
/*++
  Copyright (c) 2015 Microsoft Corporation

  --*/

#include<vector>
#include"z3++.h"

using namespace z3;




/**
  Demonstration of how Z3 can be used to prove validity of
  De Morgan's Duality Law: {e not(x and y) <-> (not x) or ( not y) }
  */
void demorgan() {
  std::cout << "de-Morgan example\n";

  context c;

  expr x = c.bool_const("x");
  expr y = c.bool_const("y");
  expr conjecture = (!(x && y)) == (!x || !y);

  solver s(c);
  // adding the negation of the conjecture as a constraint.
  s.add(!conjecture);
  std::cout << s << "\n";
  std::cout << s.to_smt2() << "\n";
  switch (s.check()) {
    case unsat:   std::cout << "de-Morgan is valid\n"; break;
    case sat:     std::cout << "de-Morgan is not valid\n"; break;
    case unknown: std::cout << "unknown\n"; break;
  }
}

/**
  \brief Find a model for <tt>x >= 1 and y < x + 3</tt>.
  */
void find_model_example1() {
  std::cout << "find_model_example1\n";
  context c;
  expr x = c.int_const("x");
  expr y = c.int_const("y");
  solver s(c);

  s.add(x >= 1);
  s.add(y < x + 3);
  std::cout << s.check() << "\n";

  model m = s.get_model();
  std::cout << m << "\n";
  // traversing the model
  for (unsigned i = 0; i < m.size(); i++) {
    func_decl v = m[i];
    // this problem contains only constants
    assert(v.arity() == 0); 
    std::cout << v.name() << " = " << m.get_const_interp(v) << "\n";
  }
  // we can evaluate expressions in the model.
  std::cout << "x + y + 1 = " << m.eval(x + y + 1) << "\n";
}


/**
  \brief Prove <tt>x = y implies g(x) = g(y)</tt>, and
  disprove <tt>x = y implies g(g(x)) = g(y)</tt>.

  This function demonstrates how to create uninterpreted types and
  functions.
  */
void prove_example1() {
  std::cout << "prove_example1\n";

  context c;
  expr x      = c.int_const("x");
  expr y      = c.int_const("y");
  sort I      = c.int_sort();
  func_decl g = function("g", I, I);

  solver s(c);
  expr conjecture1 = implies(x == y, g(x) == g(y));
  std::cout << "conjecture 1\n" << conjecture1 << "\n";
  s.add(!conjecture1);
  if (s.check() == unsat) 
    std::cout << "proved" << "\n";
  else
    std::cout << "failed to prove" << "\n";

  s.reset(); // remove all assertions from solver s

  expr conjecture2 = implies(x == y, g(g(x)) == g(y));
  std::cout << "conjecture 2\n" << conjecture2 << "\n";
  s.add(!conjecture2);
  if (s.check() == unsat) {
    std::cout << "proved" << "\n";
  }
  else {
    std::cout << "failed to prove" << "\n";
    model m = s.get_model();
    std::cout << "counterexample:\n" << m << "\n";
    std::cout << "g(g(x)) = " << m.eval(g(g(x))) << "\n";
    std::cout << "g(y)    = " << m.eval(g(y)) << "\n";
  }
}

/**
  \brief Prove <tt>not(g(g(x) - g(y)) = g(z)), x + z <= y <= x implies z < 0 </tt>.
  Then, show that <tt>z < -1</tt> is not implied.

  This example demonstrates how to combine uninterpreted functions and arithmetic.
  */
void prove_example2() {
  std::cout << "prove_example1\n";

  context c;
  expr x      = c.int_const("x");
  expr y      = c.int_const("y");
  expr z      = c.int_const("z");
  sort I      = c.int_sort();
  func_decl g = function("g", I, I);

  expr conjecture1 = implies(g(g(x) - g(y)) != g(z) && x + z <= y && y <= x,
      z < 0);

  solver s(c);
  s.add(!conjecture1);
  std::cout << "conjecture 1:\n" << conjecture1 << "\n";
  if (s.check() == unsat) 
    std::cout << "proved" << "\n";
  else
    std::cout << "failed to prove" << "\n";

  expr conjecture2 = implies(g(g(x) - g(y)) != g(z) && x + z <= y && y <= x,
      z < -1);
  s.reset();
  s.add(!conjecture2);
  std::cout << "conjecture 2:\n" << conjecture2 << "\n";
  if (s.check() == unsat) {
    std::cout << "proved" << "\n";
  }
  else {
    std::cout << "failed to prove" << "\n";
    std::cout << "counterexample:\n" << s.get_model() << "\n";
  }
}

