/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.

  0 : [] --> o 
  a!6220!6220div : [o * o] --> o 
  a!6220!6220geq : [o * o] --> o 
  a!6220!6220if : [o * o * o] --> o 
  a!6220!6220minus : [o * o] --> o 
  div : [o * o] --> o 
  false : [] --> o 
  geq : [o * o] --> o 
  if : [o * o * o] --> o 
  mark : [o] --> o 
  minus : [o * o] --> o 
  s : [o] --> o 
  true : [] --> o 

  a!6220!6220minus(0, X) => 0 
  a!6220!6220minus(s(X), s(Y)) => a!6220!6220minus(X, Y) 
  a!6220!6220geq(X, 0) => true 
  a!6220!6220geq(0, s(X)) => false 
  a!6220!6220geq(s(X), s(Y)) => a!6220!6220geq(X, Y) 
  a!6220!6220div(0, s(X)) => 0 
  a!6220!6220div(s(X), s(Y)) => a!6220!6220if(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 
  a!6220!6220if(true, X, Y) => mark(X) 
  a!6220!6220if(false, X, Y) => mark(Y) 
  mark(minus(X, Y)) => a!6220!6220minus(X, Y) 
  mark(geq(X, Y)) => a!6220!6220geq(X, Y) 
  mark(div(X, Y)) => a!6220!6220div(mark(X), Y) 
  mark(if(X, Y, Z)) => a!6220!6220if(mark(X), Y, Z) 
  mark(0) => 0 
  mark(s(X)) => s(mark(X)) 
  mark(true) => true 
  mark(false) => false 
  a!6220!6220minus(X, Y) => minus(X, Y) 
  a!6220!6220geq(X, Y) => geq(X, Y) 
  a!6220!6220div(X, Y) => div(X, Y) 
  a!6220!6220if(X, Y, Z) => if(X, Y, Z) 

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] a!6220!6220minus#(s(X), s(Y)) =#> a!6220!6220minus#(X, Y)   
    1] a!6220!6220geq#(s(X), s(Y)) =#> a!6220!6220geq#(X, Y)   
    2] a!6220!6220div#(s(X), s(Y)) =#> a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0)   
    3] a!6220!6220div#(s(X), s(Y)) =#> a!6220!6220geq#(X, Y)   
    4] a!6220!6220if#(true, X, Y) =#> mark#(X)   
    5] a!6220!6220if#(false, X, Y) =#> mark#(Y)   
    6] mark#(minus(X, Y)) =#> a!6220!6220minus#(X, Y)   
    7] mark#(geq(X, Y)) =#> a!6220!6220geq#(X, Y)   
    8] mark#(div(X, Y)) =#> a!6220!6220div#(mark(X), Y)   
    9] mark#(div(X, Y)) =#> mark#(X)   
    10] mark#(if(X, Y, Z)) =#> a!6220!6220if#(mark(X), Y, Z)   
    11] mark#(if(X, Y, Z)) =#> mark#(X)   
    12] mark#(s(X)) =#> mark#(X)   

  Rules R_0:

    a!6220!6220minus(0, X) => 0 
    a!6220!6220minus(s(X), s(Y)) => a!6220!6220minus(X, Y) 
    a!6220!6220geq(X, 0) => true 
    a!6220!6220geq(0, s(X)) => false 
    a!6220!6220geq(s(X), s(Y)) => a!6220!6220geq(X, Y) 
    a!6220!6220div(0, s(X)) => 0 
    a!6220!6220div(s(X), s(Y)) => a!6220!6220if(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 
    a!6220!6220if(true, X, Y) => mark(X) 
    a!6220!6220if(false, X, Y) => mark(Y) 
    mark(minus(X, Y)) => a!6220!6220minus(X, Y) 
    mark(geq(X, Y)) => a!6220!6220geq(X, Y) 
    mark(div(X, Y)) => a!6220!6220div(mark(X), Y) 
    mark(if(X, Y, Z)) => a!6220!6220if(mark(X), Y, Z) 
    mark(0) => 0 
    mark(s(X)) => s(mark(X)) 
    mark(true) => true 
    mark(false) => false 
    a!6220!6220minus(X, Y) => minus(X, Y) 
    a!6220!6220geq(X, Y) => geq(X, Y) 
    a!6220!6220div(X, Y) => div(X, Y) 
    a!6220!6220if(X, Y, Z) => if(X, Y, Z) 

