Post on 01-Jan-2016
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Derivation Schemes for Topological Logics
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Derived Logics
What Are They? Why Do We Need Them? How Can We Use Them? Colleague: Michael Westmoreland
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History
1936 Von Neumann and Birkhoff a lattice of propositions based on the closed
subspaces of Hilbert space now known as “quantum logic” based on measurement non-Boolean (fails to meet distributive
properties) No satisfying way to do implication
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A Topological Logic
A proposition is an equivalence class of sets S S’ iff int(S) = int(S’) [S] = [ (int S)c ] [S] [S’] = [ (int S) (int S’) ] [S] [S’] = [ (int S) (int S’) ] Most Boolean properties hold Law of noncontradiction:
[S] [S] = [int S] [ (int S)c ]= [S (int Sc) ] choosing canonical representation= []
But not all: [S] =? [S]No!
[S] = [ (int S)c ] So [S] [int( (int Sc )c ] = [ int Sc ] = [S]
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Logic Properties
No tertium non datur:[S] [S] [U] where U is the universal set.
What about truth assignment?A measurement (open set) m verifies a proposition P iff m Pi Pi P.
Example: the real line with the standard topology. P = [ (-3, 5) ]. m = (0, 4) verifies P since (0,4) (-3, 5), [-3, 5), [-3,5], (-3,5]
We speak of “verification” rather than truth. Rationale:
Let S be a classical system and P a proposition about S with P0 as the canonical representative of [P]. Then P0 = int Pj
Pj [P]. A measurement m that contains points of P0 but does not lie entirely in P0 would not verify P.
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More Properties
P = (-3, 5). m = (0,6) does not verify P. Should we conclude P is false? The state of S could lie in P0
and still be consistent with the result of the measurement m. In fact, there is a more precise measurement, say m’ that lies entirely in P0 and the result of m. Hence, we cannot conclude that P is false.
New concept for assigning truth values: associated with a given measurement (set) , three possibilities: verifiability set, falsifiability set, indeterminate.
Twin Open Set Phase Space Logic (TOSPS) A measurement m verifies P if m P0 where P0 is the
canonical rep of P. A measurement falsifies a proposition if m Cl(P0)c.
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Twin Open Set Phase Space Logic
Definition: P is a proposition in TOSPS logic if P = ( [V0], [F0] ), where V0 and F0 are disjoint open sets.
Definition: Let P = ( [V0], [F0] ) be a proposition in TOSPS logic and m be a measurement. P will be assigned the truth value
true if m V0;false if m F0;indeterminate otherwise.
Logical OperatorsP = ( [PV], [PF] )Q = ( [QV], [QF] )PQ = ([int PV int QV], [int PF int QF] )PQ = ( [int PV int QV], [int PF int QF] )P = ( [int PF], [int PV] )
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Properties
P = ([PF], [QF]) = ( [PV], [PF] ) = P P P = ( [PV], [PF] ) ( [PF], [PV] )
= ( [int PV int PF], [int PF int PV] )= ([], [U])
P P = ([U], []) DeMorgan’s laws Ditributivity All Boolean properties, but tertium non datur.
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Note: fails to be truth functional
P = [(-1,2), (5,9)] Q = [(1,3), (8,11)] P Q = [(-1,3), (8,9)]
The measurement m = (0, 2.5) assigns I to P, I to Q, and T to P Q, since m PV QV
m = (0,4) assigns I to P Q and I to P and I to Q, since m ⊈ PV QV
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Twin Open Set Logic Based on Exact Measurement (Discrete)
P Q PQT T T
T I T
T F T
I T T
I I I
I F I
F T T
F I I
F F F
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Truth Tables for TOPSL
P Q PQT T T
T I T
T F T
I T T
I I T or I
I F I
F T T
F I I
F F F
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P Q P QT T T
T I I
T F F
I T I
I I F or I
I F F
F T F
F I F
F F F
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P P
T F
I I
F T
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Example
Example to illustrate lack of truth functionality for disjunctionP = [(-2, 2), (5,9) ]Q = [(1, 3), (8, 11) ]P Q = [(-2, 3), (8,9) ]Suppose m = (0, 2.5) m assigns “I” to each of P and Q, “T” to
P Q since m PV QV
Now suppose m = ( 0, 4)m assigns “I” to P Q as well as to P and to Q
since m (PV QV)
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P Q P Q = P QT T T
T I I
T F F
I T T
I I T or I
I F I
F T T
F I T
F F T
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Applications to Billiard Ball Model of Computation
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OR-Gate
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AND-Gate
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NOT-Gate
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Derivation Gate
Input the value of P and
the value of P → Q
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Derivation
For a Boolean lattice, define P ≤ Q when
P Q is valid where ≤ is the lattice ordering Modus Ponens
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Three Questions to Consider
1. What is a proper ordering for the propositions in twin open set logic?
2. What is a proper implication operator in twin open set logic?
3. What derivation method can be implemented given the answers to 1 and 2?
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Characterization Theorem
Let (A, , , ) be a DeMorgan algebra. If we define an ordering ≼ on the algebra by
P ≼ Q def P Q = Q,
then P (P Q) ≼ Q iff (A, , , ) is a boolean algebra .