/**
  \brief Nonlinear arithmetic example 1
  */
void nonlinear_example1() {
  std::cout << "nonlinear example 1\n";
  config cfg;
  cfg.set("auto_config", true);
  context c(cfg);

  expr x = c.real_const("x");
  expr y = c.real_const("y");
  expr z = c.real_const("z");

  solver s(c);

  s.add(x*x + y*y == 1);                     // x^2 + y^2 == 1
  s.add(x*x*x + z*z*z < c.real_val("1/2"));  // x^3 + z^3 < 1/2
  s.add(z != 0);
  std::cout << s.check() << "\n";
  model m = s.get_model();
  std::cout << m << "\n";
  set_param("pp.decimal", true); // set decimal notation
  std::cout << "model in decimal notation\n";
  std::cout << m << "\n";
  set_param("pp.decimal-precision", 50); // increase number of decimal places to 50.
  std::cout << "model using 50 decimal places\n";
  std::cout << m << "\n";
  set_param("pp.decimal", false); // disable decimal notation
}

/**
  \brief Simple function that tries to prove the given conjecture using the following steps:
  1- create a solver
  2- assert the negation of the conjecture
  3- checks if the result is unsat.
  */
void prove(expr conjecture) {
  context & c = conjecture.ctx();
  solver s(c);
  s.add(!conjecture);
  std::cout << "conjecture:\n" << conjecture << "\n";
  if (s.check() == unsat) {
    std::cout << "proved" << "\n";
  }
  else {
    std::cout << "failed to prove" << "\n";
    std::cout << "counterexample:\n" << s.get_model() << "\n";
  }
}

/**
  \brief Simple bit-vector example. This example disproves that x - 10 <= 0 IFF x <= 10 for (32-bit) machine integers
  */
void bitvector_example1() {
  std::cout << "bitvector example 1\n";
  context c;
  expr x = c.bv_const("x", 32);

  // using signed <=
  prove((x - 10 <= 0) == (x <= 10));

  // using unsigned <=
  prove(ule(x - 10, 0) == ule(x, 10));

  expr y = c.bv_const("y", 32);
  prove(implies(concat(x, y) == concat(y, x), x == y));
}

/**
  \brief Find x and y such that: x ^ y - 103 == x * y
  */
void bitvector_example2() {
  std::cout << "bitvector example 2\n";
  context c;
  expr x = c.bv_const("x", 32);
  expr y = c.bv_const("y", 32);
  solver s(c);
  // In C++, the operator == has higher precedence than ^.
  s.add((x ^ y) - 103 == x * y);
  std::cout << s << "\n";
  std::cout << s.check() << "\n";
  std::cout << s.get_model() << "\n";
}

/**
  \brief Mixing C and C++ APIs.
  */
void capi_example() {
  std::cout << "capi example\n";
  context c;
  expr x = c.bv_const("x", 32);
  expr y = c.bv_const("y", 32);
  // Invoking a C API function, and wrapping the result using an expr object.
  expr r = to_expr(c, Z3_mk_bvsrem(c, x, y));
  std::cout << "r: " << r << "\n";
}

/**
  \brief Demonstrate how to evaluate expressions in a model.
  */
void eval_example1() {
  std::cout << "eval example 1\n";
  context c;
  expr x = c.int_const("x");
  expr y = c.int_const("y");
  solver s(c);

  /* assert x < y */
  s.add(x < y);
  /* assert x > 2 */
  s.add(x > 2);

  std::cout << s.check() << "\n";

  model m = s.get_model();
  std::cout << "Model:\n" << m << "\n";
  std::cout << "x+y = " << m.eval(x+y) << "\n";
}

/**
  \brief Several contexts can be used simultaneously.
  */
void two_contexts_example1() {
  std::cout << "two contexts example 1\n";
  context c1, c2;

  expr x = c1.int_const("x");
  expr n = x + 1;
  // We cannot mix expressions from different contexts, but we can copy
  // an expression from one context to another.
  // The following statement copies the expression n from c1 to c2.
  expr n1 = to_expr(c2, Z3_translate(c1, n, c2));
  std::cout << n1 << "\n";
}

/**
  \brief Demonstrates how to catch API usage errors.
  */
void error_example() {
  std::cout << "error example\n";

  context c;
  expr x = c.bool_const("x");

  // Error using the C API can be detected using Z3_get_error_code.
  // The next call fails because x is a constant.
  //Z3_ast arg = Z3_get_app_arg(c, x, 0);
  if (Z3_get_error_code(c) != Z3_OK) {
    std::cout << "last call failed.\n";
  }

  // The C++ layer converts API usage errors into exceptions.
  try {
    // The next call fails because x is a Boolean.
    expr n = x + 1;
  }
  catch (exception ex) {
    std::cout << "failed: " << ex << "\n";
  }

  // The functions to_expr, to_sort and to_func_decl also convert C API errors into exceptions.
  try {
    expr arg = to_expr(c, Z3_get_app_arg(c, x, 0));
  }
  catch (exception ex) {
    std::cout << "failed: " << ex << "\n";
  }
}

/**
  \brief Demonstrate different ways of creating rational numbers: decimal and fractional representations.
  */
void numeral_example() {
  std::cout << "numeral example\n";
  context c;

  expr n1 = c.real_val("1/2");
  expr n2 = c.real_val("0.5");
  expr n3 = c.real_val(1, 2);
  std::cout << n1 << " " << n2 << " " << n3 << "\n";
  prove(n1 == n2 && n1 == n3);

  n1 = c.real_val("-1/3");
  n2 = c.real_val("-0.3333333333333333333333333333333333");
  std::cout << n1 << " " << n2 << "\n";
  prove(n1 != n2);
}

