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\begin{document}
\title{Quantum superposition justified in a new non-Aristotelian
finitary logic}
\author{Radhakrishnan Srinivasan\thanks{R \& D Group -- Exports services,
IBM Global Services India Pvt.\ Ltd., 5th floor, Golden Enclave,
Airport road, Bangalore~560017, India.
E-Mail:~\texttt{sradhakr@in.ibm.com}}}
\date{}
\maketitle
\begin{abstract}
A new non-Aristotelian finitary logic (NAFL) is proposed in which
it is postulated that the truth or falseness of an undecidable
proposition in a theory T is meaningful only when asserted
axiomatically; there is no truth other than axiomatic truth.
It is shown that under this hypothesis, the law of the
excluded middle and the law of non-contradiction for such undecidable
propositions must fail to be theorems of T. The phenomenon
of quantum superposition is thus explained in NAFL. It is also
shown that infinite sets cannot exist in any consistent theory
of NAFL, which makes it a very restrictive logic. Implications
for some modern mathematical and physical theories are analyzed
from the point of view of NAFL.
\end{abstract}
\section{Introduction}
Let $\Psi$ be an undecidable proposition in a theory T
formulated in classical first order predicate logic, i.e., neither $\Psi$
nor $\neg \Psi$ is provable in T. If T is consistent (equivalently in
first order logic, if models for T exist) then the truth or falseness
of $\Psi$ in T is defined semantically via model theory, that
is, there exists a model for T in which $\Psi$ is true and another
model in which $\Psi$ is false. Syntactically (or in a proof theoretic
sense), however, the truth or falseness of $\Psi$ in T is not defined
\emph{constructively}, and amounts to a Platonic belief expressed by the
law of the excluded middle $\Psi \vee \neg \Psi$.
In this paper, a different, novel constructivist viewpoint
is formulated that makes the syntactic and semantic definitions
of the truth or falseness of $\Psi$ in T equivalent. The main
new postulate (and point of departure from classical logic),
whose plausibility follows from elementary classical model
theory, is that the truth or falseness of $\Psi$ in T must be
\emph{axiomatic} in nature, if T is consistent; that is,
$\Psi$ can be true or false in T if and only if so asserted axiomatically,
and the truth (falseness) of $\Psi$ in T \emph{is} its truth (falseness)
in the theory T+$\Psi$ (T+$\neg \Psi$).
In the resulting new logic, it is meaningless to talk of the
truth of a formal proposition \emph{outside} of axiomatic
theories. For example, the notion of truth is not definable in
just the \emph{language} of T, without reference to any formal system;
in this sense, all \emph{formalizable} truths are axiomatic. We
argue later that the notions of \emph{undecidability} in T of a
proposition and the \emph{consistency} of T are Platonic truths that
cannot be formalized in any consistent theory T; this is another
point of departure from classical logic.
It is demonstrated that this new postulate leads to the conclusion that
the law of the excluded middle ($\Psi \vee \neg \Psi$) is not a
theorem of T as in intuitionistic logic. However, in contrast to
intuitionistic logic, it is shown that the law of non-contradiction
($\neg(\Psi \& \neg \Psi)$)
also fails to be a theorem of T by the above postulate and therefore
there must exist a non-classical model for T in which both
$\Psi$ and $\neg \Psi$
are asserted as axioms. It is argued that this does \emph{not} make
T inconsistent; inconsistency only follows if both $\Psi$ and
$\neg \Psi$ are \emph{deducible} in T. Thus this new, non-Aristotelian
logic provides a simple justification of the phenomenon of quantum
superposition in which, for example, a single photon seemingly takes
two different paths at the same time. However, the law of the excluded
middle does indeed apply to \emph{Platonic} notions such as consistency
and undecidability. Thus T is either consistent or inconsistent and
a proposition is either decidable or undecidable in T, although,
in both cases, we may never know which.
It is shown that the proposed logic is also finitary.
In a suitable set theory with classes formulated in this logic,
infinite classes are permitted, but not infinite sets if the
consistency constraint is to be satisfied.
Thus consistency demands that all infinite classes must necessarily
be proper classes (that is, they cannot belong to any class); in other
words, the axiom of infinity for sets must be dropped. Therefore
modern mathematics, theoretical physics
and a whole lot of other scientific theories
formulated using Cantorian assumptions cannot be justified in the
framework of this new logic. Some other implications
for quantum mechanics, elementary Peano arithmetic, special theory
of relativity, theoretical computer science and non-Euclidean
geometries are also considered.
\section{The proposed non-Aristotelian finitary logic}\label{nafl}
In the rest of this paper, the proposed new
non-Aristotelian logic is abbreviated as NAFL. The logic NAFL is
basically a modified version of the classical first order predicate
logic. The modification mainly affects the treatment of undecidable
propositions in a theory T, in that the law of the excluded middle
and the law of non-contradiction are not required to hold for
such propositions. Instead, truth for propositions undecidable in T
is governed by the main postulate of NAFL, to be described below.
The constraint that T be consistent will then force additional
points of departure from classical first order predicate logic, mainly
in the form of restrictions. Some of these are
stated in this section, in the remarks
following the main postulate and the two metatheorems that follow
from it. Additional restrictions are elaborated upon in the ensuing
sections where Cantor's set theory and other well known theories are
considered from the point of view of NAFL.
It should be emphasized that NAFL must be viewed as a series of
metamathematical assertions \emph{about} axiomatic theories.
These assertions are to be viewed as Platonic truths; as noted earlier,
the formal propositions of NAFL theories do not have any truth values in
NAFL, which by itself is not a theory.
NAFL is the sum total of all that is described progressively
(and informally) in the rest of this paper. Throughout this paper,
consistency of T is equated with the existence of models for T;
this holds in any first order logic.
