An LFI with a transparent truth predicate
Eduardo Barrio Damin Szmuc Federico Pailos
Abstract. We will present an LFI based on Coniglio &
Silvestrini [2014] first order three-valued matrix logic called
MPT1.There are two main differences with between MPT1 and the
matrix that here presented, that we will called MPTTT*. The first
one is that MPTTT* admits a transparent truth predicate. The second
one is that the conditionals of the two matrices treat differently
conditionals like, where v()=1/2 and v(). In MPT1, those
conditionals receive value 0. In MPTTT*, they are valued with .
This also has important consequences in the way they treat
biconditionals, and those consequences helps MPTTT* dealwith
self-reference sentences. In particular, with biconditionals that
can be read as expressing in the language The Liar or a Curry
sentence. But MPTTT* matrix is non-monotonic, and this makes harder
finding a fixed-point interpretation of the truth predicate, and
thus proving the non-triviality of the theory. In order to prove
this, and also prove the completeness of it, we will use a
three-side disjunctive sequent system, based on the one that Ripley
[2012] use to prove the completeness of his paraconsistent truth
theory STTT. We will present a semantics for the disjunctive
sequents that traduces MPTTT into a disjunctive sequent language,
and then show that:
MPTTT* iff MPTTT** is valid.
The, we will show that MPTTT** is non-trivial, and that will
show that MPTTT* is it, also is non-trivial. This proof will
involve a cut-elimination proof for MPTTT*. This is an induction
proof over the index of a cutproof, and adopts notions developed in
Paoli [2013]. Finally, we will prove that MPTTT** is complete. The
strategy employed will be similar to the one Ripley [2012] uses to
show STTTs completeness.
I-INTRODUCTION:
The logics of formal inconsistency (LFIs) are powerful
paraconsistent logics that encode classical logic and allow us to
fix an interesting distinction between contradictions and
inconsistencies. These systems, introduced by Carnielli and Marcos
[2002], internalize the metatheoretical notion of consistency,
expressing it in the object language. Hence one can isolate
contradictions in such a way that the application of the principle
of explosion is restricted to consistent sentences only, thus
avoiding triviality. This is achieved by means of adding to a
collection of appropriate axioms and rules already accepted in
classical propositional logic a restricted principle of
explosion,
oA, A, A B
where oA means that A is consistent. If A is not consistent,
explosion cannot be applied. Classical reasoning can be restored
into LFIs. The inferential behavior in consistent fragment of the
language of the LFIs is completely classical. But, can we add a
truth predicate to aLFI? Most paraconsistent logics cannot handle
at the same time a transparent truth predicate and a consistent (or
inconsistent) operator. XXXXX AC FALTARAN EJEMPLOS XXXXX We are
going to explore the possibility of doing exactly that with a
slight modification of MPT1, LPT1s semantic counterpart, one of the
most well known LFIs. We will introduce a new conditional that is
in principle capable to avoid triviality even in presence of
transparent truth and consistency. Modifying the 3-valued matrices
of MPT1, we will get stable valuations for the Strengthened Liar
and Currys sentence. The main difference with MPT1s conditional is
that when the antecedent receive value , and the consequent
receives value 0, then the conditional receives (the designated
value) . In this way, we may recover biconditionals that mimic
problematic instances of the diagonal lemma, even the ones with the
truth predicate or the consistent operator.
II.- Reasoning with truth and inconsistencies
Para-complete and para-consistent theories wind up with a
non-classical material conditional, where a material conditional A
B is defined as A v B.
Para-complete: A A.
Para-consistent: A, A B B.
Hence, in either case, the resulting material conditional is
often thought to be inadequate. In the para-complete case, the
given conditional detaches (i.e., satisfies Modus Ponens) but fails
to support all instances of the given (material) T-schema: Tr() A
and its converse can fail. In the para-consistent case, all
instances of the given (material) T-schema hold; however, the given
conditional fails to detach. As a result of these apparent
deficiencies, much of the work in para-consistent and para-complete
responses to paradox has focused on supplementing such theories
with a suitable conditional, one that both detaches and validates
all T-biconditionals [Beall, 2009, Field, 2008, Priest, 2006]. But
the task is complicated. What makes the task particularly difficult
is Currys paradox, which involves (conditional) sentences that say
of themselves that if they are true then absurdity is true (e.g.,
that everything is true). We need a conditional detachable but
Curry-paradoxical-safe.
An alternative strategy could be found adopting the Logics of
Formal Inconsistency. These kind of systemsare logics able to
internalize, in a precise sense, the notions of consistency and
inconsistency at the object-language level. They do it by
introducing primitive unary connective. As the strategy followed by
[Priest, 2006] and [Beall, 2009], such logics are para-consistent
in the following sense: given a contradiction of the form (A A), it
is not possible in general to deduce an arbitrary formula B from
the contradiction. That is, such logics do not fall into deductive
triviality when exposed to a contradiction. An LFI explodes if A, A
and oA occur simultaneously, for some arbitrary A, such that oA
expresses the fact that A is consistent. Thus, contradictions can
be isolated in such a way that the application of the principle of
explosion is restricted to consistent sentences only, thus avoiding
triviality. This is done in different ways. We are going to explore
oneoption: the system LPT introduced by Coniglio&Silvestrini
[2014] in which the semantics will be given by 3-valued matrices.
This matrix logic will be denoted by MPT. A first-order version of
LPT, LPT1, is also presented adding axioms and inferential rules
for quantifiers. LPT may be axiomatized by the following schemas of
a Hilbert calculus. Consider the positive intuitionistic sentential
logic (PISL):
Axiom Schemas
(A1) A(BA)
(A2) (AB)((A(BC))(AC))
(A3) A(B(AB))
(A4) (AB)A
(A4) (AB)B
(A5) A(AB)
(A5) B(AB)
(A6) (AC)((BC)((AB)C))
plus the following axiom-schemas:
(A7) A v (A B)
(A8) A v A
Now, adding the following axioms:
(A9) AA
(A10) oA(A(AB))
(A11) oA(AA)
(A12) o(AB)
(A13) (oAoB)o(AB)
(A14) (AAB)((AB)(BA))
and the Rule of inference:
(MP) infer B from A and AB.
The resulting axiomatic system will be called LPT.
Coniglio&Silvestrini [2014] present a complete and sound
semantics for LPT. The truth-tables of MPT can be constructed
straightforwardly:
The truth-tables of the defined connectives are given below:
It can also be shown that the LPT is sound and complete with
respect to a paraconsistentbivaluation semantics, i.e. truth
functions (not truth-functional) that assign, for each sentence of
the language, a truth-value 1 or 0. The system LPT1, the
first-order version of LPT, will be defined adding axioms and
inferential rules for the quantifiers. XXXX CULES?XXX
Could we use LPT1 to talk about truth? More specifically,
assuming the diagonal lemma, could we add the unrestricted validity
of the T-schema to LPT1 into a semantically self-sufficiency
languages without leading to a trivialization? We are going to show
that the response in negative: this logic is trivial in presence of
the truth predicate.