/**
  \brief Test ite-term (if-then-else terms).
  */
void ite_example() {
  std::cout << "if-then-else example\n";
  context c;

  expr f    = c.bool_val(false);
  expr one  = c.int_val(1);
  expr zero = c.int_val(0);
  expr ite  = to_expr(c, Z3_mk_ite(c, f, one, zero));

  std::cout << "term: " << ite << "\n";
}

void ite_example2() {
  std::cout << "if-then-else example2\n";
  context c;
  expr b = c.bool_const("b");
  expr x = c.int_const("x");
  expr y = c.int_const("y");
  std::cout << (ite(b, x, y) > 0) << "\n";
}

/**
  \brief Small example using quantifiers.
  */
void quantifier_example() {
  std::cout << "quantifier example\n";
  context c;

  expr x = c.int_const("x");
  expr y = c.int_const("y");
  sort I = c.int_sort();
  func_decl f = function("f", I, I, I);

  solver s(c);

  // making sure model based quantifier instantiation is enabled.
  params p(c);
  p.set("mbqi", true);
  s.set(p);

  s.add(forall(x, y, f(x, y) >= 0));
  expr a = c.int_const("a");
  s.add(f(a, a) < a);
  std::cout << s << "\n";
  std::cout << s.check() << "\n";
  std::cout << s.get_model() << "\n";
  s.add(a < 0);
  std::cout << s.check() << "\n";
}

/**
  \brief Unsat core example
  */
void unsat_core_example1() {
  std::cout << "unsat core example1\n";
  context c;
  // We use answer literals to track assertions.
  // An answer literal is essentially a fresh Boolean marker
  // that is used to track an assertion.
  // For example, if we want to track assertion F, we 
  // create a fresh Boolean variable p and assert (p => F)
  // Then we provide p as an argument for the check method.
  expr p1 = c.bool_const("p1");
  expr p2 = c.bool_const("p2");
  expr p3 = c.bool_const("p3");
  expr x  = c.int_const("x");
  expr y  = c.int_const("y");
  solver s(c);
  s.add(implies(p1, x > 10));
  s.add(implies(p1, y > x));
  s.add(implies(p2, y < 5));
  s.add(implies(p3, y > 0));
  expr assumptions[3] = { p1, p2, p3 };
  std::cout << s.check(3, assumptions) << "\n";
  expr_vector core = s.unsat_core();
  std::cout << core << "\n";
  std::cout << "size: " << core.size() << "\n";
  for (unsigned i = 0; i < core.size(); i++) {
    std::cout << core[i] << "\n";
  }
  // Trying again without p2
  expr assumptions2[2] = { p1, p3 };
  std::cout << s.check(2, assumptions2) << "\n";
}

/**
  \brief Unsat core example 2
  */
void unsat_core_example2() {
  std::cout << "unsat core example 2\n";
  context c;
  // The answer literal mechanism, described in the previous example,
  // tracks assertions. An assertion can be a complicated
  // formula containing containing the conjunction of many subformulas.
  expr p1 = c.bool_const("p1");
  expr x  = c.int_const("x");
  expr y  = c.int_const("y");
  solver s(c);
  expr F  = x > 10 && y > x && y < 5 && y > 0;
  s.add(implies(p1, F));
  expr assumptions[1] = { p1 };
  std::cout << s.check(1, assumptions) << "\n";
  expr_vector core = s.unsat_core();
  std::cout << core << "\n";
  std::cout << "size: " << core.size() << "\n";
  for (unsigned i = 0; i < core.size(); i++) {
    std::cout << core[i] << "\n";
  }
  // The core is not very informative, since p1 is tracking the formula F
  // that is a conjunction of subformulas.
  // Now, we use the following piece of code to break this conjunction
  // into individual subformulas. First, we flat the conjunctions by
  // using the method simplify.
  std::vector<expr> qs; // auxiliary vector used to store new answer literals.
  assert(F.is_app()); // I'm assuming F is an application.
  if (F.decl().decl_kind() == Z3_OP_AND) {
    // F is a conjunction
    std::cout << "F num. args (before simplify): " << F.num_args() << "\n";
    F = F.simplify();
    std::cout << "F num. args (after simplify):  " << F.num_args() << "\n";
    for (unsigned i = 0; i < F.num_args(); i++) {
      std::cout << "Creating answer literal q" << i << " for " << F.arg(i) << "\n";
      std::stringstream qname; qname << "q" << i;
      expr qi = c.bool_const(qname.str().c_str()); // create a new answer literal
      s.add(implies(qi, F.arg(i)));
      qs.push_back(qi);
    }
  }
  // The solver s already contains p1 => F
  // To disable F, we add (not p1) as an additional assumption
  qs.push_back(!p1);
  std::cout << s.check((unsigned)qs.size(), &qs[0]) << "\n";
  expr_vector core2 = s.unsat_core();
  std::cout << core2 << "\n";
  std::cout << "size: " << core2.size() << "\n";
  for (unsigned i = 0; i < core2.size(); i++) {
    std::cout << core2[i] << "\n";
  }
}

