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A detailed logical resolution process showcasing how every horse can outrun every rabbit using given premises and background knowledge. Learn about linear resolution, Theo theorem prover, and Clause Curiosity in an engaging way.
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Example Everybody loves somebody. Everybody loves a lover. Show that everybody loves everybody. "x.$y.loves( x, y) "u."v."w.(loves(v, w) Þ loves(u, v)) Ø"x."y.loves( x, y) negated conclusion In clausal form they become: {loves(x, f( x))} {Øloves(v, w), loves(u, v)} {Øloves(jack, jill)}
{loves(x, f(x))} Premise • {Øloves(v, w), loves(u, v)} Premise • {Øloves(jack, jill)} Goal • {loves(u, x)} 1,2 • {} 4,3
Horses, dogs, and rabbits Every horse can outrun every dog. Some greyhounds can outrun every rabbit. Show that every horse can outrun every rabbit. "x."y.(Horse( x) Ù Dog( y) Þ Faster( x, y)) $y.(Greyhound(y) Ù "z.(Rabbit( z) Þ Faster( y, z))) "y.(Greyhound( y) Þ Dog( y)) (background knowledge) "x."y."z.(Faster(x, y) Ù Faster( y, z) Þ Faster( x, z)) (background knowledge) Ø"x."y.(Horse(x) Ù Rabbit(y) Þ Faster(x, y)) negated conclusion
"x."y.(Horse( x) Ù Dog( y) Þ Faster( x, y)) $y.(Greyhound(y) Ù "z.(Rabbit( z) Þ Faster( y, z))) "y.(Greyhound( y) Þ Dog( y)) "x."y."z.(Faster(x, y) Ù Faster( y, z) Þ Faster( x, z)) Ø"x."y.(Horse(x) Ù Rabbit(y) Þ Faster(x, y)) In clausal form they become: {ØHorse( x), ØDog(y), Faster( x, y)} {Greyhound(Greg)} {ØRabbit(z), Faster(Greg, z)} {ØGreyhound(y), Dog(y)} {ØFaster(x, y), ØFaster(y, z), Faster(x, z)} …continued
Let’s transform the goal into clausal form: Ø"x."y.(Horse(x) Ù Rabbit(y) Þ Faster(x, y)) IØ"x."y.(Ø(Horse(x) Ù Rabbit(y)) ÚFaster(x, y)) NØ"x."y.(ØHorse(x) Ú ØRabbit(y) ÚFaster(x, y)) x. y.(Horse(x) Ù Rabbit(y) Ù ØFaster(x, y)) Sx. y.(Horse(x) Ù Rabbit(y) Ù ØFaster(x, y)) E Horse(Jack) Ù Rabbit(Smith) Ù ØFaster(Jack, Smith)) A Horse(Jack) Ù Rabbit(Smith) Ù ØFaster(Jack, Smith)) D Horse(Jack) Ù Rabbit(Smith) Ù ØFaster(Jack, Smith)) O {Horse(Jack)} {Rabbit(Smith)} {ØFaster(Jack, Smith)}
Let’s try to infer the {} using resolution: • {ØHorse(x), ØDog(y), Faster(x, y)} • {Greyhound(Greg)} • {ØRabbit(z), Faster(Greg, z)} • {ØGreyhound(y), Dog(y)} • {ØFaster(x, y), ØFaster(y, z), Faster(x, z)} • {Horse(Jack)} • {Rabbit(Smith)} • {ØFaster(Jack, Smith)} • {ØDog(y), Faster(Jack, y)} 1,6 • {Faster(Jack, y), ØGreyhound(y)} 4,9 • {Faster(Jack, Greg)} 2,10 • {Faster(Greg, Smith)} 3,7 • {ØFaster(Greg, z), Faster(x, z)} 5,11 • {Faster(x, Smith)} 12,13 • {} 8,14
Linear Restriction Strategy • A linear resolution is one in which one of the parents is in the initial database or an ancestor of the other parent. • Example. 1. {p, q} 2. {p,Øq} 3. {Øp, q} 4. {Øp,Øq} 5. {p} 1,2 6. {q} 3,5 7. {Øp} 4,6 8. {} 5,7 Linear Resolution is refutation complete!!
Theo • Theo is theorem prover based on linear resolution (as well as other resolution strategies besides linearity) • Accepts input in clausal form. • Uses: ~ for | for (doesn’t use the set notation for clauses, but the OR, i.e. | notation, e.g. p(x) | q(y) instead of {p(x), q(y)}) • Distinguish variables from constants, by adding parenthesis. E.g. tuna(). • If you omit parenthesis and write instead just tuna, it will be considered a variable.