Theorem 1: Adding the instance of the diagonal lemma known as
the Curry sentence C(T(C))- to LPT + CAPTURE + RELEASE, lead to
trivialization.
Proof:
1)C(T(C))The Curry Sentence
2)C(C)Release, 1
3)(C(C))(C)Absorption
4)CModus Ponens, 2, 3
5)T(C)Capture, 4
6)CModus Ponens, 1, 5
7)Modus Ponens, 4, 6
Modus Ponens is a valid rule of LPT. What about Absorption? Its
also valid, as Absorption is a tautology of MPT, and LPT is
complete with respect to MPT.
Corollary 1: Adding all instance of the diagonal lemma with the
truth predicate to LPT + CAPTURE + RELEASE, lead to
trivialization.
Proof: The Curry sentence is one of those instances.
Now we are going to consider LPT1 and MPT1.
Let ~ be the strong negation operator that is defined in
Coniglio& Da Cruz 2013. Can we add to the MTP1 they defined, a
transparent truth operator? Lets see why not. Well start with the
instance of the Diagonal Lemma known that represents (or is) the
Liar sentence:
LT(L).
Its obvious that it cannot get a classical value. Can it get the
value ? If that were the case, then well get and equivalence such
that the first member of it gets value , but the second one
(because of the way the strong negation behave) will get value 0,
and so the equivalence will get value 0. But then well have an
instance of the Diagonal Lemma that doesnt receive a designated
value, which seems an undesirable result.
If we want to avoid this result, some changes need to be made.
One inmediate option is to change the meaning of (at least) one of
the operators involved in the sentence. We dont want to touch the
strong negation, at least if we want something like a consistent
operator in the language. But maybe we can do something with the
new conditional. In particular, lets see what happens when we
define a new one exactly like that one, but such that if the
antecedent gets value and the consequent gets value 0, then the
conditional receives the value -a designated one. This is how the
new new conditional will behave.
1
0
1
1
1
0
1
1
0
1
1
1
It replaces Coniglio& Da Cruz operator, that functions this
way:
1
0
1
1
1
0
1
1
0
0
1
1
1
If one adopts this matrix, with this new conditional, one thing
you lose is that the strong negation operator can no longer be
defined with the (new) conditional and . But one can still add it
as a primitive constant, and specify its behavior in terms of the
truth table already shown.
Lets see now what will happen with Currys paradox. It begins
with an instance of the diagonal lemma like the one shown below,
known as the Curry sentence:
C(T(C))
What can its value be? Once again, it cant receive a classical
value. But if its value is , then the conditional that is the
second term of the equivalence will get the value , and so the
biconditional will get the value 1.
Will this matrix, with this new conditional, be a LFI? Yes, it
will. As its build up to Coniglio & Da Cruzs one, then validity
is understood as designated-value preservation, and 1 and are both
designated values. One might think otherwise, because the Explosion
Principle will be valid in this new matrix. This is the Explosion
Principle: A(AB). It will have no counterexamples. If v(A)= 0, then
the whole conditional will have value 1. If v(A)=1, v(A)=0, then
v(AB)=1, and then v(A(AB))=1. If v(A)=1/2, v(A)=1/2, then v(AB)=1,
1/2, and then v(A(AB))=1.
Nevertheless, this is not a big issue. According to Coniglio y
Da Cruzs definition 3.1, a logic is a LFI if two clauses are
satisfied. The second one demands that the gentle explosion
principle is satisfied. This is the GEP:
A, A, A B
And in this matrix, it is. A possible counterexample must give a
designated value to every premise. But if v(A)=1/2, then v(A)=0. So
v(A)=1, 0. But then, either v(A)=1, and so v(A)=0, or v(A)=1, but
then v(A)=0. So there is no valuation such that every premise
receives a designated value, and the inference will have no
counterexamples.
The first condition demands that the Explosion RULE must be
invalid. So,
A, A B
We havent still build a proof system for this matrix. Anyway,
the idea will be that the proof system reflects what happens in the
matrix. And in it, this happens:
A, A B
Theres no stable truth-value assignment that gives 1 to all the
premises, nor one that gives 1 to one of them and to the other.
But, naturally, there are valuations that assigns to every premise.
And, of course, nothing forbids that the conclusion receive value 0
(unless it is a tautology).
As weve seen, the weak negation and the conjunction will be
define in the new matrix lets call it MTP*- in the same way as in
MTP. The only variation in the basic constants is found in the
conditional, and in particular in just one case: where the
antecedent receives value and the consequent receives value 0. The
new truth-conditions will give the (new) conditional (the
designated) value , where in the previous case it receives value
0.
With all these constants, its possible to define disjunction,
top, bottom and the biconditional. This last one will, of course,
behave different in some cases. Lets take alook at them. The
following correspond to the new biconditional:
1
0
1
1
1
0
1
1
0
0
1
1
This will replace the Coniglio& Da Cruz one:
1
0
1
1
1
0
1
1
0
0
0
0
1
On the other side, its no longer possible to define the strong
negation in terms of a conditional that has the negated formula has
the antecedent, and bottom as the consequent, as Coniglio and Da
Cruz did, because that conditional, if the negated formula receives
value , will no longer get value 0. Nevertheless, one can specify
its meaning in a direct fashion, in this way:
=-, +u,
With this strong negation operator, we can define a consistent
operator in the same way as Coniglio y Da Cruz did, using both
negations.
Will this new matrix make valid all the axioms of LTP the logic
that is correct and complete with respect to the matrix logic MPT?
LTP has sixteen axioms, A1-A16. This new semantic will make valid
all of them, but two: A2 y A12.
This results are expected, because the two of them involved the
(new) conditional, that is such that there are combinations of the
values of the antecedent and the consequent that make the
conditional false. So there will be less valuations that make false
a conditional. And this is the key feature of the conditional, that
allows MPT* to get rid of Currys paradox.
On the other side, the failure of A2 corresponds to a failure of
Modus Ponens A, ABB- in the new semantics. In order to see it, just
take a valuation v such that v(A)=1/2 and v(B)=0. A12 is invalid
because now there will be more conditional like the previous one-
that get a designated value -1/2, to be more specific. So now not
every conditional will be true of false, but, why should they? What
may be special with the conditionals? In this new framework, is not
the case that every conditional receives value 1 or 0 as many of
the formulae of the language.
Anyway, the goal is to build an analogue of LPT lets call it
LPT*- that is correct and complete with respect to MPT* -which is
analogue to MPT in the way already specified.Unfortunately, MPT*
doesnt validate Modus Ponens neither.(Counterexample: L, L).
Let MPTTT be a semantics that works as MPT1s one, but, instead
of MPT1s conditional, has MPT*s one. Also, MPTTT will work with a
language without the identity predicate, so this is another
difference from MPT1. The goal will be to prove that we can have a
LFI such that has both a consistent operator and a transparent
truth predicate, without getting into trouble. We will see how much
close to it we might get.