Let T be a consistent theory formulated in classical first-order
predicate logic (with the specific points of departure to be elaborated
below in the definition of NAFL) and
suppose that $\Psi$ is an undecidable proposition
in T. Note that the undecidability is established by the rules
of inference and principles of proof in the classical first order predicate
calculus. An important restriction in NAFL is that $\Psi$ itself does
not contain any component of the form $\phi \vee \neg \phi$
where $\phi$ is undecidable in T. As will be asserted shortly,
propositions like ``$\phi \vee \neg \phi$, but we do not know which'' are
not valid in NAFL. However, if $\phi$ is \emph{not}
undecidable in T, $\phi \vee \neg \phi$ may
be \emph{assumed} in a \emph{proof} of either $\phi$ or of $\neg \phi$
in T. Further, if $\psi$ and $\phi$ are both
known to be undecidable in T, then it may happen (under certain
conditions, but not always) that $\psi \Leftrightarrow \phi$,
$\psi \Rightarrow \phi$ and $\phi \Rightarrow \psi$ are not legitimate
propositions in T (see Remark~\ref{rmrk} below for details). Note that
these are disjunctions involving undecidable propositions. It is assumed
that $\Psi$ does not include any such component as well, when
it is deemed illegal.
Define the theories $\mbox{T}^{+}$ and $\mbox{T}^{-}$ by:
\[
\mbox{T}^{+} \equiv \mbox{T}+\Psi \quad \& \quad
\mbox{T}^{-} \equiv \mbox{T}+\neg \Psi,
\]
where addition in the above notation means addition as an axiom.
Note that the undecidability of $\Psi$ in T (assumed consistent) makes
each of the theories $\mbox{T}^{+}$ and $\mbox{T}^{-}$ consistent
and distinct from T itself. The main postulate, which
is the single assumption from which essentially all the other
features of NAFL may be deduced, is stated as follows.
\begin{proposition}[NAFL]
Under the assumptions that \emph{T} is consistent and that
$\Psi$ is undecidable in \emph{T},
the truth or falseness of $\Psi$ in \emph{T}
is indistinguishable from its truth in $\emph{T}^{+}$
or falseness in $\emph{T}^{-}$ respectively. To say that
$\Psi$ is interpreted as true (false) in \emph{T} is therefore the
\emph{same} as saying that \emph{T} is to be interpreted
as $\emph{T}^{+}$ ($\emph{T}^{-}$) respectively. Equivalently,
the truth or falseness of $\Psi$ in \emph{T} has meaning \emph{only}
when asserted \emph{axiomatically}, i.e., by an act of human
will. In NAFL, it is not meaningful to talk of the truth or falseness of
formal propositions outside of axiomatic theories.
\end{proposition}
The meaning of the main postulate is summarized as follows. Given the
axioms of a consistent T, the question
``Is $\Psi$ true (false) in T?'' is answered by ``Yes, if it has been
so asserted axiomatically''; in other words, the
truth (falseness) of $\Psi$ in T is equated with its truth (falseness)
in $\mbox{T}^{+}$ ($\mbox{T}^{-}$). It should be kept in mind that,
unlike classical logic, there are \emph{two} sets of axioms to be
specified when T is being interpreted. These are the axioms of the
theory being interpreted (T) and the axioms of the theory that
interprets T (say, T*). If T*=T, then every undecidable proposition
in T, including $\Psi$, is being interpreted as neither true nor false
(as will become clear from metatheorem~\ref{mt2} below). If (and only if)
T*=$\mbox{T}^{+}$ ($\mbox{T}^{-}$), then $\Psi$ is being interpreted
as true (false) in T.
As a justification for the main postulate, we offer the following
plausible argument. According to the modern
\emph{structuralist} ideology, a theory can be
meaningfully interpreted only in \emph{structures},
in particular, \emph{models}~\cite{Sh}. Every valid
metamathematical statement about a consistent theory
is in fact a statement about the class of all of its
models. In particular, two consistent theories $\mbox{T}_{1}$ and
$\mbox{T}_{2}$ are identical if and only if every model for
$\mbox{T}_{1}$ is a model for $\mbox{T}_{2}$ and conversely,
every model for $\mbox{T}_{2}$ is a model for $\mbox{T}_{1}$.
The said justification follows by noting
that every model for T in which $\Psi$ is true (false) is a model
for $\mbox{T}^+$ ($\mbox{T}^-$) and conversely, every model
for $\mbox{T}^+$ ($\mbox{T}^-$) is a model for T in which
$\Psi$ is true (false).
It should be emphasized that this argument is not necessarily a ``proof''
of the main postulate. The standard (classical) interpretation denies
this postulate by maintaining a distinction between the truth
or falseness of $\Psi$ in T versus its truth in $\mbox{T}^+$ or
falseness in $\mbox{T}^-$, respectively; however, the point of the
main postulate is that such a distinction cannot be justified
\emph{constructively} (if one views a model as a construction)
and so must be dropped in NAFL. It is extemely important to note
that the main postulate essentially rejects the Platonistic
philosophy inherent in classical logic. Every assertion like
``Let $\Psi$ be true (false) in T'' (which is used to create models
for T in classical logic) must necessarily be
interpreted in NAFL as amounting to an axiomatic declaration of
such truth (falseness) of $\Psi$ and so T is being interpreted
as $\mbox{T}^+$ ($\mbox{T}^-$). In other words,
``Let $\Psi$ be true (false) in T, but I am not axiomatically
declaring it as true (false)'' is a contradiction in NAFL; the
second part of this assertion contradicts the first. In NAFL (unlike
classical logic), there is no Platonic world in which such a
truth holds independent of an axiomatic declaration (by the human
mind that interprets T).
The classical interpretation also asserts that the law of the
excluded middle
\begin{equation}\label{excm}
\Psi \vee \neg \Psi
\end{equation}
must hold in T despite the fact that $\Psi$ is undecidable in T.
This is denied in NAFL as a consequence of the main postulate,
as stated below.
\begin{lemma}[NAFL]\label{mt1}
Suppose that $\Psi$ is undecidable in a consistent theory \emph{T}.
Then if the main postulate holds, it is not valid to assert
$\Psi \vee \neg \Psi$ as a theorem of \emph{T}.
\end{lemma}
The proof is trivial and is as follows.
\begin{proof}
Suppose that (\ref{excm}) is a theorem of T.
