Automated Reasoning Systems

1. Automated Reasoning Systems For first order Predicate Logic

2. Is F a logical consequence of T ? AR: general context. • Given is a knowledge base in predicate logic: T • is a set of formulae in first order logic • formally also called: a “Theory’’ • Given is also an additional first order formula: F • Notation: T |=F (T implies F) • Find reasoning techniques that allow to decide on this for EACH F and T. • Requirements: • correctness -- completeness -- efficiency

3. There CANNOT EXIST AN ALGORITHMthat decides whether T|=F, for any theory T and any formula F. • There exists a reasoning technique, such that for any theory T and formula F, such that T|=F, the reasoning technique provesT |= F. AR: decidability. • Theorem Church ‘36: • BUT: semi-decidable ! • Completeness Theorem of Goedel ‘31: • SO: if F follows from T, then we find a proof, else it is possible that the procedure doesn’t terminate.

4. The theorems of Church and Goedel are contradicting: We can try to prove F and ~F in parallel, and according to Goedel’s theorem one of these must succeed after a finite time. Wait a second ... • Wrong ! • Let: T ={smart(Kelly)} en F =strong(Kelly) • Although strong(Kelly)  ~strong(Kelly) is always true, we have: • neither: {smart(Kelly)} |= strong(Kelly) • nor: {smart(Kelly)} |= ~strong(Kelly)

5. backward resolution AR: general outline (1). • First we sketch the most generally used approach for automated reasoning in first-order logic: • The different technical componentswill only be explained in full detail in a second pass (outline (2)).

6. ground Horn clause logic AR: general outline (2) . • We study different subsets of predicate logic: • Horn clause logic • Clausal logic • full predicate logic • In each case we study semi-deciding procedures. • Each extension requires the introduction of new techniques.

7. Backward Reasoning Resolution … in a nutshell

8. 0) The task: an example. • 1) Proof by inconsistency. • 2) Conversion to clausal form. • 3) Unification. • 4) The resolution step. • 5) (Backward) resolution proofs. Backward resolution:

9. T p= parent f= father r= rich q= old z ~ q(z) y p(f(y)) x p(x)  q(x)  r(x) F • In this world, is always true ? u r(f(u)) 0) The TASK (example): • Are axioms: describe knowledge about some world. • How to prove such theorems in general?

10. F • Don’t prove F directly: u r(f(u)) • NEW TASK: ~ u r(f(u)) z ~ q(z) y p(f(y)) x p(x)  q(x)  r(x) • is inconsistent. 1) Proof by inconsistency • But add the negation of F to the axioms and prove that this extension is inconsistent. • the 4 axioms are never true in 1 same interpretation.

11. Each set of axioms can be transformed into a new set of formulae, that contains only formulae of the form: xy…zp(…)  q(…)  …r(…)  t(…)  s(…)  …  u(…) 2) Clausal form: = Normalize the formulae to a (more simple) standard form. •  only left;  only right • no ~ ; no  which is inconsistentif and only ifthe original set was inconsistent. Notice: “x y … z“can be dropped.

12. u false  r(f(u)) z false  q(z) x q(x)  r(x)  p(x) Example: • ~ u r(f(u)) • z ~ q(z) • y p(f(y)) : is already in clausal form: ( P  P  true) • x p(x)  q(x)  r(x) Ps: usually requires much more work!

13. Ex.: p(f(A),y) p(x, g(x)) • Ex.: x must become: f(A) • g(x) must become: g(f(A)) • y must become: g(f(A)) p(f(A), g(f(A))) 3) Unification: • Given 2 atomic formulae: • find their most general common instance. • Most general unifier (mgu) : x -> f(A) • y -> g(f(A))

14. Modus tollens Modus ponens P  Q P Q P  Q ~Q ~P Q  P P  true Q true Q  P false  Q false  P Resolution P1 P2  …  PnQ1  ...  Qm R1  …  Rk P1 S1  …  Sl P2 …  Pn  R1  …  RkQ1  ...  Qm  S1  …  Sl 4) The resolution step • Proposition logic:

15. p(x,f(A)) q(g(x)) r(z) p(B,z) r(f(A))  q(g(B)) mgu(p(x,f(A)),p(B,z)) = x-> B z-> f(A) De resolutie step (2): • Predicate logic: • Example: = mgu applied tor(z) q(g(x)) Clauses on which resolution is performed must not have any variables in common.