Reminder: TOPSL is a DeMorgan algebra.
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Proof
Need: (A, , , ) satisfies the law of non contradiction.
In any DeMorgan algebra satisfying our hypothesis, 0 ≼ P P.
Substituting Q = 0 in the modus ponens scheme,
P (P 0) ≼0
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Using distributivity, P (P 0) ≼ 0(P P) (P 0) ≼ 0Since P 0 = 0 and Q 0 = Q for any Q,
P P ≼ 0By antisymmetry of ≼, P P = 0and so (A, , , ) is boolean.
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Implications of the theorem:
Any DeMorgan algebra in which
1. Entailment is given by (),2. The implication operator is given by P Q,
and
3. Modus ponens is satisfied
must be a Boolean algebra.
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Non Standard Derivation
TOSPL is a DeMorgan algebra, but not a boolean algebra.
At least one of the three properties above must fail.
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Modus Ponens Fails
Ordering for TOSL (suggested by or )P Q PV QV and QF PF
motivated by either
P Q = P PV QV and QF PF
P Q = Q PV QV and QF PF
Q is more readily verified and less easily falsified than P.
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Implication
P→Q def P Q
So P→Q = P Q
= [(PF QV),(PV QF)]
Previous Theorem tells us that modus ponens fails. Why does it?
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Theorem: With the ordering given by , it is not the case that P (P→Q ) Q
Proof: P (P→Q ) = P (P Q )
= (P P) (P Q)
= [(PV PF), (PVPF)] [(PV QV), (PF QF)]
= [, (PVPF)] [(PV QV), (PF QF)]
= [(PV QV),, ((PVPF) (PF QF))]
= [(PV QV), ((PV QF) PF)]
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For P (P→Q ) ≼ Q, QF (PV QF) PF
But whenever PV PF X (the whole space), the containment fails.
In any nondiscrete topology we have disjoint open sets PV and PF such that PV PF X
and the claim is established.
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Need: a proposition that contains
[(PV QV), ((PV QF) PF)]
One possibility
[QV, ]
Given P and P→Q [QV, ]
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Good Point: It works.
Not so good: So does any proposition of the form [QV, Y] where Y is any open subset of
int(QVC)
Cannot falsify
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Modus Tollens
P→Q = P Q = (Q) P P
= Q → P
Consider Q (P→Q )
= [(PF QF), ((PV QF) QV)]
Analog to modus ponens: [PF, ]
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Another Possibility
For Modus Ponens: Given P and P→Q,Conclude [QV, PV QF] =def QP
For Modus Tollens: Given Q (P→Q ),Conclude [PF, PV QF] =def PQ
Now P (P→Q ) ≼ QP
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Moreover,
PV QV QV and
PV QF (PV QF) PF
thereby respecting entailment
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Non Standard Entailment
P Q def Pv Qv
Not antisymmetric, but is reflexive and transitive (a quasi ordering relation)
Theorem: satisfies: P (P→Q ) Q
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What about falsifiability?
P Q def QF PF
Does not give a valid modus ponens!
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Both Verifiability and Falsifiability
Quasi ordering:
Reminder: PS = PV PF
P ≤ Q def V V
F F F S
P Q and
Q P = Q P
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theorem
The quasi ordering defined gives
P (P→Q ) ≤ Q
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Non Standard Implication
Instead of P→Q = P Q
P ↪ Q =def [PV QV, QF\ ]FP
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Motivation
sup(X | P X ≤ Q) well defined for any orthonormal lattice.
Propositions in TOSL make a lattice, but not orthonormal
sup(X | P X ≲ Q) where X = [XV, XF] and
XV = sup(Y | PV Y QV) and
XF = inf(Y | QF PF Y)
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To get existence
need: PF QV XV
This blocks inf(Y | QF PF Y)
Leading to XF = (QF \ )FP
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Theorem: P (P ↪ Q) ≴ Q
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Why ?
1. We get the usual implication operator when considering the discrete twin logic.
2. Natural interpretation of implication when measurement P verifies P and P ↪ Q, whatever form ↪ may take.
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Discoveries
In any derivation scheme that is given by the lattice theoretic entailment, an implication P Q that is equivalent to P Q must be Boolean.
Define P Q = P Q = [(PF QV), (PV QF)] m will assign a value of true to P Q iff m assigns a value of true to either P or Q. i.e., m (PF QV).
Alternately, m will assign a value of false to
P Q iff m (PV QF). m assigns indeterminate to P Q iff
m (PV QF) and m (PF QV)
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Derivation in Collision Models
Replace modus ponens by
P and P → Q yield
[QV, PV QF]
Replace modus tollens by
Q and P → Q yield
[QV, PV QF]