What (\ref{excm}) asserts is that ``$\Psi$ is either true
or false in T, but we do not know which'' (because
of the undecidability of $\Psi$ in T). But from
the main postulate, this assertion is exactly equivalent
to ``T is to be interpreted as either
$\mbox{T}^{+}$ or $\mbox{T}^{-}$, but we do not know
which''. This in turn means that ``either $\Psi$ or
$\neg \Psi$ is present in T as an \emph{axiom},
but we do not know which''; in other words, T has
been defined as equivalent to the disjunction
\[
\mbox{T}=\mbox{T}^{+} \vee \mbox{T}^{-}.
\]
But this definition of T is illegal! Since
axioms are added to theories by human beings, we \emph{ought}
to know which of the two propositions $\Psi$ or $\neg \Psi$ is
present in T as an axiom! It is not legal to assert
that one of these two propositions has been \emph{axiomatically}
declared to be true, but we do not know which one. In short,
``$\mbox{T}^{+} \vee \mbox{T}^{-}$'' is not a legitimately
defined theory, but the law of the excluded middle forces
T to be equivalent to it, under the main postulate. Formally
(or syntactically), this disjunction merely asserts the truth
of the propositions belonging to (i.e., provable in)
either $\mbox{T}^{+}$ or $\mbox{T}^{-}$. Clearly, this is how T
has been defined; syntactically, (\ref{excm}) cannot mean that
``(\ref{excm}) holds in all interpretations of T'', for the
formal definition of T cannot include the word ``interpretation''.
We conclude that from (\ref{excm}) and the main postulate,
T becomes indistinguishable from the illegal definition
$\mbox{T}^{+} \vee \mbox{T}^{-}$ and is therefore a non-existent theory.
The only possible conclusion that one can arrive at
is that the assumption that (\ref{excm}) is a theorem
of T, which forced this contradiction, must be false if T is consistent.
\end{proof}
\begin{remark}
If one insists on the validity of (\ref{excm}) in NAFL, then the
conclusion from the main postulate
would be that no consistent theory can admit undecidable propositions.
However, this would render inconsistent even a theory T containing
no axioms, for clearly, every proposition is undecidable with
respect to the null set of axioms. This in turn means that
consistent theories do not exist, which is not a very useful
conclusion. It is tempting to say that this conclusion would
violate G\"odel's first incompleteness theorem~\cite{Gd}, but G\"odel's
theorems and his notion of consistency are derived from different
assumptions than those made in NAFL and so do not apply in this logic.
\end{remark}
We next demonstrate that the main postulate implies that the
law of non-contradiction must also fail to be a theorem of T.
\begin{lemma}[NAFL]\label{mt2}
Assuming that the main postulate holds, metatheorem~\ref{mt1}
implies that there must exist a non-classical model for \emph{T} in which
both $\Psi$ and $\neg \Psi$ are asserted as (non-classical) axioms.
\end{lemma}
Obviously both $\mathrm{T}^{+}$
and $\mathrm{T}^{-}$ would be invalid in such a non-classical model,
which would give $\mathrm{T}$ the required distinct identity.
\begin{proof}
One can deduce the contradiction
$\Psi \& \neg \Psi$ in such a model, but this is a deduction
from precisely the above two (non-classical) axioms, neither of which is
refutable in T. In NAFL, under the assumptions of the main postulate,
the contradiction is in fact
refutable in each of $\mbox{T}^{+}$ and $\mbox{T}^{-}$,
but not in T. Note that the classical refutation of the contradiction
in T is equivalent to the following
argument: ``If $\Psi$ ($\neg \Psi$) is true in T, then
$\neg \Psi$ ($\Psi$) must be false because (\ref{excm}) is
a theorem of T''. But this argument, by the main postulate in
NAFL, really amounts to refuting
$\Psi \& \neg \Psi$ in $\mbox{T}^{+}$ ($\mbox{T}^{-}$). The refutation
works in classical logic because the truth or falseness of
$\Psi$ in T is postulated to be \emph{not} axiomatic by the law
of the excluded middle. Note that the truth
of $\Psi$ ($\neg \Psi$) in T is interpreted in classical logic as
a Platonic (and \emph{not} axiomatic) truth, when the
refutation is executed semantically in models for T.
In NAFL metatheorem~\ref{mt1} implies that the law of the excluded
middle is not a theorem of T and secondly, the refutation of
$\Psi \& \neg \Psi$ in T fails because it must necessarily
appeal to axiomatic truths outside of T (both syntactically
and semantically). The metatheorem follows from these two facts.
\end{proof}
\begin{remark}[Quantum superposition justified]
Metatheorem~\ref{mt2} explicitly justifies the quantum superposition
principle. What it asserts is that if $\Psi$ is undecidable
in a consistent theory T in NAFL, then
$\Psi \& \neg \Psi$ must obtain in T if neither $\Psi$ nor
$\neg \Psi$ has been declared to be true. Here $\Psi \& \neg \Psi$
must be interpreted as not that ``$\Psi$ is true and false'', that is,
not as a contradiction, but in the sense that ``neither $\Psi$ nor
$\neg \Psi$ is true in (or an axiom of) T''. In other words, ``$\Psi$'' in
$\Psi \& \neg \Psi$ asserts that ``$\neg \Psi$ is not an axiom of
T'' whereas ``$\neg \Psi$'' asserts that ``$\Psi$ is not an axiom
of T''. Seen in this light, there is really no contradiction, because
clearly this is how T has been defined. Note that there is no
distinction between $\Psi$ and $\neg \neg \Psi$ in NAFL
(unlike intuitionism). Similarly, if in
quantum theory it is undecidable as to whether a photon took
path~A or path~B (but is restricted to these paths), then the
proposition that ``the photon took path A'' (with the negation
``the photon took path B'') is undecidable when suitably formalized,
and the superposition follows. According to
NAFL, the superposition only means that the photon
took neither path A nor path B. Thus NAFL is more in tune with
the Copenhagen interpretation than the Many Worlds interpretation
of quantum mechanics. The role of ``measurement'' in quantum
theory must be equated with the axiomatic assertion of the truth
of the measured result (e.g., ``The photon took path~A'').