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 : 4, 5 
    * 3 : 1 
    * 4 : 6, 7, 8, 9, 10, 11, 12 
    * 5 : 6, 7, 8, 9, 10, 11, 12 
    * 6 : 0 
    * 7 : 1 
    * 8 : 2, 3 
    * 9 : 6, 7, 8, 9, 10, 11, 12 
    * 10 : 4, 5 
    * 11 : 6, 7, 8, 9, 10, 11, 12 
    * 12 : 6, 7, 8, 9, 10, 11, 12 

This graph has the following strongly connected components:

  P_1:

    a!6220!6220minus#(s(X), s(Y)) =#> a!6220!6220minus#(X, Y)   

  P_2:

    a!6220!6220geq#(s(X), s(Y)) =#> a!6220!6220geq#(X, Y)   

  P_3:

    a!6220!6220div#(s(X), s(Y)) =#> a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0)   
    a!6220!6220if#(true, X, Y) =#> mark#(X)   
    a!6220!6220if#(false, X, Y) =#> mark#(Y)   
    mark#(div(X, Y)) =#> a!6220!6220div#(mark(X), Y)   
    mark#(div(X, Y)) =#> mark#(X)   
    mark#(if(X, Y, Z)) =#> a!6220!6220if#(mark(X), Y, Z)   
    mark#(if(X, Y, Z)) =#> mark#(X)   
    mark#(s(X)) =#> mark#(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) and (P_3, 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) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, 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:

  a!6220!6220div#(s(X), s(Y)) >? a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 
  a!6220!6220if#(true, X, Y) >? mark#(X) 
  a!6220!6220if#(false, X, Y) >? mark#(Y) 
  mark#(div(X, Y)) >? a!6220!6220div#(mark(X), Y) 
  mark#(div(X, Y)) >? mark#(X) 
  mark#(if(X, Y, Z)) >? a!6220!6220if#(mark(X), Y, Z) 
  mark#(if(X, Y, Z)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  a!6220!6220minus(0, X) >= 0 
  a!6220!6220minus(s(X), s(Y)) >= a!6220!6220minus(X, Y) 
  a!6220!6220geq(X, 0) >= true 
  a!6220!6220geq(0, s(X)) >= false 
  a!6220!6220geq(s(X), s(Y)) >= a!6220!6220geq(X, Y) 
  a!6220!6220div(0, s(X)) >= 0 
  a!6220!6220div(s(X), s(Y)) >= a!6220!6220if(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 
  a!6220!6220if(true, X, Y) >= mark(X) 
  a!6220!6220if(false, X, Y) >= mark(Y) 
  mark(minus(X, Y)) >= a!6220!6220minus(X, Y) 
  mark(geq(X, Y)) >= a!6220!6220geq(X, Y) 
  mark(div(X, Y)) >= a!6220!6220div(mark(X), Y) 
  mark(if(X, Y, Z)) >= a!6220!6220if(mark(X), Y, Z) 
  mark(0) >= 0 
  mark(s(X)) >= s(mark(X)) 
  mark(true) >= true 
  mark(false) >= false 
  a!6220!6220minus(X, Y) >= minus(X, Y) 
  a!6220!6220geq(X, Y) >= geq(X, Y) 
  a!6220!6220div(X, Y) >= div(X, Y) 
  a!6220!6220if(X, Y, Z) >= if(X, Y, Z) 

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

We consider usable_rules with respect to the following argument filtering:

  a!6220!6220div#(x_1,x_2) = a!6220!6220div#(x_2) 
  a!6220!6220if#(x_1,x_2,x_3) = a!6220!6220if#(x_2x_3) 

This leaves the following ordering requirements:

  a!6220!6220div#(s(X), s(Y)) >= a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 
  a!6220!6220if#(true, X, Y) >= mark#(X) 
  a!6220!6220if#(false, X, Y) >= mark#(Y) 
  mark#(div(X, Y)) >= a!6220!6220div#(mark(X), Y) 
  mark#(div(X, Y)) >= mark#(X) 
  mark#(if(X, Y, Z)) >= a!6220!6220if#(mark(X), Y, Z) 
  mark#(if(X, Y, Z)) > mark#(X) 
  mark#(s(X)) >= mark#(X) 

