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Automated Theorem Proving Lecture 5

Automated Theorem Proving Lecture 5. Theory of lists.  Formula := A |    A  Atom := t = t | t  t t  Term := c | car(t) | cdr(t) | cons(t,t) c  SymConst. Axioms: x,y. car(cons(x,y)) = x x,y. cdr(cons(x,y)) = y Extend congruence closure to deal with these axioms. Car axiom.

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Automated Theorem Proving Lecture 5

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  1. Automated Theorem ProvingLecture 5

  2. Theory of lists •  Formula := A |    A  Atom := t = t | t  t t  Term := c | car(t) | cdr(t) | cons(t,t) c  SymConst • Axioms: • x,y. car(cons(x,y)) = x • x,y. cdr(cons(x,y)) = y • Extend congruence closure to deal with these axioms

  3. Car axiom Cdr axiom x = cons(u,v)  cons(car(x), cdr(x))  x cons cons cdr car u v x • Car axiom: x,y. car(cons(x,y)) = x • Cdr axiom: x,y. cdr(cons(x,y)) = y

  4. cons(u,v) = cons(x,y)  u  x cons cons u v x y • Car axiom: x,y. car(cons(x,y)) = x • Cdr axiom: x,y. cdr(cons(x,y)) = y Suppose cons(x,y) = cons(u,v) = n. Then car(n) = x and car(n) = u, which contradicts u  x. Hence, our current algorithm is incomplete.

  5. Problem: There are not enough terms in the e-graph. Solution: Whenever the term cons(u,v) exists in the e-graph, add the term car(cons(u,v)) to the e-graph. cons(u,v) = cons(x,y)  u  x car car cons cons u v x y

  6. cons(u,v) = cons(x,y)  v  y cons cons u v x y Problem: There are not enough terms in the e-graph. Solution: Whenever the term cons(u,v) exists in the e-graph, add the term cdr(cons(u,v)) to the e-graph.

  7. Algorithm • Add terms to the e-graph as follows: if cons(u,v) • exists, add car(cons(u,v)) and cdr(cons(u,v)) • 2. Close the graph under congruence and the car and • cdr axioms • 3. If there is a disequality t1  t2 and an equivalence • class containing both t1 and t2, return unsatisfiable. • 4. Otherwise, return satisfiable.

  8. An e-graph G defines a partial interpretation I over the set U of equivalence classes of G. I(c) = EC(c) For f  {cons, car, cdr}: I(f)() = EC(f(u)), if f()  G I(f)() is undefined, otherwise Completeness • Definition: Suppose  and  are equivalence classes of G. • car()  G iff u s.t. car(u) is a term in G. • cdr()  G iff u s.t. cdr(u) is a term in G. • cons(,)  G iff u,v  s.t. cons(u,v) is a term in G. Definition: Suppose t  G. Then EC(t) is the equivalence class containing t.

  9. G0 = e-graph at the termination of the algorithm Gi+1 is obtained from Gi by adding terms as follows: Case I: car cdr cons u v If there are equivalence classes , of Gi such that cons(,)  Gi

  10. Case II: car car car car cdr u If there is an equivalence class  of Gi such that car()  Gi cdr cdr cdr car cdr u If there is an equivalence class  of Gi such that cdr()  Gi

  11. Model Suppose Ik is the partial interpretation corresponding to the e-graph Gk. Then Ik+1 extends Ik for all k  0. The model is given by the infinite union Uk0Ik.

  12. Theory of arrays •  Formula := A |    A  Atom := t = t | t  t | m = m | m  m t  Term := c | Select(m,t) m  MemTerm := f | Update(m,t,t) c  SymConst for all objects o and o’, and memories m:  o = o’  Select(Update(m,o,v),o’) = v o  o’  Select(Update(m,o,v),o’) = Select(m,o’) Extend congruence closure with these axioms

  13. b = Update(a,i,v)  Select(b,i) = u  u  v Select Update u b a i v

  14. b = Update(a,i,e)  b = Update(a’,i,e’)  e  e’ Update b Update a a’ i e e’ Select(b,i) = e and Select(b,i) = e’, which contradicts e  e’. Hence, algorithm is incomplete.

  15. Select Rule 1 Update a i e

  16. Select b = Update(a,i,e)  b = Update(a’,i,e’)  e  e’ Update b Update a a’ i e e’

  17. x = Select(Update(a,i,e),j)  i  j  y = Select(Update(a,i’,e’),j)  i’  j  x  y x Select Select y Update Update a i e i’ e’ j

  18. Select Rule 2 Select Update j a i e

  19. Select x = Select(Update(a,i,e),j)  i  j  y = Select(Update(a,i’,e’),j)  i’  j  x  y x Select Select y Update Update a i e i’ e’ j

  20. Update(a,i,e) = Update(a’,i’,e’)  Select(a,j)  Select(a’,j)  i  j  i’  j Select Select Update Update a i e a’ i’ e’ j Suppose Update(a,i,e) = Update(a’,i’,e’) = n Select(n,j) = Select(a,j) since i  j Select(n,j) = Select(a’,j) since i’  j Hence, we get a contradiction

  21. Select Rule 3 Select Update j a i e

  22. Select Update(a,i,e) = Update(a’,i’,e’)  Select(a,j)  Select(a’,j)  i  j  i’  j Select Select Update Update a i e a’ i’ e’ j

  23. Algorithm • Construct e-graph G for initial set of constraints • For each Update(a,b,c)  G, add to G: (Rule 1) • - term Select(Update(a,b,c),b) • constraint Select(Update(a,b,c),b) = c • Maintain e-graph G and set of disequalities D • Iterate: • - Generate case split • Add new terms, equalities, and disequalities • to satisfy Rules 2 and 3 • - If satisfiable return else backtrack

  24. Case I G  i = j, D Case II  (G  Select(a,j))  Select(Update(a,i,v),j) = Select(a,j) , D  i  j G, D i j  Select(Update(a,i,v),j)  G Case split:

  25. Case I G  i = j, D Case II  (G  Select(Update(a,i,v),j))  Select(Update(a,i,v),j) = Select(a,j) , D  i  j G, D i j  Update(a,i,v)  G  Select(a,j)  G Case split:

  26. Completeness Similar to the theory of lists

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