Z3 Ocaml API递归函数

Question

假设我想检查公式x + y = z（x，y，z整数）是否可满足。使用Z3，我可以输入类似以下内容：

(declare-fun x () Int)
(declare-fun y () Int)
(declare-fun z () Int)
(assert (= z (+ x y)))
(check-sat)

我可以等效地使用Ocaml api并编写以下代码：

let ctx=Z3.mk_context  [("model", "true"); ("proof", "false")] in 
let v1=(Z3.Arithmetic.Integer.mk_const_s ctx "x") in
let v2=(Z3.Arithmetic.Integer.mk_const_s ctx "y") in
let res=(Z3.Arithmetic.Integer.mk_const_s ctx "z") in
let sum=Z3.Arithmetic.mk_add ctx [ v1 ; v2] in 
let phi=Z3.Boolean.mk_eq ctx sum res in
let solver = (Z3.Solver.mk_solver ctx None) in
let _= Z3.Solver.add solver [phi] in
let is_sat=Z3.Solver.check solver [] in 
match is_sat with 
      | UNSATISFIABLE -> Printf.printf "unsat";
      | SATISFIABLE -> Printf.printf "sat";
      | UNKNOWN -> Printf.printf "unknown";

我想使用z3的ocaml api来检查以下阶乘实现，以便获得x的值（如果存在）。

(declare-fun x () Int)
(define-fun-rec f ((x Int)) Int
                  (ite (> x 1)
                       (* (f (- x 1))
                            x)
                       1))
(assert (= x (f 10)))
(check-sat)
(get-model)

不幸的是，我找不到使用z3的ml api进行递归定义的示例。

Answer 1

它运行我的计算机

let ctx=Z3.mk_context  [("model", "true"); ("proof", "false")] in 
let v1=(Z3.Arithmetic.Integer.mk_const_s ctx "x") in
let v2=(Z3.Arithmetic.Integer.mk_const_s ctx "y") in
let res=(Z3.Arithmetic.Integer.mk_const_s ctx "z") in
let sum=Z3.Arithmetic.mk_add ctx [ v1 ; v2] in 
let phi=Z3.Boolean.mk_eq ctx sum res in
let solver = (Z3.Solver.mk_solver ctx None) in
let _= Z3.Solver.add solver [phi] in
let is_sat=Z3.Solver.check solver [] in 
match is_sat with 
      | UNSATISFIABLE -> Printf.printf "unsat";
      | SATISFIABLE -> let m = (get_model solver) in    
                        match m with 
                        | None -> raise (TestFailedException "")
                        | Some (m) -> 
                          Printf.printf "Model: \n%s\n" (Model.to_string m);

********output*********
    Model: 
    (define-fun x () Int
      0)
    (define-fun y () Int
      0)
    (define-fun z () Int
      0)