/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 
  any : [o] --> o 
  gcd : [o * o] --> o 
  max : [o * o] --> o 
  min : [o * o] --> o 
  minus : [o * o] --> o 
  s : [o] --> o 

  min(X, 0) => 0 
  min(0, X) => 0 
  min(s(X), s(Y)) => s(min(X, Y)) 
  max(X, 0) => X 
  max(0, X) => X 
  max(s(X), s(Y)) => s(max(X, Y)) 
  minus(X, 0) => X 
  minus(s(X), s(Y)) => s(minus(X, any(Y))) 
  gcd(s(X), s(Y)) => gcd(minus(max(X, Y), min(X, Y)), s(min(X, Y))) 
  any(s(X)) => s(s(any(X))) 
  any(X) => X 

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] min#(s(X), s(Y)) =#> min#(X, Y)   
    1] max#(s(X), s(Y)) =#> max#(X, Y)   
    2] minus#(s(X), s(Y)) =#> minus#(X, any(Y))   
    3] minus#(s(X), s(Y)) =#> any#(Y)   
    4] gcd#(s(X), s(Y)) =#> gcd#(minus(max(X, Y), min(X, Y)), s(min(X, Y)))   
    5] gcd#(s(X), s(Y)) =#> minus#(max(X, Y), min(X, Y))   
    6] gcd#(s(X), s(Y)) =#> max#(X, Y)   
    7] gcd#(s(X), s(Y)) =#> min#(X, Y)   
    8] gcd#(s(X), s(Y)) =#> min#(X, Y)   
    9] any#(s(X)) =#> any#(X)   

  Rules R_0:

    min(X, 0) => 0 
    min(0, X) => 0 
    min(s(X), s(Y)) => s(min(X, Y)) 
    max(X, 0) => X 
    max(0, X) => X 
    max(s(X), s(Y)) => s(max(X, Y)) 
    minus(X, 0) => X 
    minus(s(X), s(Y)) => s(minus(X, any(Y))) 
    gcd(s(X), s(Y)) => gcd(minus(max(X, Y), min(X, Y)), s(min(X, Y))) 
    any(s(X)) => s(s(any(X))) 
    any(X) => 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 : 1 
    * 2 : 2, 3 
    * 3 : 9 
    * 4 : 4, 5, 6, 7, 8 
    * 5 : 2, 3 
    * 6 : 1 
    * 7 : 0 
    * 8 : 0 
    * 9 : 9 

This graph has the following strongly connected components:

  P_1:

    min#(s(X), s(Y)) =#> min#(X, Y)   

  P_2:

    max#(s(X), s(Y)) =#> max#(X, Y)   

  P_3:

    minus#(s(X), s(Y)) =#> minus#(X, any(Y))   

  P_4:

    gcd#(s(X), s(Y)) =#> gcd#(minus(max(X, Y), min(X, Y)), s(min(X, Y)))   

  P_5:

    any#(s(X)) =#> any#(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), (P_2, R_0, m, f), (P_3, R_0, m, f), (P_4, R_0, m, f) and (P_5, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_5, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(any#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(any#(s(X))) = s(X) |> X = nu(any#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_0, minimal, formative).

The formative rules of (P_4, R_0) are R_1 ::=

  min(X, 0) => 0 
  min(0, X) => 0 
  min(s(X), s(Y)) => s(min(X, Y)) 
  max(X, 0) => X 
  max(0, X) => X 
  max(s(X), s(Y)) => s(max(X, Y)) 
  minus(X, 0) => X 
  minus(s(X), s(Y)) => s(minus(X, any(Y))) 
  any(s(X)) => s(s(any(X))) 
  any(X) => X 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_4, R_0, minimal, formative) by (P_4, R_1, minimal, formative).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  gcd#(s(X), s(Y)) >? gcd#(minus(max(X, Y), min(X, Y)), s(min(X, Y))) 
  min(X, 0) >= 0 
  min(0, X) >= 0 
  min(s(X), s(Y)) >= s(min(X, Y)) 
  max(X, 0) >= X 
  max(0, X) >= X 
  max(s(X), s(Y)) >= s(max(X, Y)) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= s(minus(X, any(Y))) 
  any(s(X)) >= s(s(any(X))) 
  any(X) >= X 

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

We consider usable_rules with respect to the following argument filtering:

  minus(x_1,x_2) = minus(x_1) 

This leaves the following ordering requirements:

  gcd#(s(X), s(Y)) > gcd#(minus(max(X, Y), min(X, Y)), s(min(X, Y))) 
  min(X, 0) >= 0 
  min(0, X) >= 0 
  min(s(X), s(Y)) >= s(min(X, Y)) 
  max(X, 0) >= X 
  max(0, X) >= X 
  max(s(X), s(Y)) >= s(max(X, Y)) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= s(minus(X, any(Y))) 

The following interpretation satisfies the requirements:

  0 = 0 
  any = \y0.0 
  gcd# = \y0y1.y1 + 2y0 
  max = \y0y1.y0 + y1 
  min = \y0y1.y0 
  minus = \y0y1.y0 
  s = \y0.2 + 2y0 

Using this interpretation, the requirements translate to:

  [[gcd#(s(_x0), s(_x1))]] = 6 + 2x1 + 4x0 > 2 + 2x1 + 4x0 = [[gcd#(minus(max(_x0, _x1), min(_x0, _x1)), s(min(_x0, _x1)))]] 
  [[min(_x0, 0)]] = x0 >= 0 = [[0]] 
  [[min(0, _x0)]] = 0 >= 0 = [[0]] 
  [[min(s(_x0), s(_x1))]] = 2 + 2x0 >= 2 + 2x0 = [[s(min(_x0, _x1))]] 
  [[max(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[max(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[max(s(_x0), s(_x1))]] = 4 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[s(max(_x0, _x1))]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 2 + 2x0 >= 2 + 2x0 = [[s(minus(_x0, any(_x1)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_1) by ({}, R_1).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, 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 apply the subterm criterion with the following projection function:

  nu(minus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, any(Y))) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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(max#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(max#(s(X), s(Y))) = s(X) |> X = nu(max#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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(min#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(min#(s(X), s(Y))) = s(X) |> X = nu(min#(X, 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.