The Suppression Task Steffen H olldobler International Center for - - PowerPoint PPT Presentation

◮ The Suppression Task – Part I ◮ Three-Valued Łukasiewicz Logic ◮ Logic Programs ◮ Completion Semantics and Least Models ◮ Semantic Operators and Least Fixed Points ◮ Contractions ◮ The Suppression Task – Part II ◮ Abduction

Steffen H¨

• lldobler

The Suppression Task 1

The Suppression Task

Steffen H¨

• lldobler

International Center for Computational Logic Technische Universit¨ at Dresden Germany

The Suppression Task – Forward Reasoning

◮ Byrne: Suppressing Valid Inferences with Conditionals.

Cognition 31, 61-83: 1989

◮ If she has an essay to write then she will study late in the library.

She has an essay to write. ⊲ Modus Ponens (MP) in classical logic ⊲ 96% of subjects conlude that she will study late in the library.

◮ If she has an essay to write then she will study late in the library.

She has an essay to write. If she has a textbook to read she will study late in the library. ⊲ Alternative Arguments ⊲ 96% of subjects conlude that she will study late in the library.

◮ If she has an essay to write then she will study late in the library.

She has an essay to write. If the library stays open she will study late in the library ⊲ 38% of subjects conlude that she will study late in the library. ⊲ Additional arguments lead to suppression of earlier conclusions.

Steffen H¨

• lldobler

The Suppression Task 2

Reasoning Towards an Appropriate Logical Form

◮ Context independent rules

⊲ If she has an essay to write and the library is open then she will study late in the library. If the library is open and she has a reason for studying in the library then she will study late in the library.

◮ Context dependent rule plus exception

⊲ If she has an essay to write then she will study late in the library. However, if the library is not open, then she will not study late in the library. ⊲ The last sentence is the contrapositive of the converse of the original sentence!

Steffen H¨

• lldobler

The Suppression Task 3

The Suppression Task – Denial of Antecedent (DA)

◮ Byrne: Suppressing Valid Inferences with Conditionals.

Cognition 31, 61-83: 1989

◮ If she has an essay to write then she will study late in the library.

She does not have an essay to write. ⊲ 46% of subjects conlude that she will not study late in the library.

◮ If she has an essay to write then she will study late in the library.

She does not have an essay to write. If she has a textbook to read she will study late in the library. ⊲ 4% of subjects conlude that she will not study late in the library.

◮ If she has an essay to write then she will study late in the library.

She does not have an essay to write. If the library stays open she will study late in the library. ⊲ 63% of subjects conlude that she will not study late in the library.

Steffen H¨

• lldobler

The Suppression Task 4

Human Reasoning – The Search for Models

◮ Goal Find a logic which adequately models human reasoning. ◮ How about classical two-valued propositional logic? ◮ Let’s consider a direct encoding:

{l ← e, e} {l ← e, e, l ← t} {l ← e, e, l ← o} {l ← e, ¬e} {l ← e, ¬e, l ← t} {l ← e, ¬e, l ← o}

Steffen H¨

• lldobler

The Suppression Task 5

Two-Valued Interpretations

◮ Let L be a language of propositional logic. ◮ A (two-valued) interpretation is a mapping L → {⊤, ⊥} represented by I,

where I is a set containing all atoms which are mapped to ⊤. ⊲ All atoms which do not occur in I are mapped to ⊥.

◮ Let I denote the set of all interpretations.

⊲ (I, ⊆) is a lattice.

◮ An interpretation I is a model for a program P, in symbols I |

= P, iff I(P) = ⊤. ∅ | = {l ← e, e} {e} | = {l ← e, e} {l} | = {l ← e, e} {e, l} | = {l ← e, e} ∅ {e} {l} {e, l}

Steffen H¨

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The Suppression Task 6

Logical Consequence (1)

◮ A formula G is a logical consequence of a set of formulas F, in symbols

F | = G, iff all models for F are also models for G. {l ← e, e} | = l {l ← e, e} | = e ∅ {e} {l} {e, l}

Steffen H¨

• lldobler

The Suppression Task 7

Logical Consequence (2)

◮ A formula G is a logical consequence of a set of formulas F, in symbols

F | = G, iff all models for F are also models for G. {l ← e, ¬e} | = ¬e {l ← e, ¬e} | = l {l ← e, ¬e} | = ¬l ∅ {e} {l} {e, l}

Steffen H¨

• lldobler

The Suppression Task 8

The Suppression Task – A Classical Logic Approach

◮ Recall the examples:

{l ← e, e} | = l modus ponens {l ← e, e, l ← t} | = l classical logic is monoton {l ← e, e, l ← o} | = l upps, humans don’t do this {l ← e, ¬e} | = ¬l denial of antecendent {l ← e, ¬e, l ← t} | = ¬l {l ← e, ¬e, l ← o} | = ¬l

◮ Conclusion classical logic is inadequate.

⊲ Often mistakenly generalized to “logic is inadequate”.

Steffen H¨

• lldobler

The Suppression Task 9

The Suppression Task – A Computational Logic Approach

◮ Goal Find a logic which adequately models human reasoning. ◮ Solution I propose the following:

⊲ Logic programs under (weak) completion semantics

◮ ◮ Non-monotonicity

⊲ Reasoning towards an appropriate logical form

◮ ◮ Logic programs

⊲ Three-valued Łukasiewicz logic

◮ ◮ Least models

⊲ An appropriate semantic operator

◮ ◮ Least fixed points are least models ◮ ◮ Least fixed points can be computed by iterating the operator

⊲ Reasoning with respect to the least models ⊲ A connectionist realization

Steffen H¨

• lldobler

The Suppression Task 10

Adequateness

◮ When is a logic adequate? ◮ In this talk If it qualitatively gives the same answers as subjects in the

corresponding experiments.