3-A COMPLETENESS PROOF FOR A LFI WITH A TRANSPARENT TRUTH
PREDICATE
We will work with two languages: L and L+. L is the base
language, an ordinary first order language without identity plus
and , and a consistency operator . L+ is L plus a distinguished
unary truth predicate T. To ensure that there will be paradoxes
around, the interpretation of L and L+ will be partially
constrained. In particular, the individual constants will come
divided into two countable sets. The members of first set will
function as the usual sort of individual constant, receiving their
denotation from each model. We will call them ordinary names. The
members of the other set will receive denotations independently. To
simplify things, we will not allow the ordinary names to name any
formula.[footnoteRef:2]We will fix a quote-name-forming deviceQsuch
that Q(A), for any formula A, is a singular term that will denote A
in any model. Well write A for the distinguished name of
A.[footnoteRef:3]All this restrictions will have as a result that
L+ will be a subset of the domain of every model. So we will work
only with infinite models. [2: This is just to avoid the problems
that contingent self-reference might generate. A slight
modification of the system we are about to defend will handle those
problems, but also will result in a system a little bit more
obscure. Since contingent self-reference is not the special target
of this work, we will just keep things as simple as possible, and
restrict contingent self-reference the way we have specify.] [3: A
key advantage of this name-forming device over the one that Ripley
uses in xxxx, is that this kind of procedure is purely syntactic.
Ripleys one, on the other hand, depends on a particular
meta-linguistic function. But, why not use arithmetic? It has some
obvious advantages, but also other major disadvantages. The main
one is that such a setting would be too rich to allow for a
completeness proof, which is what we ultimately are looking for.
]
Still, if we want to achieve self-reference in S, we need to
make some adjustments. Specifically, we will expand the vocabulary
with propositional constants. This will also be a key feature in
the self-reference sentence formation procedure.In particular, we
will have instances of following sentences as formulas:
p(p)
Where p is a variable that range over the new set of
propositional variables introduced. The bicondicional sign , of
course, represent just a conjunction of
conditionals.[footnoteRef:4](p) is a metavariable of a formula of
any complexity, which includes an atomic subformula of the form
B(p). This guarantee that something like The Liar sentence or
Currys sentence are represented in the language. But we need a
further restriction. That restriction will be semantical, and is
the following. We will select, for each (x) where (x) is a formula
with one free variable-, a biconditional of the form [4: Those
biconditionals can be read as a way to mimic instances of the
diagonal lemma, that are themselves supposed to be a way to achieve
self-reference. But we are not especially interested in that
particular way to get self-reference, just to achieve it in some
way.]
p(p)
Lets call that set of biconditional, Z. And we will restrict our
set of models to the ones that assign a designated value to each
one of the members of Z.
The following sentence, for example, may belong to that set:
pT( p )
So we will only consider models that assign a designated value
to it there will be such, as we have seen. Just assign to p, and
then the biconditional will get value 1.If this sentence is part of
that set that expresses the Liar sentence-, then the following one
wont be:
pT( p )
But of course, there will be, for example, another one that sort
of expresses the Truthteller. For example, this one:
q T( q )
In the rest of the paper, we will prove the correctness and
completeness of MPTTT.W will use as the target proof theory, a
special disjuntive sequent system S. Lets make the following
specification with respect to the metalanguage that will be
used.Let A and B be any formulas, and , , , , , , , , be any set of
formulas. Also, for the -rules, let t be any term, and let a be a
variable not occurring in the rules conclusion sequent. Now we need
to specify what disjunctive sequents are.
Definition. A disjunctive sequent is satisfied by a model M=D, I
iff either I()=0 for some or I()=1/2 for some or I()=1 for some . A
sequent is valid iff it is satisfied by every model. A model is a
counterexample to a sequent iff it does not satisfy the
sequent.
The disjuntivesequents will be finite e.g., it will only include
a finite number of formulas. So the correctness and completeness
proofs will be relative to the inferences that only involved finite
sequents of MPTTT.
MPTTT consequence relation is just LPs one what Ripley XXXX
calls a tolerant-tolerant notion of logical consequence. So an
inference from to will be valid in MPTTT if and only if there is no
model such that every formula in receives a designated value (1, )
and some formula in receive an undesignated value (0). But there is
a strong relation between inferences as are usually understand, as
a relation between one set of formulae and another[footnoteRef:5]
things with the following structure: MPTTT - and the
disjuntivesequents we have presented. That relation is the
following: [5: It might be defended that inferences, as are usually
understand, involved just formulas (and not sets of them) as
conclusions. Fair enough. Nothing on the approach defended depends
on that. If you think that that is the right position, just read
the conclusion as a single formula, or just singletons of a single
formula.]
MPTTT iff MPTTT MPTTT is valid[footnoteRef:6] [6: In the rest of
the article we will just give away the sub-indexes from the
disjunctive sequents.]
The proof just follows from the definition of MPTTT validity and
validity from disjunctive sequents.[footnoteRef:7] [7: Still, there
may be disjunctive valid sequents that do not have that form. For
example, if a formula A receives value in every admissible model,
the A A will be valid. But so will A . But our completeness proof,
is not about the disjuntivesequents proof system and its semantics,
but about semantically valid inferences in MPTTT and inferences
that have a proof in this disjunctive sequent system. So, more
preciselly, what we will prove is that:MPTTT iff is provable.We
will prove that by proving that:(For all sequents ) A sequent is
provable iff it is valid.So what about sequents like p (p) ? Some
of them will be valid, but they may not have a proof. Nevertheless,
if a sequent like that is valid, then also will be valid a sequent
like this: p (p) p (p) And if that is the case, then that sequent
will also have a proof, and then it will both be the case that
MPTTT p (p) and MPTTT p (p). So completeness and soundness will be
preserve. Well see all these more in detail in the next pages.
]
The proof system Swe are about to present consists on a number
of axioms and rules. A sequent is provable iff it follows from the
axioms by some number (possibly 0)of applications of the rules.As
we are working with sets, the effects of Exchange and Contraction
are built in, and Weakening is built into the axioms.
S has the following axioms and rules:
Axioms:
*For every formula A,
, A , A, A
Is an axiom.
*For every formula A,
, ~A , ~A
Is an axiom.
*For every formula A,
, A , A
Is an axiom.[footnoteRef:8] [8: These last two axiom-schema can
be read as saying that every formula receives at least one of the
three truth values, associated in turn with one of the three sides
of a sequent. ]
*For every formula p (p) that belongs to Z,
, p (p) , p (p)
Is an axiom.
Structural Rules:
Cut 1
, A, A
Cut 2
, A , A
Cut 3
, A , A
Derived Cut
, A, A , A, A , A , A
Operational rules:
Left
, A
, A
Right
, A
, A
Middle
, A
, A
Left ~
, A
, ~A
Right ~
, A, A
, ~A
Left
, A, B
, AB
Middle
, A , A , B , B, A, B
, , , , ,AB, ,
Right
, A , B
, , , , AB
Left
, A , B
, , AB , ,
Middle
, A , B
, , , AB ,
Right
, A , B , B
, AB
Left o
, A
, oA
Right o
, A, A
, oA
Left
, A(t)
, xA(x)
Middle
, A(a), A(a) , A(t)
, , , xA(x), , A
Right
, A(a)
, xA(x)
Left T
, A
, T(A)
Middle T
, A
, T(A)
Right T
, A
, T(A)
As the rest of the connectives , , the disjunction and the
biconditional- can be defined in terms of the former, we wont
specify rules for them.