/**
  \brief Unsat core example 3
  */
void unsat_core_example3() {
  // Extract unsat core using tracked assertions
  std::cout << "unsat core example 3\n";
  context c;
  expr x  = c.int_const("x");
  expr y  = c.int_const("y");
  solver s(c);

  // enabling unsat core tracking
  params p(c);
  p.set("unsat_core", true);
  s.set(p);

  // The following assertion will not be tracked.
  s.add(x > 0);

  // The following assertion will be tracked using Boolean variable p1.
  // The C++ wrapper will automatically create the Boolean variable.
  s.add(y > 0, "p1");

  // Asserting other tracked assertions.
  s.add(x < 10, "p2");
  s.add(y < 0,  "p3");

  std::cout << s.check() << "\n";
  std::cout << s.unsat_core() << "\n";
}

void tactic_example1() {
  /*
     Z3 implements a methodology for orchestrating reasoning engines where "big" symbolic
     reasoning steps are represented as functions known as tactics, and tactics are composed
     using combinators known as tacticals. Tactics process sets of formulas called Goals.

     When a tactic is applied to some goal G, four different outcomes are possible. The tactic succeeds
     in showing G to be satisfiable (i.e., feasible); succeeds in showing G to be unsatisfiable (i.e., infeasible); 
     produces a sequence of subgoals; or fails. When reducing a goal G to a sequence of subgoals G1, ..., Gn, 
     we face the problem of model conversion. A model converter construct a model for G using a model for some subgoal Gi.

     In this example, we create a goal g consisting of three formulas, and a tactic t composed of two built-in tactics: 
     simplify and solve-eqs. The tactic simplify apply transformations equivalent to the ones found in the command simplify. 
     The tactic solver-eqs eliminate variables using Gaussian elimination. Actually, solve-eqs is not restricted 
     only to linear arithmetic. It can also eliminate arbitrary variables. 
     Then, sequential composition combinator & applies simplify to the input goal and solve-eqs to each subgoal produced by simplify. 
     In this example, only one subgoal is produced.
     */
  std::cout << "tactic example 1\n";
  context c;
  expr x = c.real_const("x");
  expr y = c.real_const("y");
  goal g(c);
  g.add(x > 0);
  g.add(y > 0);
  g.add(x == y + 2);
  std::cout << g << "\n";
  tactic t1(c, "simplify");
  tactic t2(c, "solve-eqs");
  tactic t = t1 & t2;
  apply_result r = t(g);
  std::cout << r << "\n";
}

void tactic_example2() {
  /*
     In Z3, we say a clause is any constraint of the form (f_1 || ... || f_n). 
     The tactic split-clause will select a clause in the input goal, and split it n subgoals. 
     One for each subformula f_i.
     */
  std::cout << "tactic example 2\n";
  context c;
  expr x = c.real_const("x");
  expr y = c.real_const("y");
  goal g(c);
  g.add(x < 0 || x > 0);
  g.add(x == y + 1);
  g.add(y < 0);
  tactic t(c, "split-clause");
  apply_result r = t(g);
  for (unsigned i = 0; i < r.size(); i++) {
    std::cout << "subgoal " << i << "\n" << r[i] << "\n";
  }
}

void tactic_example3() {
  /*
     - The choice combinator t | s first applies t to the given goal, if it fails then returns the result of s applied to the given goal.      
     - repeat(t) Keep applying the given tactic until no subgoal is modified by it.
     - repeat(t, n) Keep applying the given tactic until no subgoal is modified by it, or the number of iterations is greater than n.
     - try_for(t, ms) Apply tactic t to the input goal, if it does not return in ms millisenconds, it fails.
     - with(t, params) Apply the given tactic using the given parameters.      
     */
  std::cout << "tactic example 3\n";
  context c;
  expr x = c.real_const("x");
  expr y = c.real_const("y");
  expr z = c.real_const("z");
  goal g(c);
  g.add(x == 0 || x == 1);
  g.add(y == 0 || y == 1);
  g.add(z == 0 || z == 1);
  g.add(x + y + z > 2);
  // split all clauses
  tactic split_all = repeat(tactic(c, "split-clause") | tactic(c, "skip"));
  std::cout << split_all(g) << "\n";
  tactic split_at_most_2 = repeat(tactic(c, "split-clause") | tactic(c, "skip"), 1);
  std::cout << split_at_most_2(g) << "\n";
  // In the tactic split_solver, the tactic solve-eqs discharges all but one goal. 
  // Note that, this tactic generates one goal: the empty goal which is trivially satisfiable (i.e., feasible)
  tactic split_solve = split_all & tactic(c, "solve-eqs");
  std::cout << split_solve(g) << "\n";
}

void tactic_example4() {
  /*
     A tactic can be converted into a solver object using the method mk_solver(). 
     If the tactic produces the empty goal, then the associated solver returns sat. 
     If the tactic produces a single goal containing False, then the solver returns unsat. 
     Otherwise, it returns unknown.