However, it is important to recognize that the probabilistic
component of quantum mechanics, which arises in the measurement
process, cannot be modelled consistently in NAFL. Thus the notion
that ``The photon will be detected either in detector~C with a probability
$p$ or in detector~D with probability $1-p$'' can only be formulated
in classical logic and not in NAFL, because the law of the excluded
middle becomes indispensable for this notion.
\end{remark}
\begin{remark}
The (non-classical) model in metatheorem~\ref{mt2} is in fact a
superposition of two or more classical models, at least one
of which is a model for $\mbox{T}^+$ and another, a model
for $\mbox{T}^-$. Note that in this non-classical model, we do permit
the inferences
\[
\Psi \& \neg \Psi \Rightarrow \Psi \quad \mbox{and} \quad
\Psi \& \neg \Psi \Rightarrow \neg \Psi,
\]
and conversely, $\Psi \& \neg \Psi$ follows from $\Psi$ and $\neg \Psi$.
But, as noted in the previous remark, ``$\Psi$'' and
``$\neg \Psi$'' in this non-classical model are \emph{not} to be
interpreted as corresponding to the truth or falseness of $\Psi$ in
T (as required by Tarski in classical logic). The classical assertion
that ``from a contradiction, any proposition can be deduced''
fails in the non-classical model of NAFL. To classically deduce a
proposition $\phi$ from $\Psi \& \neg \Psi$ in this non-classical model,
one would proceed as follows:
\begin{eqnarray}
&&\Psi \& \neg \Psi \Rightarrow \Psi, \nonumber \\
&&\Psi \Rightarrow \Psi \vee \phi, \nonumber \\
&&\Psi \& \neg \Psi \Rightarrow \neg \Psi, \nonumber \\
&&\neg \Psi \& [\Psi \vee \phi] \Rightarrow \phi. \nonumber
\end{eqnarray}
It is the last step that fails in NAFL by metatheorem~\ref{mt2}
because it presumes the law of non-contradiction. A simple application
of metatheorem~\ref{mt2} is to consider T as the null set of axioms.
Clearly $\Psi$ is undecidable in T and so there must exist a
non-classical model for T in which $\Psi \& \neg \Psi$ obtains.
That such a non-classical model indeed exists is obvious;
$\Psi \& \neg \Psi$ does not contradict any axiom of T because
T contains no axioms. This observation is important becasuse it
illustrates that NAFL does permit undecdiable propositions,
unlike some versions of intuitionism.
\end{remark}
\begin{remark}
It is now clear why, in NAFL, the assertion ``$\Psi$ is undecidable in T''
cannot be formalized in T. If T is consistent, this assertion cannot be
proved in T because, by metatheorem~\ref{mt2}, it is equivalent to
$\Psi \& \neg \Psi$. Any such proof in T is also a proof of both $\Psi$
and $\neg \Psi$, which would make T inconsistent and contradicts
our hypothesis that neither proof exists in T. Hence if T is
consistent, the formalized version of ``$\Psi$ is undecidable in T'',
if it exists, would be ``true but undecidable'' in T. The requirement of
metatheorem~\ref{mt2} that a non-classical model should exist
would then produce a contradiction. In other words, ``true
but undecidable propositions'' are not permissible in NAFL; the
``true'' part has no meaning for a formal proposition unless
asserted axiomatically.
Similarly, in NAFL, the notion ``T is consistent'' cannot be
formalized in T. Intuitively, one can see why the undecidability
of G\"{o}del's proposition con(T) follows from ``T is consistent'';
if T is indeed ``really'' consistent, the formalized version
con(T) can only express this fact by referring to
the class of all propositions in T, including itself. It is this
self-reference that causes ``true but undecidable'' propositions,
a contradiction in NAFL. Therefore the notions of undecidability
and consistency must both remain Platonic in NAFL. It follows that
G\"{o}del's theorems do not apply to NAFL theories. A very
important conclusion is that the classical equivalence of
``T is consistent'' and ``$\Psi$ is undecidable in T'', which follows
from G\"{o}del's theorems for appropriate T, does not necessarily
hold in NAFL.
\end{remark}
\begin{remark}\label{rmrk}
Consider the proposition
\begin{equation}\label{psph}
\psi \Leftrightarrow \phi,
\end{equation}
and suppose that (\ref{psph}) is a theorem of T. Then neither
$\psi$ nor $\phi$ can be undecidable in T.
To prove this assertion, suppose, to obtain a contradiction,
that one of either $\psi$ or $\phi$ were indeed undecidable
in T; it immediately follows that the other too must be undecidable.
Therefore both
$\psi \& \neg \psi$ and $\phi \& \neg \phi$ would obtain in a
(non-classical) model for T, which must exist by
metatheorem~\ref{mt2}. But then in this model one could infer
$\psi \& \neg \phi$, which is the negation of (\ref{psph}) and
a contradiction, because (\ref{psph}) is a theorem of T. This
completes the proof. Indeed, (\ref{psph}) can be true in this
non-classical model if and only if
one could formalize in T the following sentence:
``$\psi$ is undecidable in T iff $\phi$ is
undecidable in T''. But we have seen that the notion of undecidability
is \emph{not} formalizable in any NAFL theory T.
The important point to note here is that
any sentence, such as, (\ref{psph}), can be a theorem of T in NAFL
if and only if it were to be \emph{classically} true.
Here (\ref{psph}) can be classically true in NAFL iff $\psi$ and $\phi$
were both either provable or refutable in T.
To avoid confusion, we use the notation $\psi \& \neg \psi$
\emph{obtains} (rather than ``$\psi \& \neg \psi$ is true'')
in a non-classical
model for T because ``true'' always means ''classically true''
in NAFL. Here $\psi \& \neg \psi$ is \emph{not} classically
true in the said non-classical model (and therefore can never be
formally asserted as an axiom) because the notion of undecidability of
$\psi$ is not formalizable. Next consider
\begin{equation}\label{psph3}
\psi \Rightarrow \phi.