We may prove the following result:
(Soundness). If a sequent is provable, then it is valid.
Proof: The axioms are valid, and validity is preserved by the
rules, as can be checked without too much trouble.
But of course the difficult part is to prove completeness.
Following Ripley 2002 proof of the completeness of his disjunctive
sequent system with respect to ST+, we will use the method of
reduction trees, that yields, for any given sequent, either a proof
of that sequent, or a countermodel of it. (The method provides of a
way of building the eventual countermodel.)
We will introduce the notions of subsequent and sequent union,
which will be used in the proof:
(Definitions).A sequent S = is a subsequent of a sequent S =
(written S S) iff , , .
A sequent S = is the sequent union of a set of sequents {i i
i}iI (written S = iI i i i iff = iI {i}, = iI { i}, = iI { i},
The construction starts from a root sequent S0 = 0 0 0, and then
builds a tree in stages, applying at each stage all operational
rules that can be applied, plus Derived Cut in reverse, from the
conclusion sequent to the premise(s) sequent(s). We will have an
enumeration of the formulas and also an enumeration of terms, and
we will reduced at each stage all the formulas in the sequent,
starting from the one with the lowest number, then doing the same
with the one with the second lowest number, and move on in that
fashion until we finish with the one with the higher number in the
enumeration. If the formula appears in more than one side in the
sequent, then well start reducing the formula that appears on the
left side, then reduce the formula on the middle side, and finally
reduce the formula on the right. The final action of each stage n
will be an application of the rule of Derived Cut to the n-formula
in the enumeration. If we apply a multipremise rule, we will
generate more branches. If we apply single-premise rules, we just
extend the branch with one more leave. We will only only add
formulas at each stage, without erasing any of them, so every
branch will be ordered by the subsequent relation. Any branch that
has as it topmost sequent an axiom, will be closed. A branch that
is not close is open. We repeat the procedure until every branch is
closed, or until there is an infinite open branch. If every branch
is closed, then the tree itself will be a proof of the root
sequent. If there is an infinite open branch Z, we can use it to
construct a countermodel to the root sequent.
So, a little bit more formally, stage 0 will just be the root
sequent S0 = 0 0 0. If its an axiom, close the branch. For any
stage n+1, one of two things might happen:
1-For all branches in the tree after stage n, if the tip is an
axiom, close the branch.
2-For open branches: For each formula A in a sequent position in
each open branch, if A already occurred in that sequent position in
that branch (so A has not been generated during stage n+1), and if
A has not already been reduced during stage n+1, then reduced A as
follows:
If A is a negation B, then
*if A is in the (left/middle/right) position, extend the branch
by copying its current tip and adding B to the (right/middle/left)
position.
If A is a conjunction B C, then
*if A is in the left position, extend the branch by copying its
current tip and adding both B and C to the left position.
*ifA is in the middle position, split the branch in three:
extend the first by copying the current tip and adding B to both
the middle and right positions; extend the second by copying the
current tip and adding C to the middle and right positions; and
extend the third by copying the current tip and adding both B and C
to the middle position.
* if A is in the right position, split the branch in two: extend
the first by copying the current tip and adding B to the right
position; and extend the second by copying the current tip and
adding C to the right position.
If A is a universal quantification xB(x), then
*if A is in the left position, extend the branch by copying its
current tip and adding B(t) to the left position, where t is the
first term in the enumeration not already used in a reduction of A
in the left position before stage n + 1.
*if A is in the middle position, split the branch in two: extend
the first by copying the current tip and adding B(a) to both the
middle and right positions, where a is the first term in the
enumeration not to occur anywhere in the current tip; extend the
second by copying the current tip and adding B(t) to the middle
position, where t is the first term in the enumeration not already
used in a reduction of A in the middle position before stage n +
1.
*if A is in the right position, extend the branch by copying its
current tip and adding B(a) to the right position, where a is the
first term in the enumeration not to occur anywhere in the current
tip.
If A is a consistency assertion B, then
*if A is in the left position, extend the branch by copying its
current tip and adding B to the middleposition.
*if A is in the right position, extend the branch by copying its
current tip and adding B to theright and left positions.
*if A is in the middle position, then do nothing.
If A is a strong negation assertion ~B, then
*if A is in the left position, extend the branch by copying its
current tip and adding B to the right position.
*if A is in the right position, extend the branch by copying its
current tip and adding B to the middle and left positions.
*if A is in the middle position, then do nothing.
If A is a conditional B C, then
*if A is in the left position, split the branch in two: extend
the first by copying the current tip and adding B to right
position, and extend the second by copying the current tip and
adding C to the left position.
*if A is in the middle position, split the branch in two: extend
the first by copying the current tip and adding B to middle
position, and extend the second by copying the current tip and
adding C to the left position.
* if A is in the right position, extend the branch by copying
the current tip and adding B to the left position, and extend
adding C to the middle and right positions.
We will also apply the Derived Cut rule at each step. XXXX NO
ESTOY SEGURO DE QUE CUT NO SEA ELIMINABLE. PERO EN CUALQUIER CASO,
ES VLIDA, AS QUE PODEMOS USARLA A PLACER. Y ES TIL, PORQUE NOS
PERMITE DAR CON CONTRAMODELOS. XXXX Just take the nth formula in
the enumeration of formulas and call it A. Now extend each branch
using the rule of Derived Cut. So for each open branch, if its tip
is , split it in three and extend the new branches with , A , A , ,
A , A and , A , A, respectively.
Now we need to repeat this procedure until every branch is
closed, or, if that doesnt happen, until there is an infinite open
branch. If the first scenario is the actual one, then the tree
itself is a proof of the root sequent each step will just be the
result of an application of a structural or operational rule to the
previous step. If the second scenario is the actual one, we can use
the infinite open branch to build a countermodel. Lets see how to
do it.
If in fact there is an infinite open branch, then the Derived
Cut rule will have been used infinite many times, and so every
formula will appear at some (finite) point in the branch, and will
remain in every step afterwards, because no formula is lost in the
construction of the reduced tree. Also, every formula will appear
in (at least and at most) two places in S. Now, the first step will
be to collect all sequents of the infinite open branch B into one
single sequent S = = S: S is a sequent of B.
But S cannot be a sequent of our system S, because we are
working with finite sequents, and S is infinite. But it can be part
of an extension of the proof system S we are working with, that do
admit infinite sequents. This extension lets call it S*- will have
the same axioms and rules, but without the restriction that every
sequent should be finite.