     In this example, the tactic t implements a basic bit-vector solver using equation solving, 
     bit-blasting, and a propositional SAT solver. 
     We use the combinator `with` to configure our little solver. 
     We also include the tactic `aig` which tries to compress Boolean formulas using And-Inverted Graphs.
     */
  std::cout << "tactic example 4\n";
  context c;
  params p(c);
  p.set("mul2concat", true);
  tactic t = 
    with(tactic(c, "simplify"), p) &
    tactic(c, "solve-eqs") &
    tactic(c, "bit-blast") &
    tactic(c, "aig") &
    tactic(c, "sat");
  solver s = t.mk_solver();
  expr x = c.bv_const("x", 16);
  expr y = c.bv_const("y", 16);
  s.add(x*32 + y == 13);
  // In C++, the operator < has higher precedence than &.
  s.add((x & y) < 10);
  s.add(y > -100);
  std::cout << s.check() << "\n";
  model m = s.get_model();
  std::cout << m << "\n";
  std::cout << "x*32 + y = " << m.eval(x*32 + y) << "\n";
  std::cout << "x & y    = " << m.eval(x & y) << "\n";
}

void tactic_example5() {
  /*
     The tactic smt wraps the main solver in Z3 as a tactic.
     */
  std::cout << "tactic example 5\n";
  context c;
  expr x = c.int_const("x");
  expr y = c.int_const("y");
  solver s = tactic(c, "smt").mk_solver();
  s.add(x > y + 1);
  std::cout << s.check() << "\n";
  std::cout << s.get_model() << "\n";
}

void tactic_example6() {
  /*
     In this example, we show how to implement a solver for integer arithmetic using SAT. 
     The solver is complete only for problems where every variable has a lower and upper bound.
     */ 
  std::cout << "tactic example 6\n";
  context c;
  params p(c);
  p.set("arith_lhs", true);
  p.set("som", true); // sum-of-monomials normal form
  solver s = 
    (with(tactic(c, "simplify"), p) &
     tactic(c, "normalize-bounds") &
     tactic(c, "lia2pb") &
     tactic(c, "pb2bv") &
     tactic(c, "bit-blast") &
     tactic(c, "sat")).mk_solver();
  expr x = c.int_const("x");
  expr y = c.int_const("y");
  expr z = c.int_const("z");
  s.add(x > 0 && x < 10);
  s.add(y > 0 && y < 10);
  s.add(z > 0 && z < 10);
  s.add(3*y + 2*x == z);
  std::cout << s.check() << "\n";
  std::cout << s.get_model() << "\n";
  s.reset();
  s.add(3*y + 2*x == z);
  std::cout << s.check() << "\n";
}

void tactic_example7() {
  /*
     Tactics can be combined with solvers. 
     For example, we can apply a tactic to a goal, produced a set of subgoals, 
     then select one of the subgoals and solve it using a solver. 
     This example demonstrates how to do that, and 
     how to use model converters to convert a model for a subgoal into a model for the original goal.
     */
  std::cout << "tactic example 7\n";
  context c;
  tactic t = 
    tactic(c, "simplify") &
    tactic(c, "normalize-bounds") &
    tactic(c, "solve-eqs");
  expr x = c.int_const("x");
  expr y = c.int_const("y");
  expr z = c.int_const("z");
  goal g(c);
  g.add(x > 10);
  g.add(y == x + 3);
  g.add(z > y);
  apply_result r = t(g);
  // r contains only one subgoal
  std::cout << r << "\n";
  solver s(c);
  goal subgoal = r[0];
  for (unsigned i = 0; i < subgoal.size(); i++) {
    s.add(subgoal[i]);
  }
  std::cout << s.check() << "\n";
  model m = s.get_model();
  std::cout << "model for subgoal:\n" << m << "\n";
  std::cout << "model for original goal:\n" << r.convert_model(m) << "\n";
}

void tactic_example8() {
  /*
     Probes (aka formula measures) are evaluated over goals. 
     Boolean expressions over them can be built using relational operators and Boolean connectives. 
     The tactic fail_if(cond) fails if the given goal does not satisfy the condition cond.
     Many numeric and Boolean measures are available in Z3.