Steffen H¨

• lldobler

The Suppression Task 11

Logic Programs

◮ A (logic) program is a finite set of clauses.

⊲ A (program) clause is an expression of the form A ← B1 ∧ · · · ∧ Bn, where n ≥ 1, A is an atom, and each Bi, 1 ≤ i ≤ n, is either a literal, ⊤ or ⊥. ⊲ A is called head and B1 ∧ · · · ∧ Bn body of the clause. ⊲ A clause of the form A ← ⊤ is called a positive fact. ⊲ A clause of the form A ← ⊥ is called a negative fact. {l ← e, e ← ⊤} {l ← e, e ← ⊤, l ← t} {l ← e, e ← ⊥}

◮ P is definite if the bodies of all clauses of P consist only of atoms and ⊤. ◮ Here I consider only propositional programs,

but the approach extends to first-order programs.

◮ The language L underlying a program P shall contain precisely

the relation symbols occurring in P, and no others.

Steffen H¨

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The Suppression Task 12

Program Completion

◮ Let P be a program. Consider the following transformation:

1 All clauses with the same head A ← Body1, A ← Body2, . . . are replaced by A ← Body1 ∨ Body2 ∨ . . .. 2 If an atom A is not the head of any clause in P then add A ← ⊥. 3 All occurrences of ← are replaced by ↔. The resulting set is called completion of P or c P. If 2 is omitted then the resulting set is called weak completion of P or wc P.

Steffen H¨

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The Suppression Task 13

Program Completion – Example 1

◮ Consider

P1 = {l ← e, e ← ⊤} c P1 = {l ↔ e, e ↔ ⊤} wc P1 = {l ↔ e, e ↔ ⊤}

◮ The only model of c P1 and wc P1 is:

∅ {e} {l} {e, l}

◮ Hence, c P1 |

= l and wc P1 | = l.

Steffen H¨

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The Suppression Task 14

Program Completion – Example 2

◮ Consider

P2 = {l ← e, e ← ⊥} c (P2) = {l ↔ e, e ↔ ⊥} wc (P2) = {l ↔ e, e ↔ ⊥}

◮ The only model of c P2 and wc P2 is:

∅ {e} {l} {e, l}

◮ Hence, c P2 |

= ¬l and wc P2 | = ¬l.

◮ Remember, P2 |

= ¬l.

Steffen H¨

• lldobler

The Suppression Task 15

Program Completion – Example 3

◮ Consider

P3 = {l ← e, e ← ⊤, l ← t} c P3 = {l ↔ e ∨ t, e ↔ ⊤, t ↔ ⊥} wc P3 = {l ↔ e ∨ t, e ↔ ⊤}

◮ The only model of c P3 is:

{e, l}

◮ The models of wc P3 are:

{e, l} {e, l, t}

◮ Hence, c P3 |

= ¬t whereas wc P3 | = ¬t and wc P3 | = t.

Steffen H¨

• lldobler

The Suppression Task 16

Monotonicity

◮ Let F and F′ be sets of formulas and G a formula.

A logic is monotonic if the following holds: If F | = G then F ∪ F′ | = G.

◮ Classical logic is monotonic. ◮ A logic based on completion semantics is non-monotonic.

⊲ Consider P3 = {l ← e, e ← ⊤, l ← t} P′

3

= P ∪ {t ← ⊤} ⊲ Then c P3 | = ¬t c P′

3

| = ¬t

Steffen H¨

• lldobler

The Suppression Task 17

Reasoning Towards an Appropriate Logical Form (1)

◮ Stenning, van Lambalgen: Human Reasoning and Cognititve Science.

MIT Press: 2008

◮ Represent conditionals as licences for conditionals.

⊲ If she has an essay to write then she will study late in the library. She has an essay to write.

◮ ◮ P4 = {l ← e ∧ ¬ab, ab ← ⊥, e ← ⊤}

⊲ If she has an essay to write then she will study late in the library. She has an essay to write. If she has a textbook to read she will study late in the library.

◮ ◮ P5 = {l ← e ∧ ¬ab1, ab1 ← ⊥, l ← t ∧ ¬ab2, ab2 ← ⊥, e ← ⊤} ◮ Reason about additional premises.

⊲ If she has an essay to write then she will study late in the library. She has an essay to write. If the library stays open she will study late in the library.

◮ ◮ P6 = {l ← e ∧ ¬ab1, ab1 ← ¬o, l ← o ∧ ¬ab2, ab2 ← ¬e, e ← ⊤}

Steffen H¨

• lldobler

The Suppression Task 18

Reasoning Towards an Appropriate Logical Form (2)

◮ Denial of Antecedent (DA)

⊲ If she has an essay to write then she will study late in the library. She does not have an essay to write.

◮ ◮ P7 = {l ← e ∧ ¬ab, ab ← ⊥, e ← ⊥}

⊲ If she has an essay to write then she will study late in the library. She does not have an essay to write. If she has a textbook to read she will study late in the library.

◮ ◮ P8 = {l ← e ∧ ¬ab1, ab1 ← ⊥, l ← t ∧ ¬ab2, ab2 ← ⊥, e ← ⊥}

⊲ If she has an essay to write then she will study late in the library. She does not have an essay to write. If the library stays open she will study late in the library.

