/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.

  !940 : [] --> o 
  !plus : [o * o] --> o 
  !times : [o * o] --> o 
  0 : [o] --> o 
  1 : [o] --> o 
  cons : [o * o] --> o 
  nil : [] --> o 
  prod : [o] --> o 
  sum : [o] --> o 

  0(!940) => !940 
  !plus(X, !940) => X 
  !plus(!940, X) => X 
  !plus(0(X), 0(Y)) => 0(!plus(X, Y)) 
  !plus(0(X), 1(Y)) => 1(!plus(X, Y)) 
  !plus(1(X), 0(Y)) => 1(!plus(X, Y)) 
  !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) 
  !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) 
  !times(!940, X) => !940 
  !times(0(X), Y) => 0(!times(X, Y)) 
  !times(1(X), Y) => !plus(0(!times(X, Y)), Y) 
  !times(!times(X, Y), Z) => !times(X, !times(Y, Z)) 
  sum(nil) => 0(!940) 
  sum(cons(X, Y)) => !plus(X, sum(Y)) 
  prod(nil) => 1(!940) 
  prod(cons(X, Y)) => !times(X, prod(Y)) 

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] !plus#(0(X), 0(Y)) =#> 0#(!plus(X, Y))   
    1] !plus#(0(X), 0(Y)) =#> !plus#(X, Y)   
    2] !plus#(0(X), 1(Y)) =#> !plus#(X, Y)   
    3] !plus#(1(X), 0(Y)) =#> !plus#(X, Y)   
    4] !plus#(1(X), 1(Y)) =#> 0#(!plus(!plus(X, Y), 1(!940)))   
    5] !plus#(1(X), 1(Y)) =#> !plus#(!plus(X, Y), 1(!940))   
    6] !plus#(1(X), 1(Y)) =#> !plus#(X, Y)   
    7] !plus#(!plus(X, Y), Z) =#> !plus#(X, !plus(Y, Z))   
    8] !plus#(!plus(X, Y), Z) =#> !plus#(Y, Z)   
    9] !times#(0(X), Y) =#> 0#(!times(X, Y))   
    10] !times#(0(X), Y) =#> !times#(X, Y)   
    11] !times#(1(X), Y) =#> !plus#(0(!times(X, Y)), Y)   
    12] !times#(1(X), Y) =#> 0#(!times(X, Y))   
    13] !times#(1(X), Y) =#> !times#(X, Y)   
    14] !times#(!times(X, Y), Z) =#> !times#(X, !times(Y, Z))   
    15] !times#(!times(X, Y), Z) =#> !times#(Y, Z)   
    16] sum#(nil) =#> 0#(!940)   
    17] sum#(cons(X, Y)) =#> !plus#(X, sum(Y))   
    18] sum#(cons(X, Y)) =#> sum#(Y)   
    19] prod#(cons(X, Y)) =#> !times#(X, prod(Y))   
    20] prod#(cons(X, Y)) =#> prod#(Y)   

  Rules R_0:

    0(!940) => !940 
    !plus(X, !940) => X 
    !plus(!940, X) => X 
    !plus(0(X), 0(Y)) => 0(!plus(X, Y)) 
    !plus(0(X), 1(Y)) => 1(!plus(X, Y)) 
    !plus(1(X), 0(Y)) => 1(!plus(X, Y)) 
    !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) 
    !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) 
    !times(!940, X) => !940 
    !times(0(X), Y) => 0(!times(X, Y)) 
    !times(1(X), Y) => !plus(0(!times(X, Y)), Y) 
    !times(!times(X, Y), Z) => !times(X, !times(Y, Z)) 
    sum(nil) => 0(!940) 
    sum(cons(X, Y)) => !plus(X, sum(Y)) 
    prod(nil) => 1(!940) 
    prod(cons(X, Y)) => !times(X, prod(Y)) 

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 :  
    * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 4 :  
    * 5 : 2, 4, 5, 6, 7, 8 
    * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 9 :  
    * 10 : 9, 10, 11, 12, 13, 14, 15 
    * 11 : 0, 1, 2 
    * 12 :  
    * 13 : 9, 10, 11, 12, 13, 14, 15 
    * 14 : 9, 10, 11, 12, 13, 14, 15 
    * 15 : 9, 10, 11, 12, 13, 14, 15 
    * 16 :  
    * 17 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 18 : 16, 17, 18 
    * 19 : 9, 10, 11, 12, 13, 14, 15 
    * 20 : 19, 20 

This graph has the following strongly connected components:

  P_1:

    !plus#(0(X), 0(Y)) =#> !plus#(X, Y)   
    !plus#(0(X), 1(Y)) =#> !plus#(X, Y)   
    !plus#(1(X), 0(Y)) =#> !plus#(X, Y)   
    !plus#(1(X), 1(Y)) =#> !plus#(!plus(X, Y), 1(!940))   
    !plus#(1(X), 1(Y)) =#> !plus#(X, Y)   
    !plus#(!plus(X, Y), Z) =#> !plus#(X, !plus(Y, Z))   
    !plus#(!plus(X, Y), Z) =#> !plus#(Y, Z)   