     In this example, we build a simple tactic using fail_if. 
     It also shows that a probe can be applied directly to a goal.
     */
  std::cout << "tactic example 8\n";
  context c;

  expr x = c.int_const("x");
  expr y = c.int_const("y");
  expr z = c.int_const("z");

  goal g(c);
  g.add(x + y + z > 0);

  probe p(c, "num-consts");
  std::cout << "num-consts: " << p(g) << "\n";

  tactic t = fail_if(p > 2);
  try {
    t(g);
  }
  catch (exception) {
    std::cout << "tactic failed...\n";
  }
  std::cout << "trying again...\n";
  g.reset();
  g.add(x + y > 0);
  std::cout << t(g) << "\n";
}

void tactic_example9() {
  /*
     The combinator (tactical) cond(p, t1, t2) is a shorthand for:

     (fail_if(p) & t1) | t2

     The combinator when(p, t) is a shorthand for:

     cond(p, t, tactic(c, "skip"))

     The tactic skip just returns the input goal. 
     This example demonstrates how to use the cond combinator.
     */
  std::cout << "tactic example 9\n";
  context c;

  expr x = c.int_const("x");
  expr y = c.int_const("y");
  expr z = c.int_const("z");

  goal g(c);
  g.add(x*x - y*y >= 0);

  probe p(c, "num-consts");
  tactic t = cond(p > 2, tactic(c, "simplify"), tactic(c, "factor"));
  std::cout << t(g) << "\n";

  g.reset();
  g.add(x + x + y + z >= 0);
  g.add(x*x - y*y >= 0);
  std::cout << t(g) << "\n";
}

void tactic_qe() {
  std::cout << "tactic example using quantifier elimination\n";
  context c;

  // Create a solver using "qe" and "smt" tactics
  solver s = 
    (tactic(c, "qe") &
     tactic(c, "smt")).mk_solver();

  expr a = c.int_const("a");
  expr b = c.int_const("b");
  expr x = c.int_const("x");
  expr f = implies(x <= a, x < b);

  expr qf = forall(x, f);

  std::cout << qf << "\n";

  s.add(qf);
  std::cout << s.check() << "\n";
  std::cout << s.get_model() << "\n";
}

void visit(expr const & e) {
  if (e.is_app()) {
    unsigned num = e.num_args();
    for (unsigned i = 0; i < num; i++) {
      visit(e.arg(i));
    }
    // do something
    // Example: print the visited expression
    func_decl f = e.decl();
    std::cout << "application of " << f.name() << ": " << e << "\n";
  }
  else if (e.is_quantifier()) {
    visit(e.body());
    // do something
  }
  else { 
    assert(e.is_var());
    // do something
  }
}

void tst_visit() {
  std::cout << "visit example\n";
  context c;

  expr x = c.int_const("x");
  expr y = c.int_const("y");
  expr z = c.int_const("z");
  expr f = x*x - y*y >= 0;

  visit(f);
}

void incremental_example1() {
  std::cout << "incremental example1\n";
  context c;
  expr x = c.int_const("x");
  solver s(c);
  s.add(x > 0);
  std::cout << s.check() << "\n";
  // We can add more formulas to the solver
  s.add(x < 0);
  // and, invoke s.check() again...
  std::cout << s.check() << "\n";
}

void incremental_example2() {
  // In this example, we show how push() and pop() can be used
  // to remove formulas added to the solver.
  std::cout << "incremental example2\n";
  context c;
  expr x = c.int_const("x");
  solver s(c);
  s.add(x > 0);
  std::cout << s.check() << "\n";
  // push() creates a backtracking point (aka a snapshot).
  s.push();
  // We can add more formulas to the solver
  s.add(x < 0);
  // and, invoke s.check() again...
  std::cout << s.check() << "\n";
  // pop() will remove all formulas added between this pop() and the matching push()
  s.pop();
  // The context is satisfiable again
  std::cout << s.check() << "\n";
  // and contains only x > 0
  std::cout << s << "\n";
}

void incremental_example3() {
  // In this example, we show how to use assumptions to "remove" 
  // formulas added to a solver. Actually, we disable them.
  std::cout << "incremental example3\n";
  context c;
  expr x = c.int_const("x");
  solver s(c);
  s.add(x > 0);
  std::cout << s.check() << "\n";
  // Now, suppose we want to add x < 0 to the solver, but we also want
  // to be able to disable it later.
  // To do that, we create an auxiliary Boolean variable
  expr b = c.bool_const("b");
  // and, assert (b implies x < 0)
  s.add(implies(b, x < 0));
  // Now, we check whether s is satisfiable under the assumption "b" is true.
  expr_vector a1(c);
  a1.push_back(b);
  std::cout << s.check(a1) << "\n";
  // To "disable" (x > 0), we may just ask with the assumption "not b" or not provide any assumption.
  std::cout << s.check() << "\n";
  expr_vector a2(c);
  a2.push_back(!b);
  std::cout << s.check(a2) << "\n";
}

void enum_sort_example() {
  std::cout << "enumeration sort example\n";
  context ctx;
  const char * enum_names[] = { "a", "b", "c" };
  func_decl_vector enum_consts(ctx);
  func_decl_vector enum_testers(ctx);
  sort s = ctx.enumeration_sort("enumT", 3, enum_names, enum_consts, enum_testers);
  // enum_consts[0] is a func_decl of arity 0.
  // we convert it to an expression using the operator()
  expr a = enum_consts[0]();
  expr b = enum_consts[1]();
  expr x = ctx.constant("x", s);
  expr test = (x==a) && (x==b);
  std::cout << "1: " << test << std::endl;
  tactic qe(ctx, "ctx-solver-simplify");
  goal g(ctx);
  g.add(test);
  expr res(ctx);
  apply_result result_of_elimination = qe.apply(g);
  goal result_goal = result_of_elimination[0];
  std::cout << "2: " << result_goal.as_expr() << std::endl;
}