◮ ◮ P9 = {l ← e ∧ ¬ab1, ab1 ← ¬o, l ← o ∧ ¬ab2, ab2 ← ¬e, e ← ⊥}

Steffen H¨

• lldobler

The Suppression Task 19

Three-Valued Logics

◮ Consider the following truth table

¬ ∧ ∨ ←Ł ↔Ł ←K ↔K ↔C ⊤ ⊤ ⊥ ⊤ ⊤ ⊤ ⊤ ⊤ ⊤ ⊤ ⊤ ⊥ ⊥ ⊥ ⊤ ⊤ ⊥ ⊤ ⊥ ⊥ ⊤ U ⊥ U ⊤ ⊤ U ⊤ U ⊥ ⊥ ⊤ ⊤ ⊥ ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊤ ⊥ ⊥ ⊤ ⊤ ⊤ ⊤ ⊤ ⊥ U ⊤ ⊥ U U U U U ⊥ U ⊤ U U ⊤ U U U U ⊥ U ⊥ U ⊥ U ⊤ U ⊤ U ⊥ U U U U U ⊤ ⊤ U U ⊤

◮ Different three-valued logics

Łukasiewicz (Ł) semantics 1920 ¬ ∧ ∨ ←Ł ↔Ł Kleene (K) semantics 1952 ¬ ∧ ∨ ←K ↔K Fitting (F) semantics 1985 ¬ ∧ ∨ ←K ↔C (F ↔ G) ≡3Ł ≡3F

• (F ← G) ∧ (G ← F)

Steffen H¨

• lldobler

The Suppression Task 20

Principles of Łukasiewicz

◮ Łukasiewicz used 1, .5, and 0 instead of ⊤, U, and ⊥, respectively. ◮ Principles of identity and non-identity

⊲ (⊥ ↔ ⊥) ≡ (⊤ ↔ ⊤) ≡ (U ↔ U) ≡ ⊤. ⊲ (⊤ ↔ ⊥) ≡ (⊥ ← ⊤) ≡ ⊥. ⊲ (⊥ ↔ U) ≡ (U ↔ ⊥) ≡ (⊤ ↔ U) ≡ (U ↔ ⊤) ≡ U.

◮ Principles of implication

⊲ (⊥ ← ⊥) ≡ (⊤ ← ⊤) ≡ (U ← U) ≡ ⊤. (⊤ ← ⊥) ≡ (⊤ ← U) ≡ (U ← ⊥) ≡ ⊤. ⊲ (⊥ ← ⊤) ≡ ⊥. ⊲ (⊥ ← U) ≡ (U ← ⊤) ≡ U.

◮ Definitions of negation, disjunction, and conjunction

⊲ ¬F ≡ (⊥ ← F) ⊲ (F ∨ G) ≡ (G ← (G ← F)). ⊲ (F ∧ B) ≡ ¬(¬F ∨ ¬G).

Steffen H¨

• lldobler

The Suppression Task 21

Some Common Laws and Literature

◮ Laws

Ł K F Equivalence F ↔ G ≡ (F ← G) ∧ (G ← F) yes yes no Implication F → G ≡ ¬F ∨ G no yes yes Syllogism (F → G) ∧ (G → H) ≡ F → H no yes yes Excluded Middle F ∨ ¬F ≡ ⊤ no no no Contradiction F ∧ ¬F ≡ ⊥ no no no

◮ Literature

⊲ Fitting: A Kripke-Kleene Semantics for Logic Programs. Journal of Logic Programming 2, 295-312: 1985. ⊲ Kleene: Introduction to Metamathematics. North-Holland: 1952. ⊲ Łukasiewicz: O logice tr´

• jwarto´
• sciowej. Ruch Filozoficzny 5, 169-171: 1920.

English translation: On Three-Valued Logic. In: Jan Łukasiewicz Selected

• Works. (L. Borkowski, ed.), North Holland, 87-88, 1990.

Steffen H¨

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The Suppression Task 22

Three-Valued Interpretations

◮ A (three-valued) interpretation is a mapping L → {⊤, ⊥, U}

represented by

• I⊤, I⊥

, where ⊲ I⊤ contains all atoms which are mapped to ⊤, ⊲ I⊥ contains all atoms which are mapped to ⊥, ⊲ I⊤ ∩ I⊥ = ∅. ⊲ All atoms which occur neither in I⊤ nor I⊥ are mapped to U.

Steffen H¨

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The Suppression Task 23

Three-Valued Interpretations and Models

◮ Let I denote the set of all interpretations.

⊲ Fitting 1985 (I, ⊆) is a complete semi-lattice. ∅, ∅ {p}, ∅ {q}, ∅ ∅, {q} ∅, {p} {p, q}, ∅ {p}, {q} {q}, {p} ∅, {p, q}

◮ An interpretation I is a model for a program P, in symbols I |

=3 P, iff I(P) = ⊤. ∅, ∅ | =3Ł | =3F

• {p ← q}

Steffen H¨

• lldobler

The Suppression Task 24

Logic Programs under Three-Valued Ł-Semantics

◮ We consider Łukasiewicz semantics. ◮ Let P be a logic program and I = I⊤, I⊥ be an interpretation. ◮ Proposition 1 If I = I⊤, I⊥ |

=3Ł P then I′ = I⊤, ∅ | =3Ł P.

◮ Proof Suppose I = I⊤, I⊥ |

=3Ł P ⊲ Let A ← Body ∈ P. ⊲ To show I′ | =3Ł A ← Body. ⊲ We distinguish three cases 1 A ∈ I⊤ In this case, I′ | =3Ł A ← Body. 2 A ∈ I⊥ 3 A ∈ I⊤ ∪ I⊥

Steffen H¨

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The Suppression Task 25

Proof of Proposition 1 Case 2

◮ A ∈ I⊥ In this case I(A) = ⊥ and I′(A) = U.

⊲ Because I | =3Ł A ← Body we conclude I(Body) = ⊥. ⊲ Hence, we find L ∈ Body such that I(L) = ⊥.

◮ ◮ L = B In this case I(B) = ⊥ and, hence, I′(B) = I′(L) = U. ◮ ◮ L = ¬B In this case I(B) = ⊤ and, hence, I′(B) = ⊤ and I′(L) = ⊥.