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220div = \y0y1.0 
  a!6220!6220div# = \y0y1.0 
  a!6220!6220geq = \y0y1.0 
  a!6220!6220if = \y0y1y2.0 
  a!6220!6220if# = \y0y1y2.2y1 + 2y2 
  a!6220!6220minus = \y0y1.0 
  div = \y0y1.2y0 
  false = 0 
  geq = \y0y1.0 
  if = \y0y1y2.1 + y0 + y2 + 2y1 
  mark = \y0.0 
  mark# = \y0.2y0 
  minus = \y0y1.0 
  s = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220div#(s(_x0), s(_x1))]] = 0 >= 0 = [[a!6220!6220if#(a!6220!6220geq(_x0, _x1), s(div(minus(_x0, _x1), s(_x1))), 0)]] 
  [[a!6220!6220if#(true, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 = [[mark#(_x0)]] 
  [[a!6220!6220if#(false, _x0, _x1)]] = 2x0 + 2x1 >= 2x1 = [[mark#(_x1)]] 
  [[mark#(div(_x0, _x1))]] = 4x0 >= 0 = [[a!6220!6220div#(mark(_x0), _x1)]] 
  [[mark#(div(_x0, _x1))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(if(_x0, _x1, _x2))]] = 2 + 2x0 + 2x2 + 4x1 > 2x1 + 2x2 = [[a!6220!6220if#(mark(_x0), _x1, _x2)]] 
  [[mark#(if(_x0, _x1, _x2))]] = 2 + 2x0 + 2x2 + 4x1 > 2x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_4, R_0, minimal, formative), where P_4 consists of:

  a!6220!6220div#(s(X), s(Y)) =#> a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0)   
  a!6220!6220if#(true, X, Y) =#> mark#(X)   
  a!6220!6220if#(false, X, Y) =#> mark#(Y)   
  mark#(div(X, Y)) =#> a!6220!6220div#(mark(X), Y)   
  mark#(div(X, Y)) =#> mark#(X)   
  mark#(s(X)) =#> mark#(X)   