void tuple_example() {
  std::cout << "tuple example\n";
  context ctx;
  const char * names[] = { "first", "second" };
  sort sorts[2] = { ctx.int_sort(), ctx.bool_sort() };
  func_decl_vector projs(ctx);
  func_decl pair = ctx.tuple_sort("pair", 2, names, sorts, projs);
  sorts[1] = pair.range();
  func_decl pair2 = ctx.tuple_sort("pair2", 2, names, sorts, projs);

  std::cout << pair2 << "\n";
}

void expr_vector_example() {
  std::cout << "expr_vector example\n";
  context c;
  const unsigned N = 10;

  expr_vector x(c);

  for (unsigned i = 0; i < N; i++) { 
    std::stringstream x_name; 
    x_name << "x_" << i;
    x.push_back(c.int_const(x_name.str().c_str()));
  }

  solver s(c);
  for (unsigned i = 0; i < N; i++) {
    s.add(x[i] >= 1);
  }

  std::cout << s << "\n" << "solving...\n" << s.check() << "\n";
  model m = s.get_model();
  std::cout << "solution\n" << m;
}

void exists_expr_vector_example() {
  std::cout << "exists expr_vector example\n";
  context c;
  const unsigned N = 10;

  expr_vector xs(c);
  expr x(c);
  expr b(c);
  b = c.bool_val(true); 

  for (unsigned i = 0; i < N; i++) { 
    std::stringstream x_name; 
    x_name << "x_" << i;
    x = c.int_const(x_name.str().c_str());
    xs.push_back(x);
    b = b && x >= 0;
  }

  expr ex(c);
  ex = exists(xs, b);
  std::cout << ex << std::endl;
}

void substitute_example() {
  std::cout << "substitute example\n";
  context c;
  expr x(c);
  x = c.int_const("x");
  expr f(c);
  f = (x == 2) || (x == 1);
  std::cout << f << std::endl;

  expr two(c), three(c);
  two   = c.int_val(2);
  three = c.int_val(3);
  Z3_ast from[] = { two };
  Z3_ast to[]   = { three };
  expr new_f(c);
  new_f = to_expr(c, Z3_substitute(c, f, 1, from, to));

  std::cout << new_f << std::endl;
}

void opt_example() {
  context c;
  optimize opt(c);
  params p(c);
  p.set("priority",c.str_symbol("pareto"));
  opt.set(p);
  expr x = c.int_const("x");
  expr y = c.int_const("y");
  opt.add(10 >= x && x >= 0);
  opt.add(10 >= y && y >= 0);
  opt.add(x + y <= 11);
  optimize::handle h1 = opt.maximize(x);
  optimize::handle h2 = opt.maximize(y);
  while (true) {
    if (sat == opt.check()) {
      std::cout << x << ": " << opt.lower(h1) << " " << y << ": " << opt.lower(h2) << "\n";
    }
    else {
      break;
    }
  }
}

void extract_example() {
  std::cout << "extract example\n";
  context c;
  expr x(c);
  x = c.constant("x", c.bv_sort(32));
  expr y = x.extract(21, 10);
  std::cout << y << " " << y.hi() << " " << y.lo() << "\n";
}

void sudoku_example() {
  std::cout << "sudoku example\n";

  context c;

  // 9x9 matrix of integer variables 
  expr_vector x(c);
  for (unsigned i = 0; i < 9; ++i)
    for (unsigned j = 0; j < 9; ++j) {
      std::stringstream x_name;

      x_name << "x_" << i << '_' << j;
      x.push_back(c.int_const(x_name.str().c_str()));
    }

  solver s(c);

  // each cell contains a value in {1, ..., 9}
  for (unsigned i = 0; i < 9; ++i)
    for (unsigned j = 0; j < 9; ++j) {
      s.add(x[i * 9 + j] >= 1 && x[i * 9 + j] <= 9);
    }

  // each row contains a digit at most once
  for (unsigned i = 0; i < 9; ++i) {
    expr_vector t(c);
    for (unsigned j = 0; j < 9; ++j)
      t.push_back(x[i * 9 + j]);
    s.add(distinct(t));
  }

  // each column contains a digit at most once
  for (unsigned j = 0; j < 9; ++j) {
    expr_vector t(c);
    for (unsigned i = 0; i < 9; ++i)
      t.push_back(x[i * 9 + j]);
    s.add(distinct(t));
  }

  // each 3x3 square contains a digit at most once
  for (unsigned i0 = 0; i0 < 3; i0++) {
    for (unsigned j0 = 0; j0 < 3; j0++) {
      expr_vector t(c);
      for (unsigned i = 0; i < 3; i++)
        for (unsigned j = 0; j < 3; j++)
          t.push_back(x[(i0 * 3 + i) * 9 + j0 * 3 + j]);
      s.add(distinct(t));
    }
  }