⊲ Consequently, I′(Body) ∈ {U, ⊥}. ⊲ Because I′(A) = U we conclude I | =3Ł A ← Body.

Steffen H¨

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The Suppression Task 26

Proof of Proposition 1 Case 3

◮ A ∈ I⊤ ∪ I⊥ In this case I(A) = I′(A) = U.

⊲ I(Body) = ⊥ As in the previous case we find I′(Body) ∈ {⊥, U}

◮ ◮ Consequently, I′ |

=3Ł A ← Body. ⊲ I(Body) = U In this case we find L ∈ Body with I(L) = U.

◮ ◮ Then, I′(L) = U. ◮ ◮ Consequently, I′(Body) = U. ◮ ◮ Hence, I′ |

=3Ł A ← Body. qed

◮ Example Consider P = {p ← q ∧ ¬r}.

⊲ {p, q}, {r} is a model for P, and so is {p, q}, ∅. ⊲ {p, r}, {q} is a model for P, and so is {p, r}, ∅. ⊲ {r}, {q} is a model for P, and so is {r}, ∅.

Steffen H¨

• lldobler

The Suppression Task 27

Intersection of Two Models

◮ We consider Łukasiewicz semantics; let P be a logic program. ◮ Proposition 2

Let I1 = I⊤

1 , ∅ and I2 = I⊤ 2 , ∅ be two models of P.

Then, I3 = I⊤

1

∩ I⊤

2 , ∅ is also a model for P.

◮ Proof Suppose I3 |

=3Ł P. ⊲ Then, we find A ← Body ∈ P such that I3(A ← Body) = ⊤. ⊲ We distinguish the following cases: 1 I3(A) = ⊥ and I3(Body) = ⊤ Impossible, because I⊥

3

= ∅. 2 I3(A) = ⊥ and I3(Body) = U Impossible, because I⊥

3

= ∅. 3 I3(A) = U and I3(Body) = ⊤ We find j ∈ {1, 2} with Ij(A) = U. Because Ij | =3Ł A ← Body we find Ij(Body) ∈ {U, ⊥}. (∗) Because I3(Body) = ⊤ and I⊥

3

= ∅ we find for all L ∈ Body that L is an atom and L ∈ I⊤

3 .

Hence, for all L ∈ Body we find L ∈ I⊤

j , j ∈ {1, 2}.

Consequently, Ij(Body) = ⊤, j ∈ {1, 2} contradicting (∗) qed

Steffen H¨

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The Suppression Task 28

Intersection of Two Models – Example

◮ Proposition 2 does not hold for arbitrary models.

⊲ Consider P = {p ← q1 ∧ r1, p ← q2 ∧ r2}. ⊲ Let I1 = ∅, {p, q1, r2} and I2 = ∅, {p, q2, r1}. ⊲ Both, I1 and I2, are models for P. ⊲ However, I1 ∩ I2 = ∅, {p} is not a model for P.

Steffen H¨

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The Suppression Task 29

Model Intersection

◮ We consider Łukasiewicz semantics; let P be a logic program. ◮ Theorem 3 The model intersection property holds for P,

i.e., ∩{I | I | =3Ł P} | =3Ł P.

◮ Proof Follows immediately from Propositions 1 and 2

qed

◮ Let lm3Ł P denote the least model of P under Łukasiewicz semantics.

lm3Ł {p ← q} = ∅, ∅

◮ Observation Theorem 3 does not hold under Fitting semantics:

{p, q}, ∅ | =3F {p ← q} ∅, {p, q} | =3F {p ← q} but ∅, ∅ | =3F {p ← q}

Steffen H¨

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The Suppression Task 30

Weakly Completed Logic Programs under Ł-Semantics

◮ We consider Łukasiewicz semantics. ◮ Let P be a logic program. ◮ Theorem 4 The model intersection property holds for wc P as well. ◮ Theorem 5 If I |

=3Ł wc P then I | =3Ł P.

◮ Observation Theorem 5 does not hold under F-semantics.

∅, ∅ | =3F wc {p ← q} = {p ↔ q}, but ∅, ∅ | =3F {p ← q}

Steffen H¨

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The Suppression Task 31

Reasoning with Respect to the Least Model of wc P

◮ Recall our examples

wc P4 = {l ↔ e ∧ ¬ab, ab ↔ ⊥, e ↔ ⊤} lm3Ł wc P4 = {e, l}, {ab} lm3Ł wc P4(l) = ⊤ wc P5 = {l ↔ (e ∧ ¬ab1) ∨ (t ∧ ¬ab2), ab1 ↔ ⊥, ab2 ↔ ⊥, e ↔ ⊤} lm3Ł wc P5 = {e, l}, {ab1, ab2} lm3Ł wc P5(l) = ⊤ wc P6 = {l ↔ (e ∧ ¬ab1) ∨ (o ∧ ¬ab2), ab1 ↔ ¬o, ab2 ↔ ¬e, e ↔ ⊤} lm3Ł wc P6 = {e}, {ab2} lm3Ł wc P6(l) = U wc P7 = {l ↔ e ∧ ¬ab, ab ↔ ⊥, e ↔ ⊥} lm3Ł wc P7 = ∅, {e, l, ab} lm3Ł wc P7(l) = ⊥ wc P8 = {l ↔ (e ∧ ¬ab1) ∨ (t ∧ ¬ab2), ab1 ↔ ⊥, ab2 ↔ ⊥, e ↔ ⊥} lm3Ł wc P8 = ∅, {e, ab1, ab2} lm3Ł wc P8(l) = U wc P9 = {l ← (e ∧ ¬ab1) ∨ (o ∧ ¬ab2), ab1 ↔ ¬o, ab2 ↔ ¬e, e ↔ ⊥} lm3Ł wc P9 = {ab2}, {e, l} lm3Ł wc P9(l) = ⊥

◮ This logic appears to be adequate!