Clausal Curiosity (Theo format) animal(f(x1)) | loves(g(x1),x1) ~loves(x2,f(x2)) | loves(g(x2),x2) ~animal(y1) | ~kills(x3,y1) | ~loves(z,x3) ~animal(y2) | loves(jack,y2) kills(jack(),tuna()) | kills(curiosity(),tuna()) ~cat(x4) | animal(x4) cat(tuna()) negated_conclusion ~kills(curiosity(),tuna()) • {animal(f(x1)), loves(g(x1),x1)} • {loves(x2,f(x2)), loves(g(x2),x2)} • {animal(y1), kills(x3,y1), loves(z,x3)} • {animal(y2), loves(jack,y2)} • {kills(jack,tuna), kills(curiosity,tuna)} • {cat(x4), animal(x4)} • {cat(tuna)} • kills(curiosity,tuna)
Theo output Proof Found! Axioms: 1 >animalfx lovesgxx 2#>~lovesxfx lovesgxx 3 >~animalx ~killsyx ~loveszy 4 >~animalx lovesyx 5#>killsjacktuna killscuriositytuna 6 >~catx animalx 7 >cattuna Negated conclusion: 8S>~killscuriositytuna Phase 0 clauses used in proof: 9S>(8a*5b) killsjacktuna 10S>(9a*3b) ~animaltuna ~lovesxjack 11#>(10a*6b) ~lovesxjack ~cattuna 12S>(11b*7a) ~lovesxjack 15 >(4a*1a) lovesxfy lovesgyy 16 >(15a*2a) lovesgxx Phases 1 and 2 clauses used in proof: 17S>(16a,12a) []
Compile • You can create well formed formulas (WFF) i.e. FOL sentences and then convert into clausal form using “compile” • @ is forall • ! is exists • Variables and constants distinguished by quantification. Variables are quantified, constants aren’t. • Use ( and ) for functions and relations. • Use { and } for grouping sentences. • & is AND, | is OR, => is IMPLIES Example: @x: met_with_at_sea(x) & noticed_thing(x) => entered_in_log(x).
Books… ; The only books in this library, that I do not recommend for reading, are unhealthy in tone; ; The bound books are all well-written; ; All the romances are healthy in tone; ; I do not recommend that you read any of the unbound books. ; ; All the romances in this library are well-written. @x: book(x) & ~recommend(x) => ~healthy(x). @x: book(x) & ~healthy(x) => ~recommend(x). @x: book(x) & bound(x) => wellwritten(x). @x: romance(x) => healthy(x). @x: romance(x) => book(x). @x: book(x) & ~bound(x) => ~recommend(x). ; theorem (negated) ~{ @x: romance(x) => wellwritten(x) }.
Songs ; Things sold in the street are of no great value; ; Nothing but rubbish can be sold for a song; ; Eggs of the Great Auk are very valuable; ; It is only what is sold in the streets which is really rubbish. ; ; Conclusion: An egg of the Great Auk is not to be had for a song. @x: soldInStreet(x) => ~ofGreatValue(x). @x@y: exchange(y,x) & song(x) => rubbish(y). @x: greatAukEgg(x) => ofGreatValue(x). @x: rubbish(x) => soldInStreet(x). @x: soldInStreet(x) => rubbish(x). ; theorem (negated) ~{ ~{!x!y: greatAukEgg(y) & exchange(y,x) & song(x)} }.
Birds ; No birds, except ostriches, are 9 feet high; ; There are no birds in this aviary that belong to anyone but me; ; No ostrich lives on mince-pies; ; I have no birds less than nine-feet high. ; ; Conclusion: No bird in this aviary lives on mince-pies. @x: bird(x) & height(x,Nine) => ostrich(x). @x: ostrich(x) => bird(x). @x: bird(x) => belongs(x,Me). @x: ostrich(x) => ~lives(x,MincePies). @x: bird(x) & belongs(x,Me) => height(x,Nine). ;theorem (negated) ~{ @x: bird(x) => ~lives(x,MincePies) }.
British Lion ; No discussions in our Debating-Club can rouse the British Lion, so long as they are checked when they become too noisy; ; Discussions, unwisely conducted, endanger the peacefulness of our Debating-Club; ; Discussions, that go on while Tomkins is in the Chair, rouse the British Lion; ; Discussions in our Debating-Club, when wisely conducted, are always checked when they become too noisy. ~{ !x: discussion(x) & in(x, OurDebatingClub) & {noisy(x) => checked(x)} & rouse(x, BritishLion) }. @x: discussion(x) & checked(x) => ~rouse(x, BritishLion). ;background @x@y: discussion(x) & ~wiselyConducted(x) & in(x,y) => endanger(x, Peacefulness, y). @x@y: discussion(x) & in(x,y) & chair(Tomkins, y) => rouse(x, BritishLion). @x: discussion(x) & in(x, OurDebatingClub) & wiselyConducted(x) => {noisy(x) => checked(x)}.
; Conclusion: Discussions, that go on while Tomkins is in the chair, endanger the peacefulness of our Debating-Club.. ; theorem (negated) ~{@x: discussion(x) & in(x,OurDebatingClub) & chair(Tomkins, OurDebatingClub) => endanger(x, Peacefulness, OurDebatingClub) }.
Algebra ; In an associative system with left and right solutions, ; there is a right identity element. ; P(x,y,z) means x*y = z ; (x*y)*z = x*(y*z) ; x*y = u y*z = v @x@y@z@u@v@w: p(x,y,u) & p(y,z,v) & p(u,z,w) => p(x,v,w). @x@y@z@u@v@w: p(x,y,u) & p(y,z,v) & p(x,v,w) => p(u,z,w). @x@y!s: p(x,s,y). @x@y!s: p(s,x,y). ; theorem (negated) ~{!e@x: p(x,e,x)}.