As Derived Cut has been applied infinite many times in the
construction of the branch, every formula will occur in exactly two
places in S. It cannot occur in the three places, because thenthere
is some finite stage n where the formula is for the first time in
the branch in the three sides. But then that sequent will be an
axiom, and so the branch will be closed. Correspondingly, there
will be a model such that no formula in the sequent receives the
value associated with the place where it occur (0 if the formula
occur in the left, if it occur in the middle, 1 if it occur in the
right) so there will be acountermodel to it. Lets see this in
detail.
We will explain now how to design a countermodel to S. For each
formula en in the sequent, this valuation will givea different
value that the one that corresponds to its place in it. But that
includes all the formulas in the initial and finite sequent S0.
That valuation, then, will also be a countermodel to S0.
So now we need to specify a domain D and an interpretation I.
Let D = L+ -the language itself. (This is not essential for the
proof, though.)In order to build I, we need to ensure that no
formula receives the value corresponding to its location in S. And
this is how we will achieve this goal. For n-ary predicates P
(including the truth predicate T), let I(P) be as follows:
I(P)(I(s1),I(s2),...,I(sn)) = 0, , 1, respectively, i
P(s1,s2,...,sn) does not appear in //, respectively. Of course
P(s1,s2,...,sn) will appear in exactly two places in the sequent
because that will the effect of some application of the Derived Cut
rule-, and it cannot appear in the three of them, because otherwise
the sequent will be an axiom, and so the branch would eventually
have closed. One can design the interpretation of each predicate to
easily achieve this. Does P(s1,s2,...,sn) appears in exactly the
places where TP(s1,s2,...,sn) appears? Yes. As any formula in
sequent that corresponds to an infinite open branch,
P(s1,s2,...,sn) appears in exactly two places in the sequent. If
TP(s1,s2,...,sn) appears in the only place where P(s1,s2,...,sn)
does not appear, then, as TP(s1,s2,...,sn) will eventually be
reduced, P(s1,s2,...,sn) will appear in the only place where it
doesnt appear until that moment in the sequents of the branch. But
then that sequent will be an axiom, and so the branch will be
closed. This is the only possibility that we need to consider. Is
not as if TP(s1,s2,...,sn) can appear in less places that
P(s1,s2,...,sn): as any formula in a sequent corresponding to an
infinite open branch, it has to appear in exactly two places.
The rules by which we reduced formulas can be used to show by
induction that if none of the components of weak negations,
conjunctions and universal quantifications receives the value
associated with any place in which it appears in S, neither will
the compound.[footnoteRef:9]Take a weak negation A that appears on
the left side of the sequent, then A will appear in the right side
of it that will be an effect of the eventual reduction of A. So if
by inductive hypothesis,I(A) 1, then I(A) 0. Another example: if
xA(x) appears in the middle of the sequent, then for some term a,
either A(a) appears in both the middle and in the right, or else
A(t) appears in the middle of the sequent for every term t. If the
first one is the case, then I(A(a)) = 0, and so I(xA(x)) = 0. If
the second one is the case, then A(t) will appear in the middle of
the sequent for every t, no term t is such that I(A(t)) = 1/2, and
so I(xA(x)) 1/2. And either way, we have I(xA(x)) , and so does not
receive the value associated with the middle side of the sequent.
[9: As the same happens in Ripleys system, well use his
examples.]
What happens with the strong negations, the conditionals and the
consistency assertions? Letsstartwiththestrongnegations ~A. XXXX AC
FUI SUPER DETALLADO CON CADA DEMOSTRACIN, PERO POR SUPUESTO ES
DEMENTE PONER TODO ESTO EN EL ARTCULO FINAL. CON PONER UN EJEMPLO
DE CADA FRMULA CREO QUE ALCANZA.XXXX No formula like this can
appear both on the left and the right sides of the branch, because
then they would have appear at both sides in one sequent of the
branch. That sequent, then, will be an axiom, and so wont
correspond to an infinite open branch. So (i) either ~A is both in
the left and in the middle sides of the sequent, or (ii) it is both
in the middle and the right side of the sequent. Lets start with
(i). At some point, ~A will have been reduced. It appears in the
middle side of the sequent, and so nothing is supposed to be done
when this happens. But it also appears on the left side. Then A
will appear on the right side of the sequent on the next stage of
the construction. At some point, an application of Derived Cut will
also introduce A in the left or in the middle side, and so they
will appear in one of the two sides in S. (It cannot appear on the
three sides, because then the sequent will be an axiom.) If A
appears on the left and on the right side, then A will get value ,
and then ~A will get value 1. So neither of them receives a value
associated with one the sides where they appear. And the same
happens if A appears on the middle and on the right side, because
then A, by inductive hypothesis, will get value 0, and ~A will get
value 1. So once again neither of them receives a value associated
with one the sides where they appear. Now consider (ii). This case
is similar to the previous one. But one ~A is reduced, as it is on
the right side of the sequent, that will get A on both the left and
the middle sides. So by inductive hypothesis, A will get value 1,
and ~A will get value 0. So, again, none of these formulae receives
the value associated with the sides where they appear.
The cases of consistency assertions A will be very similar to
the strong negations ones. No formula like this can appear both on
the left and the right side, because then the sequent will be an
axiom, and so wont be part of an infinite open branch. So (i)
either A is both in the left and the middle sides of the sequent,
or (ii) it is both in the middle and the right side of the sequent.
Lets start with (i). At some point, A will be reduced. It appears
in the middle side of the sequent, so nothing is supposed to be
done when this happens. But it also appears on the left side. If
that happens, then A will appear on the middle side of the sequent
on the next stage of the construction. At some point, an
application of Derived Cut will also introduce A in the left or in
the right side, and so they will also appear in one of the two
sides in S. (It cannot appear on the three sides, because then the
sequent will be an axiom.) If A appears on the left, then A will
get value 1, an also will A. If it appears on the right, then it
will get value 0, and A will get value 1. So neither of them
receives a value associated with one the sides where they appear.
Now consider (ii). Once A is reduced, as it is on the right side of
the sequent, that will get A on both the left and the right sides.
So by inductive hypothesis, A will get value 1/2, and A will get
value 0. So, again, none of these formulae receives the value
associated with the sides where they appear.
Lets turn now to the cases of conditionals of the form BC. These
cases are different. We need now to consider three possible
situations: (i) either the conditional appears in both the left and
the right sides, or (ii) it appears both in the left and in the
middle sides, or (iii) it appears on the middle and the right
sides. So lets start with (i). This is a little bit tricky.
