course for someone who does not care whether the claims
he makes are true.
Post by Pete Olcott
A statement G of language F which can neither
be proved nor disproved in F, is impossible.
A theory T in which Prov_T() and sub_T() can be expressed
(one example of T would be basic arithmetic of natural
numbers without induction (ie, Q))
in which every statement G can be either proved or disproved
is provably inconsistent.
I do not doubt that you are able to invent an inconsistent
theory. But I suppose that, as long as you have impressed
yourself, that's all that matters.
[1]
Define
Q(x) :<-> Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )
[Q(x)] := [Ey:( ~Prov_T(y) & sub_T(x,[x],x) = y )]
Q[Q(x)] :<-> Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)
[Q[Q(x)]] := [Ey:(~Prov_T(y) & sub_T([Q(x)],[x],[Q(x)])=y)]
By the diagonal lemma
Q[Q(x)] <-> ~Prov_T[Q[Q(x)]]
Define G_T :<-> Q[Q(x)]
G_T <-> ~Prov_T[G_T]
If theory T is complete, then theory T is inconsistent.
He is not aware that he is doing this.
Pete Olcott
2018-02-13 06:26:47 UTC
Raw Message
Post by Pete Olcott
An {a priori fact} is an expression X
of (natural or formal) language L that
has been assigned the semantic property
of True.
A set T of {a priori facts} of language
L forms the ultimate foundation of the
notion of Truth in language L.
To verify that an expression X of
language L is True or False only requires
a syntactic logical consequence
inference chain (formal proof) from X or ~X
to one or more elements of T.
Since an expression X of language L is
not a statement of language L unless
and until it is proven to be True or
False in L, every statement of language
L can be proved or disproved in L.
∀L
∈ Formal_Systems
∀X
∈ L
Statement(X)
→ ( Provable(L, X) ∨ Refutable(L, X) )
Stanford Encyclopedia of Philosophy
Gödel’s Incompleteness Theorems
The first incompleteness theorem states
that in any consistent formal system F
within which a certain amount of
arithmetic can be carried out, there
are
statements
of the language of F which can neither be proved nor disproved
in F.
In any consistent formal system F within which a certain amount of arithmetic can be carried out, there are CLOSED WFFs of the language of F which can neither be proved nor disproved in F.
Post by Pete Olcott
And I just proved that he is necessarily incorrect.
You have to really pay attention to see this.

The syntax that he used had undetectable gaps in its inference chain that could
not be detected because the formalism could not even express these gaps.

It is like the term {ignorance squared} that I coined two decades ago, F did
not even know that it didn't know.
Pete Olcott
2018-02-14 16:52:05 UTC
Raw Message
On Monday, February 12, 2018 at 10:26:55 PM UTC-8, Pete Olcott
Post by Pete Olcott
Post by Pete Olcott
An {a priori fact} is an expression X
of (natural or formal) language L that
has been assigned the semantic property
of True.
A set T of {a priori facts} of language
L forms the ultimate foundation of the
notion of Truth in language L.
To verify that an expression X of
language L is True or False only requires
a syntactic logical consequence
inference chain (formal proof) from X or ~X
to one or more elements of T.
Since an expression X of language L is
not a statement of language L unless
and until it is proven to be True or
False in L, every statement of language
L can be proved or disproved in L.
∀L
∈ Formal_Systems
∀X
∈ L
Statement(X)
→ ( Provable(L, X) ∨ Refutable(L, X) )
Stanford Encyclopedia of Philosophy
Gödel’s Incompleteness Theorems
The first incompleteness theorem states
that in any consistent formal system F
within which a certain amount of
arithmetic can be carried out, there
are
statements
of the language of F which can neither be proved nor
disproved
in F.
Gödel knew very well that a semantic argument would open a can of
worms. Therefore he made a semantic one. It goes roughly as
In any consistent formal system F within which a certain
amount of
arithmetic can be carried out, there are CLOSED WFFs of the
language
of F which can neither be proved nor disproved in F.
Post by Pete Olcott
And I just proved that he is necessarily incorrect.
No, you attempted to refute
In any consistent formal system F within which a certain amount of
arithmetic can be carried out, there are *STATEMENTS* of the
language
of F which can neither be proved nor disproved in F.
You did not refute
In any consistent formal system F within which a certain amount of
arithmetic can be carried out, there are *CLOSED WFFs* of the
language
of F which can neither be proved nor disproved in F.
They are synonymous.
You did not prove that there are no black cats you only proved that
there are no cats that are black.
Sentence (mathematical logic)
In mathematical logic, a sentence of a predicate logic is a
boolean-valued well-formed formula with no free variables. A
sentence
can be viewed as expressing a proposition, something that must be
true
or false. The restriction of having no free variables is needed
to make
If in your opinion a closed wff must have a truth value what name do
you propose for the purely syntactical construct?
I don't think that is actually possible.
See if you can provide a concrete example.
Let us call it "humgwaard". We mean by that an uninterpreted string of
symbols subject to some syntactical constraints. An example would be
{B\$}{\$ # \$}
B = A
\$ = x
'#' = '='
{ = (
But in this case we won't.
So basically you are saying organized gibberish.
I don't see how that applies to anything.
We will get to the application in a minute. When a humgwaard is
prefixed with '^' it is called "megatotion". There is a subset of
humgwaards called schwazioms. And there are rules of inference how to
derive humgwaards from the schwazioms. Gödel proved that given a
certain set of schwazioms and certain rules of inference there is a
humgwaard such that neither it nor its megatotion can be derived. Is
there anything wrong with his proof?
Statement(F, G)  ∧ ~Provable(F, G) ∧ ~Refutable(F, G)
A David Hilbert mathematical formalism evaluates it to false.
I am not aware of that.
(EG)( SyntacticallyCorrectFormula(F, G)  ∧ ~Provable(F, G) ∧ ~Refutable(F, G) )
Do not know what they mean by "statement."
I am using the SEP language for Statement, it means closed WFF, thus necessarily
must evaluate to True or False.
One possible interpretation is the one above. I do not know if they meant to imply that statement must be either true or false.
Sentence (mathematical logic)
If they did mean that then there is of course an interpretation based on a given model. You may have issues with these models, but this is what they meant.
Anyway, as I said, Gödel knew this led to complications, so he made a purely syntactic proof. Do you understand that?
I understand that he did not understand that he was incorrect.
--
*∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) *
Jan
2018-02-13 20:43:12 UTC
Raw Message
Post by Pete Olcott
This is redundant and looks silly. The posts are automatically copyrighted.

