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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]