Derivation Schemes for Topological Logics

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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 Operators
• P ( PV, PF )
• Q ( QV, QF )
• P?Q (int PV ? int QV, int PF? int QF )
• P?Q ( 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, ?)
• DeMorgans 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 P?Q
T 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 P?Q
T 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 ? Q
T 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 disjunction
• P (-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 ? Q
T 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) ? 0
• Since P ? 0 0 and Q ? 0 Q for any Q,
• P ? ?P ? 0
• By antisymmetry of ?, P ? ?P 0
• and so (A, ?, ?, ?) is boolean.

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Implications of the theorem

• Any DeMorgan algebra in which
• Entailment is given by ? (?),
• The implication operator is given by ?P ? Q, and
• 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), (PV?PF) ? (PV ? QV), (PF ? QF)
• ?, (PV?PF) ? (PV ? QV), (PF ? QF)
• (PV ? QV),, ((PV?PF) ? (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 P?Q
• 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

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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\

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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 \ )

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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