Steffen H¨

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The Suppression Task 32

Completion versus Weak Completion

◮ Recall P8

wc P8 = {l ↔ (e ∧ ¬ab1) ∨ (t ∧ ¬ab2), ab1 ↔ ⊥, ab2 ↔ ⊥, e ↔ ⊥} lm3Ł wc P8 = ∅, {e, ab1, ab2} lm3Ł wc P8(l) = U c P8 = {l ↔ (e ∧ ¬ab1) ∨ (t ∧ ¬ab2), ab1 ↔ ⊥, ab2 ↔ ⊥, e ↔ ⊥, t ↔ ⊥} lm3Ł c P8 = ∅, {e, t, l, ab1, ab2} lm3Ł c P8(l) = ⊥

◮ A logic using completion instead of weak completion is not adequate.

Steffen H¨

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The Suppression Task 33

Computing the Least Models of Weakly Completed Programs

◮ How can we compute the least models of weakly completed programs? ◮ A first candidate:

Fitting’s immediate consequence operator ΦF,P(I) =

• J⊤, J⊥

, where J⊤ = {A | there exists A ← Body ∈ P with I(Body) = ⊤} and J⊥ = {A | for all A ← Body ∈ P we find I(Body) = ⊥}.

◮ Uses F-semantics. ◮ Some well-known results mostly due to Fitting 1985:

1 ΦF,P is monotone on (I, ⊆). 2 ΦF,P is continuous and, hence, admits a least fixed point denoted by lfp ΦF,P. 3 lfp ΦF,P can be computed by iterating ΦF,P on ∅, ∅. 4 The least F-model of c P is the least fixed point of ΦF,P.

◮ Inadequate for human reasoning.

Steffen H¨

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The Suppression Task 34

The Stenning and van Lambalgen Operator

◮ Stenning and van Lambalgen’s operator ΦSvL,P(I) =

• J⊤, J⊥

, where J⊤ = {A | there exists A ← Body ∈ P with I(Body) = ⊤} and J⊥ = {A | there exists A ← Body ∈ P and for all A ← Body ∈ P we find I(Body) = ⊥}.

◮ Theorem 6

1 ΦSvL,P is monotone on (I, ⊆). 2 ΦSvL,P is continuous and, hence, admits a least fixed point denoted by lfp ΦSvL,P. 3 lfp ΦSvL,P can be computed by iterating ΦSvL,P on ∅, ∅. 4 lm3Ł wc P = lfp ΦSvL,P.

Steffen H¨

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The Suppression Task 35

Computing Least Fixed Points

◮ Recall some of our examples

P4 = {l ← e ∧ ¬ab, ab ← ⊥, e ← ⊤} wc P4 = {l ↔ e ∧ ¬ab, ab ↔ ⊥, e ↔ ⊤} ΦSvL,P4(∅, ∅) = {e}, {ab} ΦSvL,P4({e}, {ab}) = {e, l}, {ab} = lfp ΦSvL,P4 = lm3Ł wc P4 P9 = {l ← e ∧ ¬ab1, ab1 ← ¬o, l ← o ∧ ¬ab2, ab2 ← ¬e, e ← ⊥} wc P9 = {l ↔ (e ∧ ¬ab1) ∨ (o ∧ ¬ab2), ab1 ↔ ¬o, ab2 ↔ ¬e, e ↔ ⊥} ΦSvL,P9(∅, ∅) = ∅, {e} ΦSvL,P9(∅, {e}) = {ab2}, {e} ΦSvL,P9({ab2}, {e}) = {ab2}, {e, l} = lfp ΦSvL,P9 = lm3Ł wc P9

Steffen H¨

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The Suppression Task 36

Summary

◮ Under Łukasiewicz semantics we obtain

Byrne 1989 Program lm3Ł wc Pi(l) Modus Ponens l (96%) P4 ⊤ Alternative Arguments l (96%) P5 ⊤ Additional Arguments l (38%) P6 U Modus Ponens and DA ¬l (46%) P7 ⊥ Alternative Arguments and DA ¬l (4%) P8 U Additional Arguments and DA ¬l (63%) P9 ⊥

◮ The approach appears to be adequate. ◮ Fitting semantics or completion is inadequate.

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The Suppression Task 37

Contraction Mappings

◮ Do we have to initialize the computation of lfp P with the empty interpretation? ◮ Banach’s Contraction Mapping Theorem

A contraction mapping f on a complete metric space has a unique fixed point; the sequence x, f(x), f(f(x)), . . . converges to this fixed point, where x is an arbitrary element from the metric space.

◮ Let P be a program. A level mapping is a mapping l from the set of atoms to N.

It is extended to negative atoms by defining l(¬A) = l(A) for each atom A.

◮ Let I be the set of all interpretations and I, J ∈ I.

dl(I, J) =   

1 2n

if I = J and I(A) = J(A) = U for all A with l(A) < n and I(A) = J(A) or I(A) = J(A) = U for some A with l(A) = n,

• therwise.

◮ Proposition 7 (Kencana Ramli 2009) (I, dl) is a complete metric space.

Steffen H¨

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The Suppression Task 38

Contraction Properties of the Semantic Operators

◮ Fitting 1985 If P is an acceptable program then ΦF,P is a contraction.

⊲ This does not hold for ΦSvL,P. ⊲ Consider P = {r ∧ q → p, r ∧ p → q}. ⊲ P is acceptable. ⊲ Both, ∅, ∅ and ∅, {p, q}, are fixed points of ΦSvL,P. ⊲ By Banach’s contraction mapping theorem ΦSvL,P is not a contraction.

