SNePS: A Logic for Natural Language Understanding and Commonsense Reasoning

SNePS: A Logic for Natural Language Understanding and Commonsense Reasoning Stuart C. Shapiro Department of Computer Science and Engineering and Center for Cognitive Science State University of New York at Buffalo shapiro@cse.buffalo.edu

Based on Stuart C. Shapiro, “SNePS: A Logic for Natural Language Understanding and Commonsense Reasoning,” in Lucja Iwanska and Stuart C. Shapiro, Eds., Natural Language Processing and Knowledge Representation: Language for Knowledge and Knowledge for Language, AAAI Press/The MIT Press, 2000.

Abstract Design a better logic Than FOPL For NLU and CSR SNePS

Presentation Approach • Problem • Difficulties of FOPL • Solution in SNePS • Use SNePSLOG.

Outline • Set-Oriented Logical Connectives • The Unique Variable Binding Rule • Set Arguments • “Higher-Order” Logic • Intensional Representation • The Numerical Quantifiers • Contexts • Relevance Logic • Circular and Recursive Rules • Practical Donkey Sentences

Twenty Questions • Everything is an animal, a vegetable, or a mineral. • Squash is a vegetable • Is squash an animal? • a mineral? • Marble is neither an animal nor a vegetable. • Is marble a mineral?

+ + Twenty Questions in FOPL? x[Animal(x)  Vegetable(x)  Mineral(x)] but don’t want inclusive or x[Animal(x) Vegetable(x) Mineral(x)] T T T F T So don’t want exclusive or either

andor andor(i, j){Pi, ..., Pn} True iff at least i, and at most j of the Pi are True

Twenty Questions in SNePSLOG : all(x)(andor(1,1){animal(x), vegetable(x), mineral(x)}). : vegetable(squash)! VEGETABLE(SQUASH) ~ANIMAL(SQUASH) ~MINERAL(SQUASH)

Twenty Questions in SNePSLOG II : andor(0,0){animal(marble), vegetable(marble)}! ~ANIMAL(MARBLE) ~VEGETABLE(MARBLE) MINERAL(MARBLE)

Equivalent Statements • For every object, the following statements are equivalent: • It is human. It is a featherless biped. It is a rational animal. • Socrates is human. • Is Socrates a featherless biped? A rational animal? • Snoopy is not a featherless biped. • Is Snoopy a rational animal? A human?

Equivalent Statements in FOPL? x[Human(x)  Featherless-Biped(x) Rational-Animal(x)] wrong: F  F  T T T

thresh thresh(i, j){Pi, ..., Pn} True iff either fewer than i, or more than j of the Pi are True Note: thresh(i, j) ~andor(i, j)

thresh Abbreviation thresh(i){Pi, ..., Pn} for thresh(i, n-1){Pi, ..., Pn}

Equivalent Statementsin SNePSLOG 1 : all(x)(thresh(1){human(x), featherless-biped(x), rational-animal(x)}). : human(Socrates)! HUMAN(SOCRATES) FEATHERLESS-BIPED(SOCRATES) RATIONAL-ANIMAL(SOCRATES)

Equivalent Statementsin SNePSLOG 2 : ~featherless-biped(Snoopy)! ~RATIONAL-ANIMAL(SNOOPY) ~FEATHERLESS-BIPED(SNOOPY) ~HUMAN(SNOOPY)

Putative Inclusive or • If Hilda is in Boston or Kathy is in Las Vegas, then Eve is in Providence. • What if Hilda is in Boston and Kathy is in Las Vegas? • Rips 1983: ___  ___  ___

or-entailment {Pi, ..., Pn} v=> {Qi, ..., Qn} True iff for all i, j Pi Qj

Hilda and Kathy in SNePSLOG : {in(Hilda, Boston), in(Kathy, Las_Vegas)} v=> {in(Eve, Providence)}. : in(Hilda, Boston)! Since {IN(HILDA,BOSTON),IN(KATHY,LAS_VEGAS)} v=> {IN(EVE,PROVIDENCE)} and IN(HILDA,BOSTON) I infer IN(EVE,PROVIDENCE)

Disappointed Voter Problem “If someone votes for X and someone votes for Y, one of them will be disappointed” : all(u,v,x,y)( {votesfor(u,x), votesfor(v,y)} &=> {andor(1,1){disappointed(u), disappointed(v)}}). : all(u,x)({votesfor(u,x),wins(x)} &=> {~disappointed{u}}).

Hillary and Elizabeth Vote : votesfor(Hillary, Bill). : votesfor(Elizabeth, Bob). : wins(Bill). : disappointed(?x)? DISAPPOINTED(ELIZABETH) ~DISAPPOINTED(HILLARY)

FOPL Disappointment {votesfor(Hillary, Bill), votesfor(Hillary, Bill)} &=> {andor(1,1) {disappointed(Hillary), disappointed(Hillary)}}  disappointed(Hillary)

Unique Variable Binding Rule (UVBR) Two variables in one wff cannot be replaced by the same term. or Two terms in an mgu cannot be equal.

