/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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.

  !minus : [o * o] --> o 
  0 : [] --> o 
  gcd : [o * o * o] --> o 
  max : [o * o] --> o 
  min : [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)) => !minus(X, Y) 
  gcd(s(X), s(Y), Z) => gcd(!minus(max(X, Y), min(X, Y)), s(min(X, Y)), Z) 
  gcd(X, s(Y), s(Z)) => gcd(X, !minus(max(Y, Z), min(Y, Z)), s(min(Y, Z))) 
  gcd(s(X), Y, s(Z)) => gcd(!minus(max(X, Z), min(X, Z)), Y, s(min(X, Z))) 
  gcd(X, 0, 0) => X 
  gcd(0, X, 0) => X 
  gcd(0, 0, 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, Y)   
    3] gcd#(s(X), s(Y), Z) =#> gcd#(!minus(max(X, Y), min(X, Y)), s(min(X, Y)), Z)   
    4] gcd#(s(X), s(Y), Z) =#> !minus#(max(X, Y), min(X, Y))   
    5] gcd#(s(X), s(Y), Z) =#> max#(X, Y)   
    6] gcd#(s(X), s(Y), Z) =#> min#(X, Y)   
    7] gcd#(s(X), s(Y), Z) =#> min#(X, Y)   
    8] gcd#(X, s(Y), s(Z)) =#> gcd#(X, !minus(max(Y, Z), min(Y, Z)), s(min(Y, Z)))   
    9] gcd#(X, s(Y), s(Z)) =#> !minus#(max(Y, Z), min(Y, Z))   
    10] gcd#(X, s(Y), s(Z)) =#> max#(Y, Z)   
    11] gcd#(X, s(Y), s(Z)) =#> min#(Y, Z)   
    12] gcd#(X, s(Y), s(Z)) =#> min#(Y, Z)   
    13] gcd#(s(X), Y, s(Z)) =#> gcd#(!minus(max(X, Z), min(X, Z)), Y, s(min(X, Z)))   
    14] gcd#(s(X), Y, s(Z)) =#> !minus#(max(X, Z), min(X, Z))   
    15] gcd#(s(X), Y, s(Z)) =#> max#(X, Z)   
    16] gcd#(s(X), Y, s(Z)) =#> min#(X, Z)   
    17] gcd#(s(X), Y, s(Z)) =#> min#(X, Z)   

  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)) => !minus(X, Y) 
    gcd(s(X), s(Y), Z) => gcd(!minus(max(X, Y), min(X, Y)), s(min(X, Y)), Z) 
    gcd(X, s(Y), s(Z)) => gcd(X, !minus(max(Y, Z), min(Y, Z)), s(min(Y, Z))) 
    gcd(s(X), Y, s(Z)) => gcd(!minus(max(X, Z), min(X, Z)), Y, s(min(X, Z))) 
    gcd(X, 0, 0) => X 
    gcd(0, X, 0) => X 
    gcd(0, 0, 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, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 
    * 4 : 2 
    * 5 : 1 
    * 6 : 0 
    * 7 : 0 
    * 8 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 
    * 9 : 2 
    * 10 : 1 
    * 11 : 0 
    * 12 : 0 
    * 13 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 
    * 14 : 2 
    * 15 : 1 
    * 16 : 0 
    * 17 : 0 

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, Y)   

  P_4:

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

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) and (P_4, 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) and (P_4, R_0, minimal, formative) is finite.

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

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_4, R_0) are:

  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)) => !minus(X, Y) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

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

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 
  0 = 0 
  gcd# = \y0y1y2.y2 + 2y1 + 3y0 
  max = \y0y1.y0 + y1 
  min = \y0y1.y0 
  s = \y0.1 + 3y0 

Using this interpretation, the requirements translate to:

  [[gcd#(s(_x0), s(_x1), _x2)]] = 5 + x2 + 6x1 + 9x0 > 2 + x2 + 3x1 + 9x0 = [[gcd#(!minus(max(_x0, _x1), min(_x0, _x1)), s(min(_x0, _x1)), _x2)]] 
  [[gcd#(_x0, s(_x1), s(_x2))]] = 3 + 3x0 + 3x2 + 6x1 > 1 + 2x2 + 3x0 + 5x1 = [[gcd#(_x0, !minus(max(_x1, _x2), min(_x1, _x2)), s(min(_x1, _x2)))]] 
  [[gcd#(s(_x0), _x1, s(_x2))]] = 4 + 2x1 + 3x2 + 9x0 > 1 + 2x1 + 3x2 + 6x0 = [[gcd#(!minus(max(_x0, _x2), min(_x0, _x2)), _x1, s(min(_x0, _x2)))]] 
  [[min(_x0, 0)]] = x0 >= 0 = [[0]] 
  [[min(0, _x0)]] = 0 >= 0 = [[0]] 
  [[min(s(_x0), s(_x1))]] = 1 + 3x0 >= 1 + 3x0 = [[s(min(_x0, _x1))]] 
  [[max(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[max(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[max(s(_x0), s(_x1))]] = 2 + 3x0 + 3x1 >= 1 + 3x0 + 3x1 = [[s(max(_x0, _x1))]] 
  [[!minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!minus(s(_x0), s(_x1))]] = 1 + 3x0 >= x0 = [[!minus(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0).  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, 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.