\end{equation}
If both $\psi$ and $\phi$ are undecidable in T, then (\ref{psph3})
cannot be a theorem of T, by an identical argument
to that given for (\ref{psph}). Indeed, if both $\psi$ and $\phi$
are undecidable in T, neither (\ref{psph3}) nor (\ref{psph})
can even be valid propositions of T under certain
conditions. We illustrate this for the case of (\ref{psph3}).
Suppose $\phi$ and $\neg \psi$ were to be theorems of T+$\psi$
and T+$\neg \phi$ respectively; both of these must happen
simultaneously by the classical rules of inference, which apply in NAFL.
Then (\ref{psph3}) is not a valid
proposition of T. This is so because under this
restriction (\ref{psph3}) can only be either ``true'' or
``neither true nor false'' in T
and can never be (classically) false. To see this, note that for
(\ref{psph3}) to be false in T, we need $\psi$ to be
true and $\phi$ to be false, but this can never happen. The moment
we assert the truth of $\psi$ ($\neg \phi$), we are interpreting T
as T+$\psi$ (T+$\neg \phi$) by the main postulate, and
$\phi$ ($\neg \psi$) cannot be false because
it is a theorem of T+$\psi$ (T+$\neg \phi$) by hypothesis.
However, if neither $\psi$ nor $\phi$ have been asserted as axioms,
then by metatheorem~\ref{mt2}, both $\psi \& \neg \psi$ and
$\phi \& \neg \phi$ obtain in T. But then we may infer both (\ref{psph3})
and its negation in such a non-classical model for T, which implies
(\ref{psph3}) is neither true nor false. Hence the conclusion
that (\ref{psph3}) is not a valid proposition of T if
$\phi$ and $\neg \psi$ are theorems
of T+$\psi$ and T+$\neg \phi$ respectively.
However, it is legitimate to use (\ref{psph3})
as a rule of inference in order to prove $\phi$ ($\neg \psi$)
in T+$\psi$ (T+$\neg \phi$).
If $\phi$ and $\neg \psi$ are \emph{not} theorems
of T+$\psi$ and T+$\neg \phi$ respectively, then (\ref{psph3})
is indeed a valid proposition of T. The conclusion is that
disjunctions involving undecidable propositions must be
handled very carefully in NAFL. Observe that if we replace
$\phi$ by $\psi$ then (\ref{psph3}) reduces to $\psi \vee \neg \psi$
which, as we have already observed, is an illegitimate proposition
for $\psi$ undecidable in T. Indeed, $\psi$ ($\neg \psi$) are
trivially theorems of T+$\psi$ (T+$\neg \psi$) and so the
criterion for such illegitimacy is satisfied. A similar analysis
for the legality of the proposition (\ref{psph}) in T applies.
Note that if $\psi$ were to be undecidable and $\phi$
provable in T, then (\ref{psph3}) is provable in T.
For another example, consider
\begin{equation}\label{psph2}
\psi \Rightarrow \phi \& \neg \phi,
\end{equation}
and suppose that $\phi$ is known to be undecidable in T. If
(\ref{psph2}) were a theorem of T, it would then follow even
in NAFL that $\psi$ is provably false in T. This is despite the
fact that $\phi \& \neg \phi$ \emph{does} obtain in T in a non-classical
model required by metatheorem~\ref{mt2}. Indeed,
under the hypothesis that $\psi$ were true in T
one could \emph{deduce} $\phi \& \neg \phi$ in T from (\ref{psph2}),
which \emph{must} be interpreted as a contradiction because the
notion of undecidability is not formalizable in T. This
automatically falsifies the said hypothesis and one obtains
a proof by contradiction of $\neg \psi$. It should be
emphasized that the rules of inference and the principles of proof
in NAFL are the same as in classical first order predicate logic
(except for certain special restrictions that will be mentioned
in the ensuing sections).
\end{remark}
\begin{remark}
An objection may be raised that $\Psi \vee \neg \Psi$ itself is
an undecidable proposition in T, but apparently the law of the
excluded middle has been applied to it; for there exist only the
classical models in which $\Psi \vee \neg \Psi$ is true, and the
non-classical models in which $\Psi \vee \neg \Psi$ is false.
Careful thought will show that it is only
``$\Psi \vee \neg \Psi$, but we do not know which'' that is
undecidable in this sense. But we have, at the beginning of this
section, banned such propositions as illegal in NAFL. Indeed, if the truth
or falseness of $\Psi$ in T is axiomatic, this proposition can never
be legal. Further, the negation of this proposition is precisely
$\Psi \& \neg \Psi$ which is also not formalizable in NAFL, as noted
earlier.
\end{remark}
\begin{remark}
Finally, we highlight an important distinction between classical
logic, intuitionistic/constructivistic logic and NAFL.
Consider the rhetorical question ``Which came first, the chicken
or the egg?''. Of course, in NAFL, this question can only be meaningfully
answered in some theory T. Let $\Psi$ formalize ``The chicken came
first'', and let $\neg \Psi$ formalize its negation ``The
egg came first''. Suppose $\Psi$ were undecidable in T. The
classical logician would then give the answer as ``Either the
chicken came first or the egg came first, but we do not know which''.
The intuitionist/constructivist, however, would consider the
question itself meaningless and would assert ``you tell me which
chicken and which egg, and I will tell you which of these came first''.
In other words, the non-constructive assertion of existence of chicken
and egg will not be permissible in intuitionism/constructivism.
In NAFL, we \emph{do} admit $\Psi$ as a proposition despite its
non-constructiveness, \emph{provided} there is no axiom of T
that implies $\Psi \vee \neg \Psi$. The answer in NAFL would be
``if T is consistent, then $\Psi \& \neg \Psi$ obtains, provided
we have not asserted either $\Psi$ or $\neg \Psi$ axiomatically''.
It is extremely important to note that the assumption
of undecidability of $\Psi$ automatically makes T consistent in
classical logic, but not in NAFL, as observed earlier.