◮ Theorem 8 If P is an acyclic program then ΦSvL,P is a contraction.

⊲ If P is acyclic then we find a level mapping l such that for each L1 ∧ . . . ∧ Ln → A ∈ P we have l(Li) < l(A). ⊲ dl(ΦSvL,P(I), ΦSvL,P(J)) ≤ 1

2 dl(I, J).

◮ Corollary 9 If P is an acyclic program then ΦSvL,P has a unique fixed point

which can be reached by iterating ΦSvL,P starting from some interpretation.

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The Suppression Task 39

Positive versus Negative Information

◮ Consider P = {p ← ⊤, p ← ⊥}. ◮ lm3Ł wc P = {p}, ∅. ◮ Positive information dominates negative information. ◮ However, we may take integrity constraints into account,

i.e. expressions of the form L1 ∧ . . . ∧ Ln → ⊥.

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The Suppression Task 40

The Suppression Task – Affirmation of the Consequent (AC)

◮ Byrne: Suppressing Valid Inferences with Conditionals.

Cognition 31, 61-83: 1989

◮ If she has an essay to write then she will study late in the library.

She will study late in the library. ⊲ 53% of subjects conlude that she has an essay to write.

◮ If she has an essay to write then she will study late in the library.

She will study late in the library. If she has a textbook to read she will study late in the library. ⊲ 16% of subjects conlude that she has an essay to write.

◮ If she has an essay to write then she will study late in the library.

She will study late in the library. If the library stays open she will study late in the library. ⊲ 55% of subjects conlude that she has an essay to write.

Steffen H¨

• lldobler

The Suppression Task 41

The Suppression Task – Modus Tollens (MT)

◮ Byrne: Suppressing Valid Inferences with Conditionals.

Cognition 31, 61-83: 1989

◮ If she has an essay to write then she will study late in the library.

She will not study late in the library. ⊲ 69% of subjects conlude that she does not have an essay to write.

◮ If she has an essay to write then she will study late in the library.

She will not study late in the library. If she has a textbook to read she will study late in the library. ⊲ 69% of subjects conlude that she does not have an essay to write.

◮ If she has an essay to write then she will study late in the library.

She will not study late in the library. If the library stays open she will study late in the library. ⊲ 44% of subjects conlude that she does not have an essay to write.

Steffen H¨

• lldobler

The Suppression Task 42

Abductive Frameworks and Observations

◮ Let K ⊆ L be a set of formulas called the knowledge base,

A ⊆ L be a set of formulas called abducibles, and | = ⊆ 2L × L a logical consequence relation. K, A, | = is called abductive framework.

◮ An observation O is a subset of the language L. ◮ Here

⊲ K is a logic program P (with negative facts). ⊲ L is the language underlying P. ⊲ RD

P = {A ∈ RP | A ← Body ∈ P} is the set of defined predicates in P.

⊲ RU

P = RP \ RD P is the set of undefined predicates in P.

⊲ A is the set {A ← ⊤ | A ∈ RU

P} ∪ {A ← ⊥ | A ∈ RU P}.

⊲ | = is | =lm wc

3Ł

, where P | =lm wc

3Ł

F iff lm3Ł wc P(F) = ⊤. ⊲ Observations are usually sets containing a single literal, in which case we simply write O = L instead of O = {L}.

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The Suppression Task 43

Explanations

◮ Let L be a language. ◮ Let K, A, |

= be an abductive framework and O an observation.

◮ O is explained by E (or E is an explanation for O) iff

⊲ E ⊆ A, ⊲ K ∪ E is satisfiable, ⊲ K ∪ E | = L for each L ∈ O.

◮ An explanation E for O is said to be a minimal

iff there is no explanation E′ ⊂ E for O.

Steffen H¨

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The Suppression Task 44

Sceptical and Credulous Reasoning

◮ Let P, A, |

=lm wc

3Ł

be an abductive framework, where ⊲ P is a logic program and ⊲ A = {A ← ⊤ | A ∈ RU

P} ∪ {A ← ⊥ | A ∈ RU P} the set of abducibles

◮ Let O be an observation and F a formula in the language underlying P. ◮ F follows sceptically by abduction from P and O (in symbols P, O |

=s

A F)

iff O can be explained and for all minimal explanations E we find P ∪ E | =lm wc

3Ł

F.

◮ F follows credulously by abduction from P and O (in symbols P, O |

=c

A F)

iff there exists a minimal explanation E such that P ∪ E | =lm wc

3Ł

F.

Steffen H¨

• lldobler

The Suppression Task 45

The Suppression Task – Modus Ponens and AC

◮ If she has an essay to write then she will study late in the library.

She will study late in the library. ⊲ 53% of subjects conlude that she has an essay to write.

◮ We obtain

P10 = {l ← e ∧ ¬ab, ab ← ⊥} A = {e ← ⊤, e ← ⊥} O = l

◮ Thus

lm3Ł wc P10 = ∅, {ab} lm3Ł wc (P10 ∪ {e ← ⊤}) = {e, l}, {ab} lm3Ł wc (P10 ∪ {e ← ⊥}) = ∅, {e, l, ab}

◮ Hence, {e ← ⊤} is the only minimal explanation and P10, O |

=s

A e.

Steffen H¨

• lldobler

The Suppression Task 46

The Suppression Task – Alternative Arguments and AC

◮ If she has an essay to write then she will study late in the library.

She will study late in the library. If she has a textbook to read she will study late in the library. ⊲ 16% of subjects conlude that she has an essay to write.