Eventually, BC will be reduced from a sequent like , BC , BC. The
reduction of the conditional on the right side will demand to copy
the current tip, and also the addition of B in the left, and the
addition of C in both the middle and the right sides of the
sequent. But as BC appears also in the left side, this demands to
split the branch in two, and to extend the first by copying the
current tip and adding B to the right position, and extend the
second by copying the current tip and adding C to the left
position. As we have established, we need to reduce first the ones
on the left, then the ones on the middle, and finally the ones of
the right side of the sequent. (Remember we are talking about
occurrences of the same formula, and which is the order to reduce
them at some particular stage n.) The result of reducing first the
sequent of the left, and then the one of the right, will be the
result of splitting the branch in two, and (1) extending the first
by copying the current tip and adding B to the left and right
position, but also add C to the middle and right position, and (2)
extending the second by copying the current tip and adding C to the
left, middle and right position, and also B to the left and right
positions. So the two new sequents will have the following
appearences:
(1) , BC, B , C , BC, B, C
(2) , BC, B, C , C , BC, C
So these are two new branches. The second one, (2), will be an
axiom, because the formula C appears in the three sides of the
sequent, and so that particular branch will be closed. But that
doesnt happen with (1). The complexity of B and C is less than the
one of BC, so the inductive hypothesis can be applied to them. So B
will get value , C will get value 0, and so BC will get value . So
none of these formulae receives in the valuation a value associated
with the sides of the sequent where they appear.
The second case is one where BC appears in both the left and the
middle sides. When we build the reduced tree, we start reducing the
occurrence of the conditional on the left side of the sequent. So
we split thebranch in two, extending the first by copying the
current tip and adding B to right position, and extending the
second by copying the current tip and adding C to the left
position. Then we need to reduce the occurrence of the conditional
on the middle side, and we will do that in each branch. So, in each
case, we split the branch in two, and extend the first by copying
the current tip and adding B to middle position, and extend the
second by copying the current tip and adding C to the left
position. So we start with a sequent like this:
, BC , BC .
Then we obtain these two new sequents:
(I), BC, C , BC
(II) , BC , BC , B .
The extension of (I) will produce these two new sequents:
(I), BC, C , BC
(I), BC, C , BC, B
On the other hand, the extension of (II) will produce these two
new sequents:
(II) , BC, C , BC , B
(II) , BC , BC, B , B
But no formula in (I), (I), (II) and (II) if any of them belong
to an infinite open branch- will receive the value associated with
any place in which it appears. Lets see why. In the (I)s case, the
infinite extension of , BC, C , BC will include occurrences of B in
two of the three sequents but, of course, not in the three of
them-, and one more occurrence of C on the middle or in the right
sequent. We need to make sure that the conditional gets value 1.
But we can guarantee that if C appears on the middle or on the
right side, that will indeed be the case. And that will just be the
case, because the branch is infinitely open. So by inductive
hypothesis, C will get value , if it also appears at some point on
the right side, or 1, if it eventually appears on the middle side.
B, also by inductive hypothesis, will get value 0//1, iff it does
not appear nowhere in the left/middle/right side of the
sequent.
In (I), the infinite extension of , BC, C, BC, B will include
occurrences of B in the left or in the right side, but not in both
of them, and will also include cases of C in the middle or in the
right, but not in both of them.We need to make sure that BC will
receive value 1, but if C in the middle, then by inductive
hypothesis, it will get value 1, and so will the conditional. If C
in appears on the right side, then by inductivehypothesis, it will
get value 1/2, and so the conditional will get value 1.
What about (II)? The infinite extension of , BC, C , BC , B will
get similar results as the infinite extension of (I), because
either C will not be in the middle, and so willget value , or it
will not be on the right. In either case, the conditional will get
value 1. Thecase with (II) is a little bit different. In , BC , BC,
B , B , by inductive hypothesis, as Bwill not be on the left side
in its infinite extension, it will get value 0, and so the
conditionalwill get value 1, no matter what value C receives.
Now we need to consider the third case, where the conditional
appears on the middle and the right sides, like this: , BC , BC. As
the conditional is in the middle side, when we reduce it we will
get two new sequents, both of them the result of copying the
current tip and adding some formulas: the first one has also B in
the middle side; the second one has C on the left side:
(1) , BC, B , BC
(2) , C , BC , BC
As the conditional is in both of them on the right side,
eventually it will be reduced. The result, in both cases, will be a
new sequent that copy the current tip, and add B to the left side,
and C to the middle and the right side of the sequent. So well
get:
(1), B , BC, B, C , BC, C
(2), B, C , BC, C , BC, C
In (1), we can apply the inductive hypothesis and obtain the
desired result. B will get value 1, because it is in the left and
the middle sides, but not in the right. And C, as it is on the
middle and the right sides, will get value 0. And so the
conditional will get value 0. C, in (2), will be in each side of
the sequent, and so the branch will be close. But this is not a
problem, because (1) will still be part of an infinite open
branch.
By completing the induction along these lines, we can show that
we have a model on which no formula receives the value associated
with any place in which it appears in S. But, as we know, that
includes of the formulas in the initial and finite sequent S0. That
valuation, then, will also be a countermodel to S0, which is what
we were searching for. So we have just establish that for any
sequent S, either it has a proof or it has a countermodel.
Conclusion
XXXX DESPUS DE ESTABLECER ESTO Y LO DEL CONDICIONAL, TENS QUE
METER LO DE DERIVED CUT, Y HABLAR ACERCA DEL ORDEN. EN ESTE
MOMENTO, LO MS SENSATO ME PARECE INTERCALAR UNA REDUCCIN NORMAL, Y
OTRA POR DERIVED CUT.
XXXXX
Notas:
Faltan: (i) Una prueba de no-trivialidad, y (ii) una prueba de
que Cut no es eliminable.
El punto (ii) no es necesario. Cut es semnticamente vlido, y si
no es eliminable, entonces es vlido tambin desde el punto de vista
sintctico. Por qu sera relevante probar que no es eliminable?
Porque ent va a haber secuentes que no puedan ser probados sin Cut,
y eso justifica este sistema ms complejo, con tres reglas
estructurales (en principio) de ms. Pero igual se puede usar Cut
donde se lo necesite, porque en cualquier caso va a ser vlido. Y lo
vamos a utilizar, en particular, para probar completitud. (Ms
especficamente, para construir el contramodelo.)
Si tuviramos prueba de (i), estaramos hechos. El modo ms
sencillo de hacer dado que la prueba de punto fijo del predicado de
verdad no parece sencilla de conseguir- es probando que hay un
secuente que no puede ser probado sintcticamente. Un candidato
razonable es el secuente vaco. Y si Cut fuera eliminable i.e., si
todo lo que se probara con Cut pudiera probarse sin l-, entonces
tendramos la deseada prueba de no-trivialidad, porque el secuente
vaco no es probable solo con reglas operacionales porque todas
ellas generan un secuente vaco como secuente-conclusin. Y como el
sistema es correcto, ese secuente no va a ser semnticamente vlido.
Lo que significa que hay un modelo que no lo satisface, lo que
significa que no todo modelo del sistema es trivial. Victoria.
