CIS 6614 meeting -*- Outline -*-

** background on automatic theorem proving
*** SAT Solvers
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     BACKGROUND: SAT SOLVERS

goal: decide if first-order predicate is
       always true (or not)

SAT Solvers:

   input: propositional formula in CNF
   output: satisfying assignment to vars
           (or failure indication)

   example: w1 && w2 && w3
     where: w1 == b || c
            w2 == !a || !d
            w3 == !b || d

   satisfying assignment is
      {a =   , b =   , c =   , d =   }
   
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    Q: What's the time complexity of this problem?
       exponential!

    However, there has been a lot of work making this run very fast
        most of the time, for small enough formulas

*** SMT Solvers
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         SMT SOLVERS

SMT = Satisfiability Modulo Theories

Combines SAT + decidable theories
 Example theories:
  - Uninterpreted functions and equality
  - Peano arithmetic (arith. without *, /)
  - Arrays, Strings
  - Bitvectors
  - Algebraic Datatypes (e.g., enums)


Uninterpreted functions and equality:

   x == x
   x == y <==> y == x
   (x == y) && (y == z) ==> (x == z)

   x == y ==> f(x) == f(y)

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**** proof process
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   PROVING A FORMULA WITH SMT SOLVER

To prove P

see if not P is satisfiable


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     Q: What does it mean if not P is satisfiable?
        that P is not always true (there are cases where it's negation holds)

     Q: What does it mean if not P is unsatisfiable?
        that P is always true (no cases where it can't be true)

     Bonus, if not P is satisfiable,
        then the satisfying assignment gives a counterexample
          which can be used to show how the formula fails

**** example
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           EXAMPLE

b+2 == c
  && f(read(write(a,b,3),c-2)) != f(c-b+1)

Which parts are arithmetic?


arrays?


uninterpreted functions?


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  ... b+2 == c, c-2, c-b+1
  ... write(a,b,3), read(write(a,b,3),c-2)
  ... f(read(write(a,b,3),c-2)) != f(c-b+1)

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       SMT PROCESS

b+2 == c
  && f(read(write(a,b,3),c-2)) != f(c-b+1)















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      ... use equality to substitute
           b+2 for c
         
          b+2 == b+2
           && f(read(write(a,b,3),(b+2)-2)) != f((b+2)-b+1)

          simplify by theory of arithmetic, b+2-2 == b and b+2-b+1 == 3
               and by Boolean: (true && A == A)

          f(read(write(a,b,3),b)) != f(3)

          simplify by theory of arrays:

          f(3) != f(3)

          so by theory of uninterpreted functions,
             this is unsatisfiable!