/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  cons : [o * o] --> o 
  int : [o * o] --> o 
  intlist : [o] --> o 
  nil : [] --> o 
  s : [o] --> o 

  intlist(nil) => nil 
  int(s(X), 0) => nil 
  int(X, X) => cons(X, nil) 
  intlist(cons(X, Y)) => cons(s(X), intlist(Y)) 
  int(s(X), s(Y)) => intlist(int(X, Y)) 
  int(0, s(X)) => cons(0, int(s(0), s(X))) 
  intlist(cons(X, nil)) => cons(s(X), nil) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  intlist(nil) >? nil 
  int(s(X), 0) >? nil 
  int(X, X) >? cons(X, nil) 
  intlist(cons(X, Y)) >? cons(s(X), intlist(Y)) 
  int(s(X), s(Y)) >? intlist(int(X, Y)) 
  int(0, s(X)) >? cons(0, int(s(0), s(X))) 
  intlist(cons(X, nil)) >? cons(s(X), nil) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  cons = \y0y1.y0 + y1 
  int = \y0y1.1 + y0 + y1 
  intlist = \y0.y0 
  nil = 0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[intlist(nil)]] = 0 >= 0 = [[nil]] 
  [[int(s(_x0), 0)]] = 1 + x0 > 0 = [[nil]] 
  [[int(_x0, _x0)]] = 1 + 2x0 > x0 = [[cons(_x0, nil)]] 
  [[intlist(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(s(_x0), intlist(_x1))]] 
  [[int(s(_x0), s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[intlist(int(_x0, _x1))]] 
  [[int(0, s(_x0))]] = 1 + x0 >= 1 + x0 = [[cons(0, int(s(0), s(_x0)))]] 
  [[intlist(cons(_x0, nil))]] = x0 >= x0 = [[cons(s(_x0), nil)]] 

We can thus remove the following rules:

  int(s(X), 0) => nil 
  int(X, X) => cons(X, nil) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  intlist(nil) >? nil 
  intlist(cons(X, Y)) >? cons(s(X), intlist(Y)) 
  int(s(X), s(Y)) >? intlist(int(X, Y)) 
  int(0, s(X)) >? cons(0, int(s(0), s(X))) 
  intlist(cons(X, nil)) >? cons(s(X), nil) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  cons = \y0y1.y0 + y1 
  int = \y0y1.y0 + y1 
  intlist = \y0.2y0 
  nil = 1 
  s = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[intlist(nil)]] = 2 > 1 = [[nil]] 
  [[intlist(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(_x0), intlist(_x1))]] 
  [[int(s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[intlist(int(_x0, _x1))]] 
  [[int(0, s(_x0))]] = 2x0 >= 2x0 = [[cons(0, int(s(0), s(_x0)))]] 
  [[intlist(cons(_x0, nil))]] = 2 + 2x0 > 1 + 2x0 = [[cons(s(_x0), nil)]] 

We can thus remove the following rules:

  intlist(nil) => nil 
  intlist(cons(X, nil)) => cons(s(X), nil) 

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] intlist#(cons(X, Y)) =#> intlist#(Y)   
    1] int#(s(X), s(Y)) =#> intlist#(int(X, Y))   
    2] int#(s(X), s(Y)) =#> int#(X, Y)   
    3] int#(0, s(X)) =#> int#(s(0), s(X))   

  Rules R_0:

    intlist(cons(X, Y)) => cons(s(X), intlist(Y)) 
    int(s(X), s(Y)) => intlist(int(X, Y)) 
    int(0, s(X)) => cons(0, int(s(0), s(X))) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 : 0 
    * 2 : 1, 2, 3 
    * 3 : 1, 2 

This graph has the following strongly connected components:

  P_1:

    intlist#(cons(X, Y)) =#> intlist#(Y)   

  P_2:

    int#(s(X), s(Y)) =#> int#(X, Y)   
    int#(0, s(X)) =#> int#(s(0), s(X))   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(int#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(int#(s(X), s(Y))) = s(Y) |> Y = nu(int#(X, Y)) 
  nu(int#(0, s(X))) = s(X) = s(X) = nu(int#(s(0), s(X))) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains:

  int#(0, s(X)) =#> int#(s(0), s(X))   

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(intlist#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(intlist#(cons(X, Y))) = cons(X, Y) |> Y = nu(intlist#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.