(ANTES: No hay que trabajar con secuentes finitos? Si el
secuente es infinito, puede que nunca termines de reducirlo, y sin
embargo, puede que no te queden todas las frmulas del lenguaje en
el infinite open branch. Una opcin: intercalar aplicaciones de
derivedcut por cada n aplicaciones del resto de las reglas. As s te
asegurs que en el infinite open branch estn todas las frmulas del
lenguaje. Otra opcin: que en cada paso apliques todas las reglas a
la vez. Pero quizs eso no sea posible. Si, por ejemplo, tens que
aplicar infinitas veces la reglas de intro del condicional a la
izquierda porque tens infinitos condicionales a la izquierda, lo
que te va a quedar (creo) es un rbol con infinitas ramas. Yo no s
si eso es un rbol.
Otra opcin: copiar el mtodo de Ripley, y despus aclarar dudas.
En particular, cmo hacer con los secuentes infinitarios. Una cosa
que se puede hacer es aplicar las reglas no en cualquier orden,
sino por el orden en la enumeracin de frmulas (intercalando, tras
una reduccin de una frmula, una aplicacin de derivedcut). Por poner
un orden, si una frmula aparece en ms de un lugar, reducir primero
las que aparecen a la derecha, despus en el medio, y a lo ltimo en
la izquierda. De esta forma, eventualmente vamos a dar con una
prueba del secuente probable. (El sistema es compacto, creo.
Verificar.)
A NON-TRIVIALITY PROOF
MPTTT is non-trivial if and only if there is one unprovable
sequent. So lets just consider this sequent:
Pa
That sequent is not an axiom, and cannot be obtained by an
application of an operational rule. Its only formula is a
consistent sentence, and we have a rule to introduce those
sentences on the left side, and also a rule to introduce them on
the right. But there is no rule that specify how to introduce it on
the middle side of a sequent. But still, in order to show that it
cannot be prove, we need to show that it cannot be obtained by an
application of neither of the three Cut rules S has.
There are several ways to go from here. But one promising route
would be to show that anything that can be proved with an
application of Cut, can also be proved without it. If we have that
Cut-Elimination proof, the we will also get the desired
non-triviality proof. So, is Cut eliminable from S?
One way to prove it is by induction on the index of the
sequents, where the index is a sequent is a order pair of the grade
and the rank of the sequent.[footnoteRef:10] In order to define
these notions, we must first introduce two other preliminary ones.
[10: We borrow these notions from Paoli XXXX]
Cutproof. A proof D in S is called a cutproofiff it contains
just one application of Cut, whose conclusion S is the endsequent
of the proof. It is called a cut-free proof iff it contains no
application of Cut at all. The formula A that appears in Cut 1, Cut
2 and Cut 3 is called the cutformula.
The idea is that if it can be prove that every cutproof can be
transformed in a cut-free proof, then every proof can be
transformed into a cut-free proof. Just take any proof P of the
sequent T. P will have a finite number n of applications of Cut.
Just take the uppermost and leftmost application of Cut. That
sub-proof will be a cutproof. If our hyphotesis is right, then it
can be turn into a cut-free proof. No take the uppermost and
leftmost application of Cut of this new proof. That sub-proof will
be a cutproof. So, by hyphotesis, one can transformed it into a
cut-free proof. So now we have a new proof, with two less
applications of Cut than the original one. Just apply this
procedure n times, and youll get a cut-free proof of the sequent
T.
Rank. Let D be a cutproof whose final inference is one of these
three:
Cut 1
, A , A
Cut 2
, A , A
Cut 3
, A , A
To define the rank of the sequent S in D --denoted by rD(S)- we
need to distinguish three subcases, one for each Cut rule. In the
case of Cut 1, the rank of S is so defined:
*If S belongs to the subproof D of D whose endsequent is [, A],
rD(S) the maximal length (diminished by one) of an upward path of
sequents S1, ,Sns.t. S1=S and Si(1in) contains A in the left
side.
*If S belongs to the subproof D of D whose endsequent is [ , A
], rD(S) the maximal length (diminished by one) of an upward path
of sequents S1, ,Sns.t. S1=S and Si(1in) contains A in the middle
side.
*rD([])=rD([, A])+rD([, A])
In the case of Cut 2, the rank of S is so defined:
*If S belongs to the subproof D of D whose endsequent is [, A],
rD(S) the maximal length (diminished by one) of an upward path of
sequents S1, ,Sns.t. S1=S and Si(1in) contains A in the right
side.
*If S belongs to the subproof D of D whose endsequent is [, A ],
rD(S) the maximal length (diminished by one) of an upward path of
sequents S1, ,Sns.t. S1=S and Si(1in) contains A in the middle
side.
*rD([])= rD([, A])+rD([, A])
In the case of Cut 3, the rank of S is so defined:
*If S belongs to the subproof D of D whose endsequent is [, A],
rD(S) the maximal length (diminished by one) of an upward path of
sequents S1, ,Sns.t. S1=S and Si(1in) contains A in the left
side.
*If S belongs to the subproof D of D whose endsequent is [ , A],
rD(S) the maximal length (diminished by one) of an upward path of
sequents S1, ,Sns.t. S1=S and Si(1in) contains A in the right
side.
*rD([])= rD([, A])+rD([, A])
Rank of a subproof in a cutproof. Let D be a cutproof and D be
any of its subproofs (possibly D itself). The rank of D in D is
denoted by rD(D) or, when the context is clear, simply r(D)-, and
coincides by definition with rD(S), where S is the endsequent of
D.
Grade of a subproof in a cutproof. Let D be a cutproof, and D be
any of its subproofs (possibly D itself). The grade of D in D is
denoted bygD(D) or, when the context is clear, simply g(D)-, and
equals the grade of the sequent prove by the proof. The grade of a
sequent S that is the conclusion of a proof D is denoted bygD(S),
and is the same as the cut-formula A. If A is not a truth assertion
(e.g., a sentence with the form T), then gD(A) is the number of
logical symbols contained in the cut-formula A. If A = T, then
gD(T) = gD().
Index of a subproof in a cutproof.Let D be a cutproof and D be
any of its subproofs (possibly D itself). The index of D in D is
denoted by iD(D) or, when the context is clear, simply i(D)-, and
is the ordered pair gD(D),rD(D). Indexes are ordered
lexicographically: that is, i, n j, m iff either ij, or nm.
So lets now look at the cut-elimination proof. We will proceed
by induction on the index of D:
[i(D)=0, 0] If the rank of D is 0, then A will be just an atomic
formula. There are three kinds of atomic formula: propositional
letters like p-, assertions like Pa, or thuth assertions, like T,
where gD(T)=0. Since rD(D), then both premise-sequents must be
axioms. So they must have one of these forms:
(i) , A , A, A
(ii) , ~A , ~A
(iii) , A , A
(iv) , p (p) , p (p)
The formula that is cut may be A from (i), but it cannot be ~A
from (ii), A from (iii), of p (p) p (p) from (iv), because the
grade of the last three formulas is different from 0. In the first
case, the two premise-sequents will be the same sequent, and will
be a sequent with a structure like (i). On the other case, then the
formula that is cut will be part of and , and , or and . And in the
three cases, the result sequent will be a formula like (i)-(iv),
and so will be an axiom and so Cut is not necessary to prove
it.
