/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.

  active : [o] --> o 
  c : [] --> o 
  f : [o] --> o 
  false : [] --> o 
  if : [o * o * o] --> o 
  mark : [o] --> o 
  true : [] --> o 

  active(f(X)) => mark(if(X, c, f(true))) 
  active(if(true, X, Y)) => mark(X) 
  active(if(false, X, Y)) => mark(Y) 
  mark(f(X)) => active(f(mark(X))) 
  mark(if(X, Y, Z)) => active(if(mark(X), mark(Y), Z)) 
  mark(c) => active(c) 
  mark(true) => active(true) 
  mark(false) => active(false) 
  f(mark(X)) => f(X) 
  f(active(X)) => f(X) 
  if(mark(X), Y, Z) => if(X, Y, Z) 
  if(X, mark(Y), Z) => if(X, Y, Z) 
  if(X, Y, mark(Z)) => if(X, Y, Z) 
  if(active(X), Y, Z) => if(X, Y, Z) 
  if(X, active(Y), Z) => if(X, Y, Z) 
  if(X, Y, active(Z)) => if(X, Y, Z) 

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

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

  active(f(X)) >? mark(if(X, c, f(true))) 
  active(if(true, X, Y)) >? mark(X) 
  active(if(false, X, Y)) >? mark(Y) 
  mark(f(X)) >? active(f(mark(X))) 
  mark(if(X, Y, Z)) >? active(if(mark(X), mark(Y), Z)) 
  mark(c) >? active(c) 
  mark(true) >? active(true) 
  mark(false) >? active(false) 
  f(mark(X)) >? f(X) 
  f(active(X)) >? f(X) 
  if(mark(X), Y, Z) >? if(X, Y, Z) 
  if(X, mark(Y), Z) >? if(X, Y, Z) 
  if(X, Y, mark(Z)) >? if(X, Y, Z) 
  if(active(X), Y, Z) >? if(X, Y, Z) 
  if(X, active(Y), Z) >? if(X, Y, Z) 
  if(X, Y, active(Z)) >? if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  c = 0 
  f = \y0.3y0 
  false = 2 
  if = \y0y1y2.y0 + 2y1 + 3y2 
  mark = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[active(f(_x0))]] = 3x0 >= 2x0 = [[mark(if(_x0, c, f(true)))]] 
  [[active(if(true, _x0, _x1))]] = 2x0 + 3x1 >= 2x0 = [[mark(_x0)]] 
  [[active(if(false, _x0, _x1))]] = 2 + 2x0 + 3x1 > 2x1 = [[mark(_x1)]] 
  [[mark(f(_x0))]] = 6x0 >= 6x0 = [[active(f(mark(_x0)))]] 
  [[mark(if(_x0, _x1, _x2))]] = 2x0 + 4x1 + 6x2 >= 2x0 + 3x2 + 4x1 = [[active(if(mark(_x0), mark(_x1), _x2))]] 
  [[mark(c)]] = 0 >= 0 = [[active(c)]] 
  [[mark(true)]] = 0 >= 0 = [[active(true)]] 
  [[mark(false)]] = 4 > 2 = [[active(false)]] 
  [[f(mark(_x0))]] = 6x0 >= 3x0 = [[f(_x0)]] 
  [[f(active(_x0))]] = 3x0 >= 3x0 = [[f(_x0)]] 
  [[if(mark(_x0), _x1, _x2)]] = 2x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, mark(_x1), _x2)]] = x0 + 3x2 + 4x1 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, _x1, mark(_x2))]] = x0 + 2x1 + 6x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] 
  [[if(active(_x0), _x1, _x2)]] = x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, active(_x1), _x2)]] = x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, _x1, active(_x2))]] = x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  active(if(false, X, Y)) => mark(Y) 
  mark(false) => active(false) 