\end{remark}
\section{Infinite sets cannot exist in theories of NAFL}
A \textit{class} is any collection of sets; here we are considering
set theory with classes and assume that the only mathematical
objects permitted to belong to classes are sets. A class is
specified as an \emph{extension} of
a definite property. Thus if $P(x)$
is one such property of sets, in
which all the bound variables range
only over sets, then we may denote the class $C$ associated
with it as $\forall x[x \in C \Leftrightarrow P(x)]$
or more concisely, $C = \{x:P(x)\}$. Set theory tells us that
certain classes are sets which may belong to other classes.
A \textit{proper class} is a collection
that is deemed to be not a set and
therefore cannot belong to any class. A necessary and sufficient
condition for a class to exist is that all of its elements should
exist as sets. It is therefore valid to speak of infinite
classes even in a theory that permits only finite sets.
The \textit{universe} of a model for a theory
will be taken here as a nonempty class $U$ over which
all the variables of the theory range.
Consider Zermelo's set theory, $\mathbf{Z}$,
in which classes are permitted.
For an elementary presentation of the axioms
of $\mathbf{Z}$ that uses classes, see
Machover~\cite{Mcr}, Ch.~1 and Ch.~5.
A rigorous justification of the notion of class is available,
for example, in G\"{o}del-Bernays set
theory ($\mathbf{GB}$) as presented in
Cohen~\cite{Coh}, pp.~73--78 and we will assume that the axioms for the
existence of classes discussed therein, on page 75, are
present in $\mathbf{Z}$. Following Cohen, we
will regard our system as having only class variables and a unary
relation, ``$X$ is a set'', which will be used whenever we wish to
indicate that the class in question, in upper case notation, is a set;
lower case notation, say $x$, is used only for classes which are also
sets and ``$x$ is a set'' is assumed to be understood. The negation
of ``$X$ is a set'' will be stated as ``$X$ is a proper class''.
The only other nonlogical symbol in our language is the membership
relation, $\in$, between sets and classes. The other notable features
of $\mathbf{Z}$ are that there is an axiom which asserts that only
sets can be members of classes, and that the axiom of
extensionality ($\mathrm{AE}$) is stated for classes, as follows:
any two classes having the same members are identical.
Upon dropping the axiom of infinity (AI) from $\mathbf{Z}$,
we obtain the theory of finite sets $\mathbf{F}$.
In what follows, we will study this theory in NAFL with the goal
of probing the existence of infinite sets.
That every property expressible in the language determines a class
is a theorem scheme of $\mathbf{GB}$ and in particular, of
$\mathbf{F}$ and $\mathbf{Z}$. We will denote this as the theorem
of comprehension for classes (theorem~TC):
\begin{theorem}
If $P(x)$ is a valid property (expressible in the language) of sets
in which all bound variables range only over sets, then the class
$\{x:P(x)\}$ exists.
\end{theorem}
See Cohen~\cite{Coh}, pp.~76--77, for a proof of this theorem in
$\mathbf{GB}$. The natural numbers in $\mathbf{F}$ are defined in the
usual set-theoretic sense, with $0$ as the empty set $\emptyset$
and $n+1=n \cup \{n\}$. The class $\mathbb{N}$ of all natural
numbers is defined by
\[
\mathbb{N}=\{x: x \; \mbox{is a natural number}\}.
\]
Note that we have admitted ``$x$ is a natural number''
(which we will henceforth abbreviate by ``$x \in \mathbb{N}$'')
as a valid property in our language. This guarantees the existence,
via theorem~TC, of $\mathbb{N}$ as an infinite class.
Consider the class $D$, defined as:
\begin{equation}\label{eq2}
D = \{x: \forall n[n \in \mathbb{N}
\Rightarrow x \notin \wp^{(n)}(\emptyset)]\}.
\end{equation}
Here $\wp^{(n)}(\emptyset)$ is the power set operation performed
$n$ times on the null set $\emptyset$; we
let $\wp^{(0)}(\emptyset)=\emptyset$. Note that $D$ is
guaranteed to exist by theorem~TC. Consider the
following proposition:
\begin{equation}\label{eq3}
D=\emptyset.
\end{equation}
According to the standard interpretation, (\ref{eq3}) is
undecidable in $\mathbf{F}$, for the following reason. Define a
model $\mathfrak{B}$ for $\mathbf{F}$ in which only finite
sets exist; the universe $B$ of $\mathfrak{B}$ is given by
\begin{equation}\label{eq4}
B = \{x: \exists n[n \in \mathbb{N} \; \& \;
x \in \wp^{(n)}(\emptyset)]\}.
\end{equation}
In the interpretation of $\mathfrak{B}$, $D=\emptyset$
holds and hence (\ref{eq3}) is true. But in a model
in which infinite sets do exist, (\ref{eq3}) is false.
Note that the undecidability of (\ref{eq3}) in $\mathbf{F}$
is a direct consequence of the fact that in the absence of
AI, the question of the existence of infinite sets
is undecidable according to the conventional wisdom which holds
that AI is indeed a valid axiom. Of course, according
to the finitists (of whom Kronecker was certainly the most
famous and uncompromising), AI \emph{is}
unambiguously false and hence (\ref{eq3})
ought to be provably true in $\mathbf{F}$. In what follows,
we demonstrate that Kronecker's vision is indeed justified
in NAFL.
Uniqueness of $D$ is deducible from AE, which implies:
\begin{equation}\label{eq5}
D=\emptyset \quad \vee \quad D \ne \emptyset.
\end{equation}
The key point is that AE \emph{requires (\ref{eq5})
to be a theorem of $\mathbf{F}$, even in NAFL.
