Hands-On: Diet Puzzle What is the secret of your long life? a man - PowerPoint PPT Presentation

Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Hands-On: “Diet Puzzle” “What is the secret of your long life?” a man was asked. How to Build an Automated Theorem Prover He replied: “I strictly follow my diet: If I don’t drink beer for dinner, then I always have fish. Any time I have both beer Part 2a: Implementing a Propositional Prover and fish for dinner, then I do without ice cream. If I have ice cream or don’t have beer, then I never eat fish.” Jens Otten Question 1: “Does the man have beer for dinner?” Question 2: “Does he have both ice cream and fish for dinner?” University of Oslo ◮ formalize this puzzle in the language of (propositional) logic! ◮ ( ( ¬ beer → fish) ∧ (beer ∧ fish → ¬ ice cream) ∧ (ice cream ∨ ¬ beer → ¬ fish) ) → beer (for Question 1) → (ice cream ∧ fish) (for Question 2) Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 1 / 33 Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 2 / 33 Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Representing First-Order Logic in Prolog Hands-On: Representing Logic in Prolog ◮ download SWI-Prolog at https://www.swi-prolog.org ◮ terms ( f ( a , x )) are represented by Prolog terms ( f(a,X) ) , i.e., ◮ open your favourite editor and express the “diet puzzle” in variables ( x , y , z ) are represented by Prolog variables ( X , Y , Z ) Prolog syntax and save it in files named ex diet1.pl and starting with a capital letter; ex diet2.pl constants ( a , b , c ) are represented by Prolog constants ( a , b , c ) ◮ ex diet1.pl : starting with a small letter fof(diet1,conjecture, ◮ atomic formulae ( p , q ( a , x )) are represented by Prolog terms, ( ∼ b => f) & ((b & f) => ∼ i) & ((i | ∼ b) => ∼ f) => b ). i.e., propositional variables and predicate symbols are represented ◮ ex diet2.pl : by Prolog constants ( p , q(a,X) ) starting with a small letter fof(diet2,conjecture, ◮ logical operators ¬ , ∧ , ∨ , → , ∀ x , ∃ x are represented by ( ∼ b => f) & ((b & f) => ∼ i) & ((i | ∼ b) => ∼ f) => (i&f) ). ∼ , & , | , => , ![X]: , ?[X]: ◮ you can find the code for this example and all of the following Prolog source code on the tutorial’s website at ◮ in a Prolog file, a formula F is represented (in TPTP syntax) by http://jens-otten.de/tutorial tableaux19 fof( formula name ,conjecture, F ). Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 3 / 33 Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 4 / 33

Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization How to Implement a Calculus in Prolog? Representing Sequents in Prolog Remember: A (proof) calculus consists of Remember: A sequent has the form ◮ axioms of the form Γ = ⇒ ∆ with Γ = { A 1 , . . . , A n } , ∆ = { B 1 , . . . , B m } w · · · w 1 w 2 w n ◮ rules of the form and is represented in Prolog by two lists: w ( w 1 , . . . , w n are the premises, w is the conclusion) Γ > ∆ with Γ= [A1,...,An] , ∆= [B1,...,Bm] ◮ the predicate select1(A,G,G’) succeeds if A is element of list G and It can be implemented in Prolog in the following way: returns G’ as G without A , e.g. select1(b,[a,b,c],L) � L=[a,c] ◮ axioms: prove(w). A ′ , Γ ′ = ⇒ ∆ ◮ rules: with Γ= { A }∪ Γ ′ is implemented by prove(w) :- prove(w1), prove(w2), ..., prove(wn). ◮ a rule of the form A , Γ ′ = ⇒ ∆ ◮ query “ is there a proof for w ′ ?” : ?- prove(w’). prove(G > D) :- select1(A,G,G’), prove([A’|G’] > D). Proof search (with backtracking) is done (implicitly) by Prolog. ◮ the proof search of formula A is invoked by ?- prove([] > [A]). ◮ for a complete proof search, iterative deepening has to be added ◮ implementation of the select1 predicate: ◮ first optimization: bottom-up proof search select1(X,L,L1) :- append(L2,[X|L3],L), append(L2,L3,L1). Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 5 / 33 Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 6 / 33 Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Hands-On: Getting Started Sequent Calculus – Axiom ◮ definition of the logical operators and the select1 predicate ◮ new operators are defined with: op(Precedence,Type,Name) ◮ the only axiom ◮ leanseq v0.pl : :- op( 500, fy, ∼ ). % negation axiom :- op(1000, xfy, &). % conjunction Γ 1 , A = ⇒ A , ∆ 1 :- op(1100, xfy, ’|’). % disjunction :- op(1110, xfy, =>). % implication :- op( 500, fy, !). % universal quantifier: ![X]: ◮ leanseq v1.pl : :- op( 500, fy, ?). % existential quantifier: ?[X]: :- op( 500,xfy, :). % axiom % ------------------------------------------------------- prove(G > D) :- member(A,G), member(A,D). prove0(F) :- prove([] > [F]). % ------------------------------------------------------- % some space for the actual prover % ------------------------------------------------------- select1(X,L,L1) :- append(L2,[X|L3],L), append(L2,L3,L1). % ------------------------------------------------------- Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 7 / 33 Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 8 / 33

Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Sequent Calculus – Rules for ∧ Sequent Calculus – Rules for ∨ ◮ rules for ∧ (conjunction) ◮ rules for ∨ (disjunction) Γ 1 , A , B = ⇒ ∆ Γ = ⇒ A , ∆ 1 Γ = ⇒ B , ∆ 1 Γ 1 , A = ⇒ ∆ Γ 1 , B = ⇒ ∆ Γ = ⇒ A , B , ∆ 1 ∧ -left ∧ -right ∨ -left ∨ -right Γ 1 , A ∧ B = ⇒ ∆ Γ = ⇒ A ∧ B , ∆ 1 Γ 1 , A ∨ B = ⇒ ∆ Γ = ⇒ A ∨ B , ∆ 1 ◮ leanseq v1.pl : ◮ leanseq v1.pl : % conjunction % disjunction prove(G > D) :- select1(A&B,G,G1), prove(G > D) :- select1(A|B,G,G1), prove([A,B|G1] > D). prove([A|G1] > D), prove([B|G1] > D). prove(G > D) :- select1(A&B,D,D1), prove(G > D) :- select1(A|B,D,D1), prove(G > [A|D1]), prove(G > [B|D1]). prove(G > [A,B|D1]). Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 9 / 33 Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 10 / 33 Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Logic in Prolog Sequent Calculus in Prolog First-Order Logic Free Variables Iterative Deepening Skolemization Sequent Calculus – Rules for → Sequent Calculus – Rules for ¬ ◮ rules for ¬ (negation) ◮ rules for → (implication) Γ 1 = ⇒ A , ∆ Γ , A = ⇒ ∆ 1 Γ 1 = ⇒ A , ∆ Γ 1 , B = ⇒ ∆ Γ , A = ⇒ B , ∆ 1 ¬ -right → -right ¬ -left → -left Γ 1 , ¬ A = ⇒ ∆ Γ = ⇒ ¬ A , ∆ 1 Γ 1 , A → B = ⇒ ∆ Γ = ⇒ A → B , ∆ 1 ◮ leanseq v1.pl : ◮ leanseq v1.pl : % implication % negation prove(G > D) :- select1(A=>B,G,G1), prove(G > D) :- select1( ∼ A,G,G1), prove(G1 > [A|D]), prove([B|G1] > D). prove(G1 > [A|D]). prove(G > D) :- select1(A=>B,D,D1), prove(G > D) :- select1( ∼ A,D,D1), prove([A|G] > [B|D1]). prove([A|G] > D1). Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 11 / 33 Jens Otten (UiO) How to Build a Theorem Prover — Part 2 TABLEAUX Tutorial 2019 12 / 33