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, 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 [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  a!6220!6220div#(s(X), s(Y)) >? a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 
  a!6220!6220if#(true, X, Y) >? mark#(X) 
  a!6220!6220if#(false, X, Y) >? mark#(Y) 
  mark#(div(X, Y)) >? a!6220!6220div#(mark(X), Y) 
  mark#(div(X, Y)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  a!6220!6220minus(0, X) >= 0 
  a!6220!6220minus(s(X), s(Y)) >= a!6220!6220minus(X, Y) 
  a!6220!6220geq(X, 0) >= true 
  a!6220!6220geq(0, s(X)) >= false 
  a!6220!6220geq(s(X), s(Y)) >= a!6220!6220geq(X, Y) 
  a!6220!6220div(0, s(X)) >= 0 
  a!6220!6220div(s(X), s(Y)) >= a!6220!6220if(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 
  a!6220!6220if(true, X, Y) >= mark(X) 
  a!6220!6220if(false, X, Y) >= mark(Y) 
  mark(minus(X, Y)) >= a!6220!6220minus(X, Y) 
  mark(geq(X, Y)) >= a!6220!6220geq(X, Y) 
  mark(div(X, Y)) >= a!6220!6220div(mark(X), Y) 
  mark(if(X, Y, Z)) >= a!6220!6220if(mark(X), Y, Z) 
  mark(0) >= 0 
  mark(s(X)) >= s(mark(X)) 
  mark(true) >= true 
  mark(false) >= false 
  a!6220!6220minus(X, Y) >= minus(X, Y) 
  a!6220!6220geq(X, Y) >= geq(X, Y) 
  a!6220!6220div(X, Y) >= div(X, Y) 
  a!6220!6220if(X, Y, Z) >= if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220div = \y0y1.2y0 
  a!6220!6220div# = \y0y1.3y0 
  a!6220!6220geq = \y0y1.0 
  a!6220!6220if = \y0y1y2.1 + y1 + y2 
  a!6220!6220if# = \y0y1y2.1 + 2y1 + 2y2 
  a!6220!6220minus = \y0y1.0 
  div = \y0y1.2y0 
  false = 0 
  geq = \y0y1.0 
  if = \y0y1y2.1 + y1 + y2 
  mark = \y0.y0 
  mark# = \y0.2y0 
  minus = \y0y1.0 
  s = \y0.1 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220div#(s(_x0), s(_x1))]] = 3 + 6x0 >= 3 = [[a!6220!6220if#(a!6220!6220geq(_x0, _x1), s(div(minus(_x0, _x1), s(_x1))), 0)]] 
  [[a!6220!6220if#(true, _x0, _x1)]] = 1 + 2x0 + 2x1 > 2x0 = [[mark#(_x0)]] 
  [[a!6220!6220if#(false, _x0, _x1)]] = 1 + 2x0 + 2x1 > 2x1 = [[mark#(_x1)]] 
  [[mark#(div(_x0, _x1))]] = 4x0 >= 3x0 = [[a!6220!6220div#(mark(_x0), _x1)]] 
  [[mark#(div(_x0, _x1))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = 2 + 4x0 > 2x0 = [[mark#(_x0)]] 
  [[a!6220!6220minus(0, _x0)]] = 0 >= 0 = [[0]] 
  [[a!6220!6220minus(s(_x0), s(_x1))]] = 0 >= 0 = [[a!6220!6220minus(_x0, _x1)]] 
  [[a!6220!6220geq(_x0, 0)]] = 0 >= 0 = [[true]] 
  [[a!6220!6220geq(0, s(_x0))]] = 0 >= 0 = [[false]] 
  [[a!6220!6220geq(s(_x0), s(_x1))]] = 0 >= 0 = [[a!6220!6220geq(_x0, _x1)]] 
  [[a!6220!6220div(0, s(_x0))]] = 0 >= 0 = [[0]] 
  [[a!6220!6220div(s(_x0), s(_x1))]] = 2 + 4x0 >= 2 = [[a!6220!6220if(a!6220!6220geq(_x0, _x1), s(div(minus(_x0, _x1), s(_x1))), 0)]] 
  [[a!6220!6220if(true, _x0, _x1)]] = 1 + x0 + x1 >= x0 = [[mark(_x0)]] 
  [[a!6220!6220if(false, _x0, _x1)]] = 1 + x0 + x1 >= x1 = [[mark(_x1)]] 
  [[mark(minus(_x0, _x1))]] = 0 >= 0 = [[a!6220!6220minus(_x0, _x1)]] 
  [[mark(geq(_x0, _x1))]] = 0 >= 0 = [[a!6220!6220geq(_x0, _x1)]] 
  [[mark(div(_x0, _x1))]] = 2x0 >= 2x0 = [[a!6220!6220div(mark(_x0), _x1)]] 
  [[mark(if(_x0, _x1, _x2))]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[a!6220!6220if(mark(_x0), _x1, _x2)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[s(mark(_x0))]] 
  [[mark(true)]] = 0 >= 0 = [[true]] 
  [[mark(false)]] = 0 >= 0 = [[false]] 
  [[a!6220!6220minus(_x0, _x1)]] = 0 >= 0 = [[minus(_x0, _x1)]] 
  [[a!6220!6220geq(_x0, _x1)]] = 0 >= 0 = [[geq(_x0, _x1)]] 
  [[a!6220!6220div(_x0, _x1)]] = 2x0 >= 2x0 = [[div(_x0, _x1)]] 
  [[a!6220!6220if(_x0, _x1, _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[if(_x0, _x1, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_0, minimal, formative) by (P_5, R_0, minimal, formative), where P_5 consists of:

  a!6220!6220div#(s(X), s(Y)) =#> a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0)   
  mark#(div(X, Y)) =#> a!6220!6220div#(mark(X), Y)   
  mark#(div(X, Y)) =#> mark#(X)   

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, 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 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 
    * 2 : 1, 2 

This graph has the following strongly connected components:

  P_6:

    mark#(div(X, Y)) =#> mark#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_5, R_0, m, f) by (P_6, 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) and (P_6, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_6, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(mark#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(mark#(div(X, Y))) = div(X, Y) |> X = nu(mark#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, 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(a!6220!6220geq#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(a!6220!6220geq#(s(X), s(Y))) = s(X) |> X = nu(a!6220!6220geq#(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(a!6220!6220minus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(a!6220!6220minus#(s(X), s(Y))) = s(X) |> X = nu(a!6220!6220minus#(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.