But by metatheorem~\ref{mt1} and metatheorem~\ref{mt2},
(\ref{eq5}) cannot be a theorem of $\mathbf{F}$, given
that $\mathbf{F}$ is consistent and that (\ref{eq3}) is
undecidable in $\mathbf{F}$}. We are forced to the conclusion
that one of these two ``givens'' must be false in NAFL. Either
(\ref{eq3}) is undecidable in $\mathbf{F}$, which is therefore
inconsistent, or else (\ref{eq3}) \emph{is}, after all,
decidable in $\mathbf{F}$. To see why AE demands (\ref{eq5}),
see the discussion of (\ref{psph}), with $\psi$ interpreted
as $D=\emptyset$ and $\phi$ as the negation of AI (i.e., infinite
sets do not exist); this interpretation makes (\ref{psph})
equivalent to an instance of AE. Hence
$D=\emptyset \; \& \; D \ne \emptyset$
does imply a violation of AE, even in NAFL.
What we have demonstrated is that if $\mathbf{F}$ is to be a
consistent theory of NAFL, then we need a principle of proof
for either $D=\emptyset$ or for $D \ne \emptyset$. Of course,
the question then arises, which one? We will first argue the
case for $D=\emptyset$. In the following paragraph, we assert
that infinite sets can only be defined impredicatively, i.e.,
by self-reference via the universal class (which must again
be defined impredicatively by reference to infinite sets).
The appropriate prinicple of proof in NAFL for $D=\emptyset$
is that such impredicative definitions must be banned if they
indeed cause a contradiction.
The axioms of $\mathbf{F}$ tacitly presume that a definite class $U$
is specified as the universe, because
these formulas contain the universal quantifiers $\forall$ and
$\exists$ which may be thought of as referring to the universe.
However, to define $U$ we need to consider the formulas
of $\mathbf{F}$ as meaningful because these assert
the existence of sets that must necessarily
belong to $U$. Thus both $U$ and all the sets of $\mathbf{F}$ are
impredicatively defined. However, finite sets may also be defined
predicatively, at least in principle, merely by listing all of
their elements; but this is not possible for infinite sets.
Note that an equivalent method for deducing a contradiction is
to consider the proposition
\begin{equation}\label{eq7}
U=B,
\end{equation}
where $U=\{x:x=x\}$ and $B$ is defined by (\ref{eq4}).
Here $U$ is the universal class which is guaranteed to
exist by theorem~TC and must be uniquely defined
by AE. In NAFL, the undecidability of (\ref{eq7}) in
$\mathbf{F}$ contradicts AE in exactly the same manner as before.
We now examine the question of whether there is a case for
deducing a principle of proof in NAFL for $D \ne \emptyset$ instead.
We assert that this is not so. Firstly, any such principle
in $\mathbf{F}$ would amount to requiring an absolute infinity.
The problem is that even if one does manage to work out such a
principle, there will always exist
other propositions in NAFL theories permitting infinite
sets whose undecidability will produce a conflict with AE.
An example is considered in the ensuing paragraph.
To kill two birds with one stone, revert to set theory with properties
instead of classes; the objective is to demonstrate that the power set
axiom leads to a contradicion similar to that described earlier
even in standard set theory, with the power set playing the
role of the universe. Of course, this contradiction is trivially
demonstrable in set theory with classes as well.
Consider conventional Zermelo-Fraenkel
set theory $\mathbf{ZF}$. Denote the axiom of
constructibility~\cite{Coh} by ACN. It is
known from G\"odel's work that if $\mathbf{ZF}$ is consistent,
then the theory $\mathbf{ZF}+\mbox{ACN}$ is consistent.
Define the set $c$ by
\[
c=\{x \in \wp(\mathbb{N}): x \mbox{ is not constructible}\},
\]
where $\mathbb{N}$ is the set of all natural numbers and the
property ``$x$ is constructible'' may be taken as short-hand
for the usual definition of constructibility (see Cohen~\cite{Coh}).
Consider the following proposition:
\begin{equation}\label{eq9}
c=\emptyset.
\end{equation}
It is easy to see that (\ref{eq9}) must be undecidable in
$\mathbf{ZF}$. If the universe is constructible (i.e., if
ACN holds), then $c=\emptyset$ and not otherwise.
Given that the axiom of subsets in $\mathbf{ZF}$ affirms the
existence of $c$, we may conclude that in NAFL, the undecidability
of (\ref{eq9}) in $\mathbf{ZF}$ constitutes a violation of
AE, now stated for sets. The arguments are identical
to those given earlier. If either (\ref{eq9}) or its negation
is taken to be provable in $\mathbf{ZF}$, then the
contradiction would be that either ACN
or its negation would also have to be provable in
$\mathbf{ZF}$ and this is known to be impossible.
The reason for this contradiction can be traced to the impredicative
nature of the power set axiom (when applied to infinite sets),
which causes the definition
of the power set to be dependent on the definition of the
universe. The existence of infinite sets via AI
also plays a crucial role in the sense that the contradiction
can only be stated in terms of power sets of infinite sets.
\section{Is Peano arithmetic consistent in NAFL?}
First note that Peano arithmetic ($\mathbf{PA}$) is strongly equivalent
to the theory $\mathbf{F}$ (when restricted to statements about sets
only). In order to argue for the consistency of either $\mathbf{PA}$
or $\mathbf{F}$ in NAFL, one needs to assert the axioms of these theories
as \emph{tautologies}, i.e., they cannot be denied; this includes the
induction axiom of $\mathbf{PA}$. Otherwise it is possible to deduce
contradictions in these theories in the same manner as in the
previous section. A proposition
that is \emph{known} to be undecidable in $\mathbf{F}$ and $\mathbf{PA}$
is the celebrated Robertson-Seymour graph minor theorem (see
Ch.~12 of Diestel~\cite{Dst}). In NAFL
however, the undecidability of the graph minor theorem does
cause a contradiction in $\mathbf{F}$ and $\mathbf{PA}$, via
metatheorem~\ref{mt2}; again, this
can be demonstrated by the methods of the previous section.