16. select 2 of them, for which resolution is possible AND inconsistency of the original set AND that F was implied by T 5) Resolution proofs: • In order to prove a set of clauses inconsistent: • apply resolution and add the result to the set • if you obtain the clause false : STOP ! This means inconsistency of the last set

17. q(x)  r(x)  p(x) false  r(f(u)) x -> f(u) q(f(u))  p(f(u)) false  q(z) z -> f(u) false  p(f(u)) p(f(y)) y -> u false  Example: So: inconsistent !

18. Ground Horn Logic Modus ponens Horn Logic Unification Clausal Logic Resolution Full Predicate Logic Normalization A deeper study:

19. All formulae in T are of the form: • x1 … xk A B1 B2 … Bn Horn clause logic • where A, B1, B2,…,Bn are atoms. • An atom is a formula of the form p(t1,…,tm), with p a predicate symbol and t1,…,tm terms. • Horn clause formulae are universally quantified over all variables that occur in them. • B1,…,Bn are called body-atoms of the Horn clause; A is the head of the Horn clause. • n may be 0: in this case we say that the Horn clause is a fact.

20. In Horn clause logic, we limit ourselves to prove formulae F of the form: • where B1, B2, …, Bn are again atoms. x1 … xk B1  B2  … Bn Wich kind of formulaecan we prove? • All variables are existentially quantified !

21. A very simple example: • Bosmans is a showmaster (1) • Showmasters are rich (2) • Rich people have big houses (3) • Big houses need a lot of maintenance (4) • Goal: automatically deduce that Bosmans’ house needs a lot of maintenance.

22. Representation in Horn logica: • Bosmans is a showmaster (1) • Showmasters are rich (2) • Rich people have big houses (3) • Big houses need a lot of maintenance (4) • To prove: showmaster(Bosmans) p rich(p)  showmaster(p) p big(house(p))  rich(p) p lot_maint(house(p))  big(house(p)) Lot_maint(house(Bosmans))

23. AR for ground Horn clause logic Backward reasoning proof procedures based on generalized Modus Ponens

24. So, for now: Horn clauses without variables: • Example: showmaster(Bosmans) rich(Bosmans)  showmaster(Bosmans) big(house(Bosmans))  rich(Bosmans) lot_maint(house(Bosmans))  big(house(Bosmans)) • Prove: lot_maint(house(Bosman)) Restricting to groundHorn clauses:

25. 3 applications of modus ponens: showm(Bos) rich(Bos)  showm(Bos) rich(Bos) big(house(Bos))  rich(Bos) big(house(Bos)) lot_maint(house(Bos))  big(house(Bos)) lot_maint(house(Bos)) • gives the desired conclusion. Easy with modus ponens !

26. Modus ponens is correct: B A  B For any interpretation making both B and A  B true(= any model of {B , A  B} ) A Modus ponens in AR: A is also true in this interpretation (see truth tables) • Problem: how to organize this into a procedure which is also complete(for ground Horn clauses)?

27. Theorem: Let T be a theory and F a formula. T implies Fif and only if T  {~F} is inconsistent. Inconsistency: • A theory T is inconsistent if it has NO model. • Proof: T implies F iff Each model of T makes F true iff Each model of T makes ~F false iffT  {~F} has no model iffT  {~F} is inconsistent

28. Prove that the theory: The extended example: showm(Bos) belg(Bos) european(Bos)  belg(Bos) rich(Bos)  showm(Bos)  european(Bos) big(house(Bos))  rich(Bos) lot_maint(house(Bos))  big(house(Bos)) ~ lot_maint(house(Bos)) • is inconsistent. • Problem: this is NOT a Horn clause theory !?

