Tableaux Implemented

1
Tableaux Implemented

• A Tableaux Prover
• Representation of tableaux
• Translate tableaux rules to Prolog
• Bring together this weeks results in a combined
  demo system to see whether our efforts have been
  worthwhile

2
Representing Signed Formulae in Prolog

• true(walk(john) v talk(john))
• false(walk(john)v talk(john))

3
Preprocessing

• Use logical equivalences
• A ? B ? ? (?A ? ?B)
• A ? B ? ? (A ? ?B)
• true(walk(john) v talk(john))
• gt true((walk(john)talk(john)))
• A simple prolog predicate does this for us
  (naonly/2).

4
The Driver Predicate

• theorem(Formula) -
• naonly(Formula,ConvFormula),
• \ tabl(false(ConvFormula),,_).
• There is no way of falsifying the converted input
  Formula.
• If tabl/3 succeeds Formula not valid.
• If tabl/3 fails Formula valid.

How is the tableaux represented?
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Representation of Tableaux
L1

• We collect literals.
• One list per branch
• L1,L2,L3
• L1,L2,L4,L5
• L1,L2,L4,L6
• no more representation of the whole tableau.

L2
L4
L3
L5
L6
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Tableaux Rules in Prolog

• Scheme
• tabl(InFormula,InBranch,OutBranch) -
• For each connective and sign
• tabl(false(A),InBranch,OutBranch) -
• tabl(true(AB),InBranch,OutBranch) -
• The tabl/3 rules are straightforward translations
  of the tableaux rules. But some details are
  tricky.

Input 1 The formula to be analyzed
Input 2 A list of literals (branch above)
Output Literals from above plus literals from
analyses of InFormula
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Conjunctive Expansion Negation

• tabl(true(A),InBranch,OutBranch) -
• !,tabl(false(A),InBranch,OutBranch).
• tabl(false(A),InBranch,OutBranch) -
• !,tabl(true(A),InBranch,OutBranch).
• Negation of input Formula is eliminated by a
  recursive tabl/3 call with reversed sign.

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Conjunctive Expansion Conjunction

• tabl(true(AB),InBranch,OutBranch) -
• !,tabl(true(A),InBranch,K),
• tabl(true(B),K,OutBranch).
• true(sleep(john)snort(john)) Asleep(john)
• InBranch true(sleep(mary)) Bsnort(john)
• K true(sleep(mary)),true(sleep(john))
• OutBranch true(sleep(mary)),true(sleep(john)),
• false(snort(john))

Two recursive calls of tabl/3. One for each
conjunct. Intermediate branch is handed over in
variable K.
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Disjunctive Expansion I

• General Idea Test branches one after another.
• Advantage Two alternatives. If we come to the
  right branch, we can forget about the left one.

F2
F4
F3
F5



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Disjunctive Expansion II

• tabl(false(A _),InBranch,OutBranch) -
• tabl(false(A),InBranch, OutBranch).
• tabl(false(_ B),InBranch,OutBranch) -
• !,tabl(false(B),InBranch, OutBranch).
• If we succeed on the left branch to prove
  false(A) without a clash, we stop (a model for
  the negation of our suspected theorem has been
  found).
• Else, we have to try the right branch.

• Use Prologs backtracking mechanism to test one
  branch and then the other.
• Two clauses for tabl(false(AB),)

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Base Case Atomic Formula

• tabl(AF,InBranch,OutBranch) -
• OutBranch AFInBranch,
• \ clash(OutBranch).
• clash(Branch) -
• member(true(A),Branch),
• member(false(A),Branch).
• If we reach a literal on some branch and there is
  no clash, we stop a counter model has been
  found.
• Else, we fail. This triggers backtracking for
  further waiting disjunctive expansions.

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Main Ideas of Our Implementation

• No explicit representation of the whole tableaux
  We always work on one branch at a time.
• Branches are represented as lists of literals
  (nothing else matters).
• Use a sequence of Prolog calls to do tableaux
  construction
• Conjunctive expansion one clause per
  conjunctive/sign with recursive call(s).
• Disjunctive expansion two clauses with a
  recursive call, backtracking if fail in first
  clause (i.e. no model found on left branch).

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• tabl(true(A),InBranch,OutBranch) -
• !,tabl(false(A),InBranch,OutBranch).
• tabl(false(A),InBranch,OutBranch) -
• !,tabl(true(A),InBranch,OutBranch).
• tabl(true(A B),InBranch,OutBranch) -
• !,tabl(true(A),InBranch,K),
• tabl(true(B),K,OutBranch).
• tabl(false(A _),InBranch,OutBranch) -
• tabl(false(A),InBranch,OutBranch).
• tabl(false(_ B),InBranch, OutBranch) -
• !,tabl(false(B),InBranch,OutBranch).
• tabl(AF,InBranch,OutBranch) -
• OutBranch AFInBranch,
• \ clash(OutBranch).

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Online Tableaux Interface

• See example Interface.htm
• Course Page
• Link to tableaux interfaced
• All Prolog code for download
• The reader
• This weeks slides
• Corrected page of reader
• Do not hesitate to email us

Try it out!
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The Proof of the Pudding is in the Eating

• Towards the end of this course
• we show a system that combines semantics
  construction with our tableaux prover.
• Given a model, we check whether a sentence is
• Informative (is the information not already
  contained in the model)
• Contradictory
• Consistent

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Conversational Maxims

• Speaker and hearer obey some rules
• H.P. Grice
• Maxim of Quality Tell the truth!
• Maxim of Manner Be relevant!
• Maximes (and violation hereof) is used in
  conversations (studied by Pragmatics)
• ADid you tell her you love her?
• BOh, nice weather today

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Our System

• Example model M
• John is a man.
• John does not smoke.
• Every man is human.
• Every human works.
• As large conjunctive Formula man(john)forall(x,h
  uman(x)gtwork(x))forall(y,man(y)gthuman(y))smoke(
  john)

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• check -
• lambda(F),
• M man(john)forall(x,human(x)gtwork(x))forall(y
  ,man
• checkFormula(F,M).
• checkFormula(F,M) -
• theorem(MgtF),
• nl, write('I know, I know...').
• checkFormula(F,M) -
• theorem((MF)),
• nl, write('Impossible!').
• checkFormula(_,_) -
• nl, write('If you say so...').

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Checking Consistency

• theorem((MF))
• Why does this ckeck whether F is inconsistent
  with M?
• M contains no contradictions.
• If ? (M ? F) is a theorem, then M ? F is a
  contradiction. This must be due to F because of
  (1).
• System Demo

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Extending our System

• Argumentation, Syllogisms
• Model extension
• Extend model with NL input
• Question answering
• Question semantics, extend DCG
• Check it out!
• Other Systems using NLP and inference
• Curt (Blackburn/Bos)
• Doris (Bos)

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Thanks for your Interest... enjoy Computational
Semantics!