[i(D)= 0, k, 1k]If the grade of D is 0, then A will be just an
atomic formula. There are three kinds of atomic formula:
propositional letters like p-, assertions like Pa, or thuth
assertions, like T, where gD(T)=0. If r(D)0, then either the rank
of the left sequent is bigger than 0, or the rank of the right
sequent is bigger than 0. There are types of cases to consider,
depending on whether we have applied Cut 1, Cut 2 or Cut 3. Lets
start with the Cut 1 case.
Cut 1
, A , A
So we need to consider two subcases: either (i) rD(, A)0, or
(ii) rD(, A )0.
(i)rD(, A)0. Thus [, A] is the conclusion of an inference where
A can either be a principal, or an auxiliary, or a side formula. It
is a side formula, the strategy that we will employ will be to push
the Cut upwards, in a way such what we will get is a new proof of
[] containing cutproofs of grade 0 and a lower rank, and hence a
lower index. This will entitle us to exploit the inductive
hypotheses. The rank of the sequent in the proof will be indicated
on the right part of the sequent. Here we have some examples:
XXXX HAY, DE HECHO, MUCHSIMOS SUBCASES, UNO POR CADA REGLA QUE
PUEDAS APLICAR PARA OBTENER EL SECUENTE [, A], con A como side
formula.
For example, let = {B}
, B, A (n), B , A (m)
, A, B (n+1) , A , B (m+1)
(n+m+2)
Turns into:
, A , B (n) , A , B (m)
, B (n+m)
, B (n+m+1)
Another example: = {~B}
, B , A, B (n) , B , A, B (m)
, A , ~B(n+1) , A , ~B(m+1)
(n+m+2)
Turns into:
, B , A, B (n) , B , A, B (m)
, B , B (n+m)
, B (n+m+1)
Another example: = {BC}
Let =1 2 3, =1 2 3, =1 2 3.
D=1,A1,B1,B(n1)
E=22,C2,C(n2)
F=33,B,C3(n3)
G=11,A,B1,B(m1)
H=22,C2,C(m2)
I=33,B,C3(m3)
D(n1)E(n2)F(n3)G(m1)H(m2)I(m3)
, A, BC (max(n1, n2, n3)+1), A,BC(max(m1, m2, m3)+1)
, BC(max(n1, n2, n3)+1)+ (max(m1, m2, m3)+1)
Turns into:
1,A1,B1,B(n1)11,A,B1,B(m1)
11, B1,B(n1+m1)22,C2,C(n2) 33,B,C3(n3)
, BC (max((m1+n1), n2, n3)+1)
If A is auxiliary, then the strategy will be basically the same.
Lets look at one example.
, A, B , A (m) , B , A , A (n)
, A , BA (m+1) , A , BA (n+1)
, BA (max((m+1),(n+1)))
Turns into:
, A, B , A (m) , B , A , A (n)
, B , A(m+n)
, BA(m+n+1)
If A is principal, then A=T(B). Lets look at one example:
, B (m) , B (n)
, T(B) (m+1) , T(B) (n+1)
(m+1)+(n+1)
Turns into:
, B (m) , B (n)
(m+n)
(ii) rD( , A )0. This subcase will be treated in a symmetric
manner than (i)
There cases that applied Cut 2 or Cut 3 are similar than the
ones we have already seen.
[i(D)= k, 0, 1k] There are types of cases to consider, depending
on whether we have applied Cut 1, Cut 2 or Cut 3. We will look at
the case that uses Cut 1. The other two cases are symmetrical.
Since r(D)=0, then the cutformula A must be principal in the
subinferences whose conclusions are [, A] and [ , A] (and also
cannot be generated by another application of Cut, because then the
rank of D wont be 0.).
Letsseethefollowingexample: XXXXX PON UN EJEMPLO DE UN
CUANTIFICADOR UNIVERSAL, Y OTRO DE UNA AFIRMACIN DE CONSISTENCIA
(no se puede con el de consistencia, porque no hay regla
operacional que te genere una afirmacin de consistencia en el
medio).XXXX
Let= , = , = .
, A(t) (n), A(a), A(a) (m) , A(t)(s)
, xA(x) (n+1) , xA(x) (max(m,s)+1)
((n+1)+(max(m,s)+1)
Turns into:
, A(t) (n), A(t) (s)[footnoteRef:11] [11: If [, A(t)] has a
proof, the so has [, A(t)]. Just start from the same axioms, but
add in each side the formulas that [, A(t)] lacks in order to
became [, A(t)]. XXXX NO ME QUEDA CLARO QUE EL RANK DEBA SER S.
PODRA SER MS. PODRA SER MS QUE MAX(M,S)? Creo que no]
(n+s)
[i(D)= k, j, 1j, k] We need to consider two subcases: either (i)
rD(, A)0, or (ii) rD( , A )0. We will just consider the case (i).
(ii) is symmetrical. Once again, we will need to consider, in each
alternative, three subcases: the ones that applied in the final
inferential step CUT 1, the ones that apply in that step CUT 2,
and, finally, the ones that apply at that stage CUT 3. Well just
look at the first subcases. The others two types are
symmetrical.
(i) rD(, A)0. Thus [, A] is the conclusion of an inference where
A can either be a principal, or an auxiliary, or a side formula. If
it is a side or an auxiliary formula, the strategy that we will
employ will be to push the Cut upwards, in a way such what we will
get is a new proof of [] containing cutproofs of grade 0 and a
lower rank, and hence a lower index. This will entitle us to
exploit the inductive hypotheses. The rank of the sequent in the
proof will be indicated on the right part of the sequent. Here we
have an example:
For example, let = {B}
, B, A , B (n) , B , A , B (m)
, A , B(n+1) , A , B(m+1)
(n+m+2)
Turns into:
, B, A , B (n) , B , A , B (m)
, B , B (n+m)
, B
What happen when A is principal? The strategy will be basically
the same. Lets see an example:
, A (m1) , B (m2), A (n1) , B (n2)
, BA (max (m1, m2)) , BA (max(n1,n2))
((max (m1, m2))+(max(n1,n2)))
Turns into:
, A , B (m2) , A , B (n2)
, A (m2+n2)) , B (m2), A (m2+ n2)) , B (n2)
, BA(max((m2+n2), m2) , BA (max((m2+n2),
XXXXXX EXPLIC BIEN ESTA TRANSFORMACIN. EXPLIC QUE CADA FORMA DE
CUT CON UN INDEX MENOR PUEDE SER REEMPLAZADO POR UNA PRUEBA DEL
SECUENTE SIN CUT, POR HIP INDUCTIVA. AS CADA VEZ, HASTA LLEGAR A LA
ULTIMA.
XXXXXXX
Cut 2
, A , A
Cut 3
, A , A
XXXX LA ESTRATEGIA SERA, CREO, RECONVERTIR TODA PRUEBA QUE USE
CUT EN EL PASO FINAL, EN UNA QUE USE CUT EN ALGN PASO ANTERIOR.
FIJATE BIEN CMO LO HACE PAOLI.
XXX P. 101 DE PAOLI.1