◮ We obtain

P11 = {l ← e ∧ ¬ab1, ab1 ← ⊥, l ← t ∧ ¬ab2, ab2 ← ⊥} A = {e ← ⊤, e ← ⊥, t ← ⊤, e ← ⊥} O = l

◮ Thus

lm3Ł wc P11 = ∅, {ab1, ab2} lm3Ł wc (P11 ∪ {e ← ⊤}) = {e, l}, {ab1, ab2} lm3Ł wc (P11 ∪ {e ← ⊥}) = ∅, {e, ab1, ab2} lm3Ł wc (P11 ∪ {t ← ⊤}) = {t, l}, {ab1, ab2} lm3Ł wc (P11 ∪ {t ← ⊥}) = ∅, {t, ab1, ab2}

◮ Hence, {e ← ⊤} and {t ← ⊤} are minimal explanations and P11, O |

=s

A e.

Steffen H¨

• lldobler

The Suppression Task 47

The Suppression Task – Additional Arguments and AC

◮ If she has an essay to write then she will study late in the library.

She will study late in the library. If the library stays open she will study late in the library. ⊲ 55% of subjects conlude that she has an essay to write.

◮ We obtain

P12 = {l ← e ∧ ¬ab1, ab1 ← ¬o, l ← o ∧ ¬ab2, ab2 ← ¬e} A = {e ← ⊤, e ← ⊥, o ← ⊤, o ← ⊥} O = l

◮ Thus

lm3Ł wc P12 = ∅, ∅ lm3Ł wc (P12 ∪ {e ← ⊤}) = {e}, {ab2} lm3Ł wc (P12 ∪ {e ← ⊥}) = {ab2}, {e, l} lm3Ł wc (P12 ∪ {o ← ⊤}) = {t}, {ab1} lm3Ł wc (P12 ∪ {o ← ⊥}) = {ab1}, {o, l} lm3Ł wc (P12 ∪ {e ← ⊤, o ← ⊤}) = {t, e, l}, {ab1, ab2}

◮ {e ← ⊤, o ← ⊤} is the only minimal explanation and P12, O |

=s

A e.

Steffen H¨

• lldobler

The Suppression Task 48

The Suppression Task – Modus Tollens (MT)

◮ If she has an essay to write then she will study late in the library.

She will not study late in the library. ⊲ 69% of subjects conlude that she does not have an essay to write.

◮ We obtain

P10 = {l ← e ∧ ¬ab, ab ← ⊥} A = {e ← ⊤, e ← ⊥} O = ¬l

◮ Thus

lm3Ł wc P10 = ∅, {ab} lm3Ł wc (P10 ∪ {e ← ⊤}) = {e, l}, {ab} lm3Ł wc (P10 ∪ {e ← ⊥}) = ∅, {e, l, ab}

◮ Hence, {e ← ⊥} is the only minimal explanation and P10, O |

=s

A ¬e.

Steffen H¨

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The Suppression Task 49

The Suppression Task – Alternative Arguments and MT

◮ If she has an essay to write then she will study late in the library.

She will not study late in the library. If she has a textbook to read she will study late in the library. ⊲ 69% of subjects conlude that she does not have an essay to write.

◮ We obtain

P11 = {l ← e ∧ ¬ab1, ab1 ← ⊥, l ← t ∧ ¬ab2, ab2 ← ⊥} A = {e ← ⊤, e ← ⊥, t ← ⊤, t ← ⊥} O = ¬l

◮ Thus

lm3Ł wc P11 = ∅, {ab1, ab2} lm3Ł wc (P11 ∪ {e ← ⊤}) = {e, l}, {ab1, ab2} lm3Ł wc (P11 ∪ {e ← ⊥}) = ∅, {e, ab1, ab2} lm3Ł wc (P11 ∪ {t ← ⊤}) = {t, l}, {ab1, ab2} lm3Ł wc (P11 ∪ {t ← ⊥}) = ∅, {t, ab1, ab2} lm3Ł wc (P11 ∪ {e ← ⊥, t ← ⊥}) = ∅, {e, l, t, ab1, ab2}

◮ {e ← ⊥, t ← ⊥} is the only minimal explanation and P11, O |

=s

A ¬e.

Steffen H¨

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The Suppression Task 50

The Suppression Task – Additional Arguments and MT

◮ If she has an essay to write then she will study late in the library.

She will not study late in the library. If the library stays open she will study late in the library. ⊲ 44% of subjects conlude that she does not have an essay to write.

◮ We obtain

P12 = {l ← e ∧ ¬ab1, ab1 ← ¬o, l ← o ∧ ¬ab2, ab2 ← ¬e} A = {e ← ⊤, e ← ⊥, o ← ⊤, o ← ⊥} O = ¬l

◮ Thus

lm3Ł wc P12 = ∅, ∅ lm3Ł wc (P12 ∪ {e ← ⊤}) = {e}, {ab2} lm3Ł wc (P12 ∪ {e ← ⊥}) = {ab2}, {e, l} lm3Ł wc (P12 ∪ {o ← ⊤}) = {t}, {ab1} lm3Ł wc (P12 ∪ {o ← ⊥}) = {ab1}, {o, l}

◮ Hence, {e ← ⊥} and {o ← ⊥} are minimal explanations and P12, O |

=s

A ¬e.

Steffen H¨

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The Suppression Task 51

Weak Completion is Needed

◮ Reconsider the case modus ponens with positive observation, i.e.

⊲ P10 = {l ← e ∧ ¬ab, ab ← ⊥}, ⊲ A = {e ← ⊤, e ← ⊥}, and ⊲ O = l.

◮ Now consider P10, A, |

=3Ł instead of P10, A, | =lm wc

3Ł

• ⊲ P10 |

=3Ł l ⊲ P10 ∪ {e ← ⊤} | =3Ł l (because ab can be mapped to ⊤). ⊲ P10 ∪ {e ← ⊥} | =3Ł l ⊲ P10 ∪ A | =3Ł l Hence, the observation can not be explained at all (in contrast to Byrne 1989).