Sisters “Mary, Sue, and Sally are sisters.”

Sisters in FOPL Sisters(Mary, Sue)  Sisters(Sue, Sally) (x,y)[Sisters(x,y)  Sisters(y,x)] (x,z)[x  z  ((y)[sisters(x,y)  sisters(y,z)]  Sisters(x,z))]

Reduction Inference P(s1,...,si,,si+1,...,sm), ’  P(s1,...,si,’,si+1,...,sm) P(s1,...,si,{t1,…,tn},si+1,...,sm) P(s1,...,si,ti,si+1,...,sm)

Sisters in SNePSLOG : sisters({Mary, Sue, Sally}). : all(x,y)(sisters({x,y}) => {likes(x,y), likes(y,x)}). : likes(?x,?y)? LIKES(SUE,MARY) LIKES(MARY,SUE) LIKES(SALLY,SUE) LIKES(SUE,SALLY) LIKES(MARY,SALLY) LIKES(SALLY,MARY)

Inadequacy of FOPL 1 • If R is a transitive relation and R(x, y) and R(y, z) then R(x, z).

Term Logic Every expression in SNePS is a term.

SNePSLOG Transitivity Rule : all(R)(Transitive(R) => all(x,y,z)({R(x,y), R(y,z)} &=> {R(x,z)})). : Transitive(bigger). : bigger(elephant,lion). : bigger(lion,mouse). : bigger(elephant,mouse)? BIGGER(ELEPHANT,MOUSE)

Inadequacy of FOPL 2 • Everything Bob believes is true. • Bob believes everything Bill believes. • Bill believes Kevin’s favorite proposition. • Kevin’s favorite proposition is that John is taller than Mary. • Is John taller than Mary?

Quantifying Over Propositions : all(p)(Believes(Bob, p) => p). : all(p)(Believes(Bill, p) => Believes(Bob, p)). : all(p) (Favorite-proposition(Kevin, p) => Believes(Bill, p)). : Favorite-proposition(Kevin, Taller(John, Mary)). : Taller(John, Mary)? TALLER(JOHN,MARY)

Opaque Contexts “George IV wished to know whether Scott was the author of Waverly” [Russell 1906]

Intensional Representation Every SNePS term represents (denotes) an intensional (mental) entity.

Uniqueness Principle No two SNePS terms denote the same entity.

McCarthy’s Telephone Number Problem : all(R)(Transparent(R) => all(a,x,y)({R(a,x), =({x,y})} &=> {R(a,y)})). : Transparent(Dial). : =({Telephone(Mike), Telephone(Mary)}). : Know(Pat, Telephone(Mike)). : Dial(Pat, Telephone(Mike)).

Correct Answer to Telephone Number Problem : ?what(Pat, ?which)? DIAL(PAT,TELEPHONE(MARY)) KNOW(PAT,TELEPHONE(MIKE)) DIAL(PAT,TELEPHONE(MIKE))

Reasoning by Elimination • No one has more than one mother. • Jane is John's mother. • Is Mary John's mother? • The committee members are Chris, Leslie, Pat, and Stevie. At least two of them are women. • Leslie and Stevie are men. • Is Pat a man or a woman?

Numerical Quantifiers nexists(i,j,k)(x) ({P1(x),..., Pn(x)}: {Q(x)})} There are k individuals that satisfy P1(x) ... Pn(x) and, of them, at least i and at most j also satisfy Q(x)

Numerical QuantifierRule of Inference 1 • If j individuals are known that satisfy P1(x)  …  Pn(x)  Q(x) then every other individual that satisfies P1(x) ... Pn(x) also satisfies ~Q(x)

Numerical QuantifierRule of Inference 2 • If k-i individuals are known that satisfy P1(x)  …  Pn(x)  ~ Q(x) then every other individual that satisfies P1(x)  …  Pn(x) also satisfies Q(x)

Numerical Quantifier Forms • nexists(i,j,k) • nexists(_,j,_) • nexists(i,_,k)

Reasoning by Eliminationin SNePSLOG 1 : all(x)(Person(x) => nexists(_,1,_)(y)({Person(y)}: {Mother(y,x)})). : Person({John, Jane, Mary}). : Mother(Jane, John). : Mother(Mary, John)? ~MOTHER(MARY,JOHN)

Reasoning by Eliminationin SNePSLOG 2 : Member({Chris, Leslie, Pat, Stevie}). : nexists(2,_,4)(x) ({Member(x)}: {Woman(x)}). : all(x)(Member(x) => andor(1,1){Man(x), Woman(x)}).

Committee Problem Answer : Man({Leslie, Stevie}). : ?What(Pat)? WOMAN(PAT) MEMBER(PAT) ~MAN(PAT)

SNePS: A Logic for Natural Language Understanding and Commonsense Reasoning