One can escape this predicament in $\mathbf{F}$, for example, by
noting that the graph minor theorem requires quantification over
infinitely many infinite (proper) classes
of finite graphs. Thus one has to assert that such
quantification should be banned in the NAFL version of
$\mathbf{F}$, so that it would not be possible to formulate the
graph minor theorem. In $\mathbf{PA}$, the corresponding principle
would have to be the banning of the nesting of universal quantifiers
that is required to formulate the graph minor theorem. Quantification
over infinitely many infinite proper classes should be interpreted
in NAFL as speaking of these infinite classes as ``completed objects'',
i.e., as infinite sets, which are illegal in NAFL. An extremely
important point to note, however, is that the classical proof of the
undecidability of the graph minor theorem in
$\mathbf{F}$ and $\mathbf{PA}$ is presumably executed in a meta-theory
that permits infinite sets and other techniques that are not acceptable
(and so would not carry conviction) in NAFL.
A convincing meta-theory for establishing undecidability results in
NAFL remains to be developed. An obvious starting point for such a
meta-theory is the observation that every proposition is undecidable
with respect to the null set of axioms.
There are many other well-known conjectures, such as, Fermat's last
theorem, the twin prime conjecture, Goldbach's conjecture, etc.,
whose undecidability in $\mathbf{F}$ and $\mathbf{PA}$ (not
established at present) would cause contradictions in NAFL which
may be routinely established as in the previous section; the requirement
of the existence of a non-classical model via metatheorem~\ref{mt2}
is the cause for such contradictions.
The intuitive reason for these contradictions is as follows.
All these conjectures (including the graph minor theorem), if indeed
undecidable in $\mathbf{F}$ and $\mathbf{PA}$, would have to be ``true''.
But in NAFL, ``true'' has no meaning independent of axiomatic
theories. Hence our intuition that these curiously ``true but
undecidable'' propositions are paradoxical is confirmed in NAFL.
However, in the absence of any metamathematical proof that these
conjectures are indeed undecidable, their status as valid/invalid
propositions in the NAFL versions of $\mathbf{F}$ and $\mathbf{PA}$
will remain unclear. The safe strategy would be to admit these
as valid propositions if and only if proofs or refutations are
found in $\mathbf{F}$ and $\mathbf{PA}$. The conclusion is that
NAFL is a very severely restricted logic, with only ultra-finitism
being more restrictive.
\section{Implications for other theories}
First consider Euclid's geometry. It is easily shown that the
undecidability of Euclid's fifth postulate (with respect to
the other four postulates) produces a contradiction in an NAFL
theory comprising the first four postulates; the non-classical
model required by metatheorem~\ref{mt2} cannot exist. Therefore all five
axioms of Euclid must be asserted as \emph{tautologies} in NAFL
and non-Euclidean geometries would become inconsistent. The reason
we must abandon non-Euclidean geometries in NAFL is that they
can be formulated only by appeal to self-reference. Thus space is
``curved'' in non-Euclidean geometries, but curvature itself is
a notion that cannot be formulated independent of space. Similarly,
in the general theory of relativity, curvature of spacetime is
defined in terms of gravity and gravity is defined in terms of
the curvature of spacetime. This self-reference causes unavoidable
contradictions (via undecidable propositions) in NAFL theories that
cannot be fixed, just as in the case of infinite sets in the
previous section. Secondly, note that Turing's undecidability
of the halting problem cannot be formalized in NAFL, for the same
reason that the graph minor theorem could not be formalized --
quantification over infinitely many infinite (proper) classes,
required to establish such undecidability,
must be banned in NAFL. A Turing machine, \emph{by definition},
must either halt or not halt and undecidability of the halting
problem in NAFL would contradict this (Aristotelian) assumption.
Once again, NAFL rejects the ``true but undecidable''
assertion that certain Turing machines do not halt. The
diagonalization principle used by Turing in his proof (and
by G\"{o}del in proving his incompleteness theorems~\cite{Gd})
must therefore be banned in NAFL. Hence G\"odel's ``true but undecidable''
propositions are illegitimate in NAFL and his incompleteness theorems
do not apply in this logic.
Finally, consider an interesting implication for the
special theory of relativity (SR). Consider a hypothetical universe
in which there are only two material objects A and B, and suppose
that both A and B measure a relative acceleration
with respect to each other. First of all, note that SR requires either
A or B to be a inertial frame of reference because (a) the existence
of an inertial frame is an axiom of SR and (b) if neither A nor B
is an inertial frame, then one must point to empty space as the
location of the origin of the inertial frame; but this would
amount to postulating the existence of absolute space, which is
denied in SR. Let $\psi$ denote ``A is an inertial frame'' and
$\neg \psi$ denote its negation ``B is an inertial frame''.
Clearly, $\psi$ is undecidable in any classical formulation
of SR. In NAFL, such undecidability means that there must
exist a non-classical model for SR in which $\psi \& \neg \psi$
must obtain (by metatheorem~\ref{mt2}). But such a non-classical
model cannot exist. By an axiom of SR, both A and B cannot be
inertial frames because they are accelerating with respect to each
other. One concludes that SR cannot be formalized in any consistent
theory of NAFL. The non-existence of infinite sets in NAFL would
inevitably force this conclusion anyway.
\section{Concluding remarks}
The non-Aristotelian finitary logic (NAFL) developed in this paper
could perhaps be described as corresponding to the best dreams
of Kronecker and the worst nightmares of Cantor and Hilbert.
Clearly, NAFL is \emph{not} the logic in which Hilbert hoped
to justify classical infinitary reasoning (in his celebrated
``Hilbert's program''). It has been conjectured that primitive
recursion arithmetic is more likely to be the finitary logic
in which Hilbert hoped to implement his program. The positive
aspect of NAFL, however, is that it justifies and de-mystifies
the phenomenon of quantum superposition. In conclusion, the
main achievement of NAFL is that the question of validity of the
vast jungle of complex infinitary reasoning has been reduced
to the question of validity of the main postulate, without which
NAFL collapses back to classical logic. An interesting idea
would be to introduce the notion of para-consistency into NAFL
with the goal of enabling it to handle infinite sets. A second
possibility is to consider a radically new formulation of set
theory in NAFL, in which, say, the axiom of extensionality is
either dropped or suitably modified to permit infinite sets.
\section*{Dedication}
The author dedicates this research to his son R.~Anand and wife
R.~Jayanti.
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\end{document}