Steffen H¨

• lldobler

The Suppression Task 52

Completion is Insufficient

◮ Reconsider the case modus ponens with positive observation, i.e.

⊲ P10 = {l ← e ∧ ¬ab, ab ← ⊥}, ⊲ A = {e ← ⊤, e ← ⊥}, and ⊲ O = l.

◮ Now consider P10, A, |

=c

3Ł instead of P10, A, |

=lm wc

3Ł

, where P | =c

3Ł F iff F holds in all models for c P.

⊲ c P10 = {l ↔ e ∧ ¬ab, ab ↔ ⊥, e ↔ ⊥} | =3Ł ¬l ⊲ c P10 | =3Ł ¬e The observation is inconsistent with the knowledge base and, thus, cannot be explained at all (in contrast to Byrne 1989).

Steffen H¨

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The Suppression Task 53

Explanations must be Completed as well

◮ Reconsider the case modus ponens with negative observation, i.e.

⊲ P10 = {l ← e ∧ ¬ab, ab ← ⊥}, ⊲ A = {e ← ⊤, e ← ⊥}, and ⊲ O = ¬l.

◮ Now (weakly) complete only the program, but not the explanations.

⊲ wc P10 = {l ↔ e ∧ ¬ab, ab ↔ ⊥}. ⊲ wc P10 | =3Ł ¬l ⊲ wc (P10) ∪ {e ← ⊤} | =3Ł ¬l ⊲ wc (P10) ∪ {e ← ⊥} | =3Ł ¬l (because e can be mapped to ⊤). ⊲ wc (P10) ∪ A | =3Ł ¬l Hence, the observation can not be explained (in contrast with Byrne 1989).

Steffen H¨

• lldobler

The Suppression Task 54

Sceptical versus Creduluos Reasoning

◮ Reconsider the case alternative arguments with positive observation, i.e.,

⊲ P11 = {l ← e ∧ ¬ab1, ab1 ← ⊥, l ← t ∧ ¬ab2, ab2 ← ⊥}, ⊲ A = {e ← ⊤, e ← ⊥}, and ⊲ O = l.

◮ Now consider P11, A, |

=lm wc

3Ł

and reason credulously: ⊲ There are two minimal explanations, viz. {e ← ⊤} and {e ← ⊥}. ⊲ Hence, P11, l | =s

A e, but P11, l |

=c

A e.

Creduluos reasoning is inconsistent with Byrne 1989.

Steffen H¨

• lldobler

The Suppression Task 55

Summary

◮ Let P10

= {l ← e ∧ ¬ab, ab ← ⊥} P11 = {l ← e ∧ ¬ab1, ab1 ← ⊥, l ← t ∧ ¬ab2, ab2 ← ⊥} P12 = {l ← e ∧ ¬ab1, ab1 ← ¬o, l ← o ∧ ¬ab2, ab2 ← ¬e}

◮ We obtain

Byrne 1989 P10, l | =s

A e

e(53%) P11, l | =s

A e

e(16%) P12, l | =s

A e

e(55%) P10, ¬l | =s

A ¬e

¬e(69%) P11, ¬l | =s

A ¬e

¬e(69%) P12, ¬l | =s

A e

¬e(44%)

Steffen H¨

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The Suppression Task 56

Summary

◮ Logic appears to be adequate for the suppression task if

⊲ weak completion, ⊲ Łukasiewicz semantics, ⊲ the Stenning and van Lambalgen semantic operator, and ⊲ abduction are used.

◮ Human Reasoning is modeled by

⊲ reasoning towards an appropriate logic program P and, thereafter, ⊲ reasoning with respect to the least Ł-model of the weak completion of P.

◮ This approach matches data from studies in human reasoning. ◮ There is a connectionist encoding.

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The Suppression Task 57

Discussion

◮ Stenning, van Lambalgen 2008 propose spreading-activation networks

like KBANN (Towell, Shavlik 1993) with two units for each propositional letter and an inhibitory link between them.

◮ Logical threshold units can be replaced by bipolar sigmoidal ones following

d’Avila Garcez, Zaverucha, Carbalho 1997. ⊲ Networks can be trained by backpropagation, ⊲ but backpropagation is not neurally plausible.

Steffen H¨

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The Suppression Task 58

Some Open Problems (1)

◮ Negation

⊲ How is negation treated in human reasoning?

◮ Errors

⊲ How can frequently made errors be explained in the proposed approach?

◮ Łukasiezicz logic

⊲ In a Łukasiezicz logic the semantic deduction theorem does not hold. ⊲ Is this adequate with respect to human reasoning?

◮ Completion

⊲ Under which conditions is human reasoning adequately modeled by completion and/or weak completion?

Steffen H¨

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The Suppression Task 59

Some Open Problems (2)

◮ Contractions

⊲ Do humans exhibit a behavior which can be adequately modeled by contractional semantic operators? ⊲ Can we generate appropriate level mappings by studying the behavior of humans?

◮ Explanations

⊲ Do humans consider minimal explanations? ⊲ In which order are (minimal) explanations generated by humans if there are several? ⊲ Does attention play a role in the selection of (minimal) explanations?

◮ Stable coalitions

⊲ Do stable coalitions occur in human reasoning? ⊲ How are they deactivated?

Steffen H¨

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The Suppression Task 60

Some Open Problems (3)

◮ Reasoning

⊲ Do humans reason sceptically or credulously? ⊲ How does a connectionist realization of sceptical reasoning looks like?

◮ Theory revision

⊲ How is theory revision modeled in human reasoning?

◮ Relation to other semantics

⊲ What is the relation between the proposed approach and well-founded and/or stable and/or projection semantics?

Steffen H¨

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The Suppression Task 61