--
Jan
a***@gmail.com
2018-02-13 22:38:19 UTC
Raw Message
Post by Jan
Post by Pete Olcott
This is redundant and looks silly. The posts are automatically copyrighted.
--
Jan
It's also silly because no one else would want to take credit for such mistaken foolishness.
Pete Olcott
2018-02-15 04:30:27 UTC
Raw Message
<html>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
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<div class="moz-cite-prefix">On 2/12/2018 10:42 AM, Pete Olcott
wrote:<br>
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cite="mid:UIGdnZNYXKRiXhzHnZ2dnUU7-***@giganews.com">
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An {a priori fact} is an expression X of (natural or formal)
language L that<br>
has been assigned the semantic property of True.<br>
<br>
A set T of {a priori facts} of language L forms the ultimate
foundation of the<br>
notion of Truth in language L.<br>
<br>
To verify that an expression X of language L is True or False only
requires<br>
a syntactic logical consequence inference chain (formal proof)
from X or ~X<br>
to one or more elements of T.<br>
<br>
Since an expression X of language L is not a statement of language
L unless<br>
and until it is proven to be True or False in L, every statement
of language<br>
L can be proved or disproved in L.<br>
<br>
<b><span style="background: #ffff00">∀L ∈ Formal_Systems</span></b><b><br>
</b><b> </b><b><span style="background: #ffff00">∀X ∈ L</span></b><b><br>
</b><b> </b><b><span style="background: #ffff00">Statement(X) → (
Provable(L, X) ∨ Refutable(L, X) )</span></b><b><br>
</b><b> </b><br>
<b>Stanford Encyclopedia of Philosophy Gödel’s Incompleteness
Theorems</b><b><br>
</b><b> </b>The first incompleteness theorem states that in any
consistent formal system F<br>
within which a certain amount of arithmetic can be carried out, <b><span
style="background: #ffff00">there are</span></b><b><br>
</b><b> </b><b><span style="background: #ffff00">statements of
the language of F which can neither be proved nor disproved in
F.</span></b><br>
<br>
<b>Copyright 2017, 2018 Pete Olcott </b><br>
<br>
<span style="background: #ffff00"></span>
<div class="moz-signature">-- <br>
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<font face="Segoe UI Symbol, sans-serif"><font size="2"><b> ∀X
True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) </b></font> </font></div>
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<br>
a Collection is defined as one or more things that have one or more
<br>
properties in common. These operations from set theory are
available:  {⊆, ∈}<br>
More may be added later, as needed. <br>
<div class="moz-signature"><br>
A close Facebook friend pointed out that my specification depends
upon<br>
the notion of set theory so I explicitly replace my reference to
set theory<br>
with a reference to a collection as defined above. <br>
<br>
This needs additional details thus is incomplete. I will probably
have two <br>
distinct types of collections:<br>
(1) Those that can only contain other  collections. <br>
(2) Those that can contain only urelements.<br>
<br>
It no case will the semantically incoherent notion of a set that
contains<br>
itself be allowed. <br>
<br>
<br>
-- <br>
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∀X True(X) ↔ ∃Γ ⊆ Axioms Provable(Γ, X) </b></font>
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Zelos Malum
2018-02-15 06:42:13 UTC
Raw Message
Since an expression X of language L is not a statement of language L unless
and until it is proven to be True or False in L

This is incorrect, a statement in a language only uses the predicates and objects of the language, along with the rules of forming them, it doesn't need to be evaluated as true or false because at its most basic level, they are just symbol manipulation and "true and false" are not relevant. There are Models/Intepritations of formal languages, including the classical ones, where true/false are meaningless.
p***@gmail.com
2018-02-16 00:05:30 UTC
Raw Message
Post by Pete Olcott
Since an expression X of language L is not a statement of language L unless
and until it is proven to be True or False in L
This is incorrect, a statement in a language only uses the predicates and objects of the language, along with the rules of forming them, it doesn't need to be evaluated as true or false because at its most basic level, they are just symbol manipulation and "true and false" are not relevant.
This is necessarily incorrect. In many formal languages the ONLY
semantic notion is ultimately grounded Boolean True and false.

Even the formalization of natural language boils everything
down to truth conditional semantics. (anchoring all semantics
in Boolean True and False).

There are Models/Intepritations of formal languages, including the classical ones, where true/false are meaningless.

This may be possible. For example Lambda Calculus.
Zelos Malum
2018-02-16 11:32:06 UTC