  P_2:

    !times#(0(X), Y) =#> !times#(X, Y)   
    !times#(1(X), Y) =#> !times#(X, Y)   
    !times#(!times(X, Y), Z) =#> !times#(X, !times(Y, Z))   
    !times#(!times(X, Y), Z) =#> !times#(Y, Z)   

  P_3:

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

  P_4:

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

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 apply the subterm criterion with the following projection function:

  nu(prod#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, 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) 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(sum#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sum#(cons(X, Y))) = cons(X, Y) |> Y = nu(sum#(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(!times#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(!times#(0(X), Y)) = 0(X) |> X = nu(!times#(X, Y)) 
  nu(!times#(1(X), Y)) = 1(X) |> X = nu(!times#(X, Y)) 
  nu(!times#(!times(X, Y), Z)) = !times(X, Y) |> X = nu(!times#(X, !times(Y, Z))) 
  nu(!times#(!times(X, Y), Z)) = !times(X, Y) |> Y = nu(!times#(Y, Z)) 

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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_0) are:

  0(!940) => !940 
  !plus(X, !940) => X 
  !plus(!940, X) => X 
  !plus(0(X), 0(Y)) => 0(!plus(X, Y)) 
  !plus(0(X), 1(Y)) => 1(!plus(X, Y)) 
  !plus(1(X), 0(Y)) => 1(!plus(X, Y)) 
  !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) 
  !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) 

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

  !plus#(0(X), 0(Y)) >? !plus#(X, Y) 
  !plus#(0(X), 1(Y)) >? !plus#(X, Y) 
  !plus#(1(X), 0(Y)) >? !plus#(X, Y) 
  !plus#(1(X), 1(Y)) >? !plus#(!plus(X, Y), 1(!940)) 
  !plus#(1(X), 1(Y)) >? !plus#(X, Y) 
  !plus#(!plus(X, Y), Z) >? !plus#(X, !plus(Y, Z)) 
  !plus#(!plus(X, Y), Z) >? !plus#(Y, Z) 
  0(!940) >= !940 
  !plus(X, !940) >= X 
  !plus(!940, X) >= X 
  !plus(0(X), 0(Y)) >= 0(!plus(X, Y)) 
  !plus(0(X), 1(Y)) >= 1(!plus(X, Y)) 
  !plus(1(X), 0(Y)) >= 1(!plus(X, Y)) 
  !plus(1(X), 1(Y)) >= 0(!plus(!plus(X, Y), 1(!940))) 
  !plus(!plus(X, Y), Z) >= !plus(X, !plus(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !940 = 0 
  !plus = \y0y1.y0 + y1 
  !plus# = \y0y1.2y0 + 2y1 
  0 = \y0.2y0 
  1 = \y0.2 + 2y0 

Using this interpretation, the requirements translate to:

  [[!plus#(0(_x0), 0(_x1))]] = 4x0 + 4x1 >= 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(0(_x0), 1(_x1))]] = 4 + 4x0 + 4x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(1(_x0), 0(_x1))]] = 4 + 4x0 + 4x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(1(_x0), 1(_x1))]] = 8 + 4x0 + 4x1 > 4 + 2x0 + 2x1 = [[!plus#(!plus(_x0, _x1), 1(!940))]] 
  [[!plus#(1(_x0), 1(_x1))]] = 8 + 4x0 + 4x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(!plus(_x0, _x1), _x2)]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[!plus#(_x0, !plus(_x1, _x2))]] 
  [[!plus#(!plus(_x0, _x1), _x2)]] = 2x0 + 2x1 + 2x2 >= 2x1 + 2x2 = [[!plus#(_x1, _x2)]] 
  [[0(!940)]] = 0 >= 0 = [[!940]] 
  [[!plus(_x0, !940)]] = x0 >= x0 = [[_x0]] 
  [[!plus(!940, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!plus(_x0, _x1))]] 
  [[!plus(0(_x0), 1(_x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] 
  [[!plus(1(_x0), 0(_x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] 
  [[!plus(1(_x0), 1(_x1))]] = 4 + 2x0 + 2x1 >= 4 + 2x0 + 2x1 = [[0(!plus(!plus(_x0, _x1), 1(!940)))]] 
  [[!plus(!plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_5, R_0, minimal, formative), where P_5 consists of:

  !plus#(0(X), 0(Y)) =#> !plus#(X, Y)   
  !plus#(!plus(X, Y), Z) =#> !plus#(X, !plus(Y, Z))   
  !plus#(!plus(X, Y), Z) =#> !plus#(Y, Z)   

Thus, the original system is terminating if (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(!plus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(!plus#(0(X), 0(Y))) = 0(X) |> X = nu(!plus#(X, Y)) 
  nu(!plus#(!plus(X, Y), Z)) = !plus(X, Y) |> X = nu(!plus#(X, !plus(Y, Z))) 
  nu(!plus#(!plus(X, Y), Z)) = !plus(X, Y) |> Y = nu(!plus#(Y, Z)) 

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.

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.