1. Automated Reasoning Systems For first order Predicate Logic

2. 2 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  Is F a logical consequence of T ?  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. 3 AR: decidability.  There CANNOT EXIST AN ALGORITHM that decides whether T |= F, for any theory T and any formula F.  Theorem Church ‘36:  There exists a reasoning technique, such that for any theory T and formula F, such that T |= F, the reasoning technique proves T |= F.  SO: if F follows from T, then we find a proof, else it is possible that the procedure doesn’t terminate.  BUT: semi-decidable !  Completeness Theorem of Goedel ‘31:

4. 4  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)  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...

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

6. 6 AR: general outline (2).  We study different subsets of predicate logic:  ground Horn clause 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. 8 Backward resolution:  0) The task: an example.  1) Proof by inconsistency.  2) Conversion to clausal form.  3) Unification.  4) The resolution step.  5) (Backward) resolution proofs.

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

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

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

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

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

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

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

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

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

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

19. 19 Horn clause logic  x1 …  xk A  B1  B2 …  Bn  All formulae in T are of the form:  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. 20 Wich kind of formulae can we prove?  x1 …  xk B1  B2  …  Bn  In Horn clause logic, we limit ourselves to prove formulae F of the form:  where B1, B2, …, Bn are again atoms.  All variables are existentially quantified !

21. 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. 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. 24 Restricting to ground Horn clauses: showmaster(Bosmans) rich(Bosmans)  showmaster(Bosmans) big(house(Bosmans))  rich(Bosmans) lot_maint(house(Bosmans))  big(house(Bosmans))  So, for now: Horn clauses without variables:  Example: lot_maint(house(Bosman))  Prove:

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

26. 26 Modus ponens in AR: B A  B A For any interpretation making both B and A  B true (= any model of {B, A  B} )  Modus ponens is correct: 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. 27 Inconsistency: T implies F iff Each model of T makes F true iff Each model of T makes ~F false iff T  {~F} has no model iff T  {~F} is inconsistent  A theory T is inconsistent if it has NO model. Let T be a theory and F a formula. T implies F if and only if T  {~F} is inconsistent.  Theorem:  Proof:

28. 28 The extended example:  Prove that the theory: 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. 29 Refutation proofs: the “false” predicate false  We introduce a new predicate symbol:  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. 30 “definite” goals:  In the Horn logic setting F has the form:  x1 …  xm B1  B2  …  Bn  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. 31 Back to the example  a ground Horn clause theory ! 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))  The extended theory (to be proven inconsistent) now is:

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

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

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

35. 35 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! Goal := false  B1  B2  …  Bn ; Repeat Select some Bi atom from the body of Goal Select some clause Bi  C1  C2  …  Cm from T Select some clause Bi  C1  C2  …  Cm from T Replace Bi in the body of Goal by C1  C2  …  Cm Until Goal = false  or no more Selections possible

36. 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. 37 p  q  r q  t q  s r  n r  o son Another example (propositional)  Prove: p  Observe: non-determinism on both atom selection and on clause selection !  we only illustrate the clause selection here

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

39. 39 The backward procedure is efficient  The proof is goal directed towards the theorem.  no exploration of irrelevant rules  Different search methodes can 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) (in case there are only finitely many ground Horn clauses)

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

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