  // sudoku instance, we use '0' for empty cells
  int instance[9][9] = {{0,0,0,0,9,4,0,3,0},
    {0,0,0,5,1,0,0,0,7},
    {0,8,9,0,0,0,0,4,0},
    {0,0,0,0,0,0,2,0,8},
    {0,6,0,2,0,1,0,5,0},
    {1,0,2,0,0,0,0,0,0},
    {0,7,0,0,0,0,5,2,0},
    {9,0,0,0,6,5,0,0,0},
    {0,4,0,9,7,0,0,0,0}};

  for (unsigned i = 0; i < 9; i++)
    for (unsigned j = 0; j < 9; j++)
      if (instance[i][j] != 0)
        s.add(x[i * 9 + j] == instance[i][j]);

  std::cout << s.check() << std::endl;
  std::cout << s << std::endl;

  model m = s.get_model();
  for (unsigned i = 0; i < 9; ++i) {
    for (unsigned j = 0; j < 9; ++j)
      std::cout << m.eval(x[i * 9 + j]);
    std::cout << '\n';
  }
}

void param_descrs_example() {
  std::cout << "parameter description example\n";
  context c;
  param_descrs p = param_descrs::simplify_param_descrs(c);
  std::cout << p << "\n";
  unsigned sz = p.size();
  for (unsigned i = 0; i < sz; ++i) {
    symbol nm = p.name(i);
    char const* kind = "other";
    Z3_param_kind k = p.kind(nm);
    if (k == Z3_PK_UINT) kind = "uint";
    if (k == Z3_PK_BOOL) kind = "bool";
    std::cout << nm << ": " << p.documentation(nm) << " kind: " << kind << "\n";
  }
}

void consequence_example() {
  std::cout << "consequence example\n";
  context c;
  expr A = c.bool_const("a");
  expr B = c.bool_const("b");
  expr C = c.bool_const("c");
  solver s(c);
  s.add(implies(A, B));
  s.add(implies(B, C));
  expr_vector assumptions(c), vars(c), consequences(c);
  assumptions.push_back(!C);
  vars.push_back(A);
  vars.push_back(B);
  vars.push_back(C);
  std::cout << s.consequences(assumptions, vars, consequences) << "\n";
  std::cout << consequences << "\n";
}

static void parse_example() {
  std::cout << "parse example\n";
  context c;
  sort_vector sorts(c);
  func_decl_vector decls(c);
  sort B = c.bool_sort();
  decls.push_back(c.function("a", 0, 0, B));
  expr a = c.parse_string("(assert a)", sorts, decls);
  std::cout << a << "\n";

  // expr b = c.parse_string("(benchmark tst :extrafuns ((x Int) (y Int)) :formula (> x y) :formula (> x 0))");
}


int main() {

  try {
    demorgan(); std::cout << "\n";
    find_model_example1(); std::cout << "\n";
    prove_example1(); std::cout << "\n";
    prove_example2(); std::cout << "\n";
    nonlinear_example1(); std::cout << "\n";
    bitvector_example1(); std::cout << "\n";
    bitvector_example2(); std::cout << "\n";
    capi_example(); std::cout << "\n";
    eval_example1(); std::cout << "\n";
    two_contexts_example1(); std::cout << "\n";
    error_example(); std::cout << "\n";
    numeral_example(); std::cout << "\n";
    ite_example(); std::cout << "\n";
    ite_example2(); std::cout << "\n";
    quantifier_example(); std::cout << "\n";
    unsat_core_example1(); std::cout << "\n";
    unsat_core_example2(); std::cout << "\n";
    unsat_core_example3(); std::cout << "\n";
    tactic_example1(); std::cout << "\n";
    tactic_example2(); std::cout << "\n";
    tactic_example3(); std::cout << "\n";
    tactic_example4(); std::cout << "\n";
    tactic_example5(); std::cout << "\n";
    tactic_example6(); std::cout << "\n";
    tactic_example7(); std::cout << "\n";
    tactic_example8(); std::cout << "\n";
    tactic_example9(); std::cout << "\n";
    tactic_qe(); std::cout << "\n";
    tst_visit(); std::cout << "\n";
    incremental_example1(); std::cout << "\n";
    incremental_example2(); std::cout << "\n";
    incremental_example3(); std::cout << "\n";
    enum_sort_example(); std::cout << "\n";
    tuple_example(); std::cout << "\n";
    expr_vector_example(); std::cout << "\n";
    exists_expr_vector_example(); std::cout << "\n";
    substitute_example(); std::cout << "\n";
    opt_example(); std::cout << "\n";
    extract_example(); std::cout << "\n";
    param_descrs_example(); std::cout << "\n";
    sudoku_example(); std::cout << "\n";
    consequence_example(); std::cout << "\n";
    parse_example(); std::cout << "\n";
    std::cout << "done\n";
  }
  catch (exception & ex) {
    std::cout << "unexpected error: " << ex << "\n";
  }
  return 0;
}