29. We introduce a new predicate symbol: false Refutation proofs:the “false” predicate • We agree that false has the truth value ‘false’ under every interpretation. • Imagine that we defined false as : false p  ~p for some predicate p

30. In the Horn logic setting F has the form: x1 … xm B1  B2  … Bn “definite” goals: • So what is the form of ~F? • ~(x1 … xm B1  B2  … Bn) •  x1 … xm ~(B1  B2  … Bn) •  x1 … xm false  ~(B1  B2  … Bn) •  x1 … xm false  B1  B2  … Bn A  ~B  A  B • Observe: ~F is again a Horn clause !!

31. The extended theory (to be proven inconsistent) now is: showm(Bos) belg(Bos) european(Bos)  belg(Bos) rich(Bos)  showm(Bos)  european(Bos) big(house(Bos))  rich(Bos) lot_maint(house(Bos))  big(house(Bos)) false  lot_maint(house(Bos)) Back to the example • a ground Horn clause theory !

32. A  B1  B2  … Bi …  Bn Bi  C1  C2  …  Cm A  B1  B2  … C1  C2  …  Cm …  Bn Modus ponensgeneralized: • Ordinary Modus ponens is the special case with: • n = i = 1 and m =0 • Correctness: via truth tables

33. false  lot_maint(house(Bos)) lot_maint(house(Bos))  big(house(Bos)) false  big(house(Bos)) big(house(Bos))  rich(Bos) Some backward reasoning steps in the example: • and so on ... false  big(house(Bos)) false  rich(Bos)

34. Convert F into a definite goal: false  B1  B2  …  Bi  …  Bn • until: is deduced. false  we have proven inconsistency ofT  {~F} The backward procedure:the idea • Apply generalized modus ponens to the body-atoms Bi of the goal, using the Horn clauses of T • Then: a false formula ia a consequence of T  {~F}

35. Goal := false  B1  B2  …  Bn; Repeat Select someBiatom from the body of Goal Select some clauseBi  C1  C2  …  CmfromT ReplaceBiin the body of Goal byC1  C2  …  Cm UntilGoal = false  or no more Selections possible Backwards procedure • On top of this you need to apply backtracking over the selected clauses and the selected body atoms. • If the algorithm stops because it has tried all these alternatives: F was not implied!

36. Back to the example Step 0: Goal :=false lot_maint(house(Bos)) select: lot_maint(house(Bos))  big(house(Bos)) Step 1: Goal :=false big(house(Bos)) select: big(house(Bos))  rich(Bos) Step 2: Goal :=false rich(Bos) select: rich(Bos)  showm(Bos)  european(Bos) Step 3: Goal :=false showm(Bos)  european(Bos) select:showm(Bos) Step 4: Goal :=false european(Bos) select:european(Bos)  belg(Bos) Step 5: Goal :=false belg(Bos) select: belg(Bos) Step 6: Goal :=false 

37. p  q  r q  t q  s r  n r  o s o n Another example (propositional) • Prove: p • Observe: non-determinism on both atom selection and on clause selection ! • we only illustrate the clause selection here

38. false q r false t r false  s r false r false n false o false  false  Search tree traversed bythe backward procedure false p p  q  r q  t q  s r  n r  o s o n

39. The backwardprocedure is efficient • The proof is goal directed towards the theorem. • no exploration of irrelevant rules • Different search methodescan be used to traverse this search tree. • Atom-selection may influence efficiency too: • ex.: by detecting a failing branch sooner • but has no impact on whether or not we find a solution (in case there are only finitely many ground Horn clauses)

40. (2) (1) false  p false  (1) false  p (1) …… Completeness: • Example: false  p p  p(1) p (2) • Possible derivations: false  p • Is only complete if the search tree is traversed using a complete search method.

41. Example: showm(Bos) showm_Bos big(house(Bos)) big_house_Bos Representation-power of ground Horn clauses • Is  a subset of propositional logic. • In general, more expressive logics are needed. • Essence: with variables, one formula may be equivalent to a very large number of propositional formulae.