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] active#(f(X)) =#> mark#(if(X, c, f(true)))   
    1] active#(f(X)) =#> if#(X, c, f(true))   
    2] active#(f(X)) =#> f#(true)   
    3] active#(if(true, X, Y)) =#> mark#(X)   
    4] mark#(f(X)) =#> active#(f(mark(X)))   
    5] mark#(f(X)) =#> f#(mark(X))   
    6] mark#(f(X)) =#> mark#(X)   
    7] mark#(if(X, Y, Z)) =#> active#(if(mark(X), mark(Y), Z))   
    8] mark#(if(X, Y, Z)) =#> if#(mark(X), mark(Y), Z)   
    9] mark#(if(X, Y, Z)) =#> mark#(X)   
    10] mark#(if(X, Y, Z)) =#> mark#(Y)   
    11] mark#(c) =#> active#(c)   
    12] mark#(true) =#> active#(true)   
    13] f#(mark(X)) =#> f#(X)   
    14] f#(active(X)) =#> f#(X)   
    15] if#(mark(X), Y, Z) =#> if#(X, Y, Z)   
    16] if#(X, mark(Y), Z) =#> if#(X, Y, Z)   
    17] if#(X, Y, mark(Z)) =#> if#(X, Y, Z)   
    18] if#(active(X), Y, Z) =#> if#(X, Y, Z)   
    19] if#(X, active(Y), Z) =#> if#(X, Y, Z)   
    20] if#(X, Y, active(Z)) =#> if#(X, Y, Z)   

  Rules R_0:

    active(f(X)) => mark(if(X, c, f(true))) 
    active(if(true, X, Y)) => mark(X) 
    mark(f(X)) => active(f(mark(X))) 
    mark(if(X, Y, Z)) => active(if(mark(X), mark(Y), Z)) 
    mark(c) => active(c) 
    mark(true) => active(true) 
    f(mark(X)) => f(X) 
    f(active(X)) => f(X) 
    if(mark(X), Y, Z) => if(X, Y, Z) 
    if(X, mark(Y), Z) => if(X, Y, Z) 
    if(X, Y, mark(Z)) => if(X, Y, Z) 
    if(active(X), Y, Z) => if(X, Y, Z) 
    if(X, active(Y), Z) => if(X, Y, Z) 
    if(X, Y, active(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 : 7, 8, 9, 10 
    * 1 : 15, 18 
    * 2 :  
    * 3 : 4, 5, 6, 7, 8, 9, 10, 11, 12 
    * 4 : 0, 1, 2 
    * 5 : 13, 14 
    * 6 : 4, 5, 6, 7, 8, 9, 10, 11, 12 
    * 7 : 3 
    * 8 : 15, 16, 17, 18, 19, 20 
    * 9 : 4, 5, 6, 7, 8, 9, 10, 11, 12 
    * 10 : 4, 5, 6, 7, 8, 9, 10, 11, 12 
    * 11 :  
    * 12 :  
    * 13 : 13, 14 
    * 14 : 13, 14 
    * 15 : 15, 16, 17, 18, 19, 20 
    * 16 : 15, 16, 17, 18, 19, 20 
    * 17 : 15, 16, 17, 18, 19, 20 
    * 18 : 15, 16, 17, 18, 19, 20 
    * 19 : 15, 16, 17, 18, 19, 20 
    * 20 : 15, 16, 17, 18, 19, 20 

This graph has the following strongly connected components:

  P_1:

    active#(f(X)) =#> mark#(if(X, c, f(true)))   
    active#(if(true, X, Y)) =#> mark#(X)   
    mark#(f(X)) =#> active#(f(mark(X)))   
    mark#(f(X)) =#> mark#(X)   
    mark#(if(X, Y, Z)) =#> active#(if(mark(X), mark(Y), Z))   
    mark#(if(X, Y, Z)) =#> mark#(X)   
    mark#(if(X, Y, Z)) =#> mark#(Y)   

  P_2:

    f#(mark(X)) =#> f#(X)   
    f#(active(X)) =#> f#(X)   

  P_3:

    if#(mark(X), Y, Z) =#> if#(X, Y, Z)   
    if#(X, mark(Y), Z) =#> if#(X, Y, Z)   
    if#(X, Y, mark(Z)) =#> if#(X, Y, Z)   
    if#(active(X), Y, Z) =#> if#(X, Y, Z)   
    if#(X, active(Y), Z) =#> if#(X, Y, Z)   
    if#(X, Y, active(Z)) =#> if#(X, Y, 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) 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 apply the subterm criterion with the following projection function:

  nu(if#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(if#(mark(X), Y, Z)) = mark(X) |> X = nu(if#(X, Y, Z)) 
  nu(if#(X, mark(Y), Z)) = X = X = nu(if#(X, Y, Z)) 
  nu(if#(X, Y, mark(Z))) = X = X = nu(if#(X, Y, Z)) 
  nu(if#(active(X), Y, Z)) = active(X) |> X = nu(if#(X, Y, Z)) 
  nu(if#(X, active(Y), Z)) = X = X = nu(if#(X, Y, Z)) 
  nu(if#(X, Y, active(Z))) = X = X = nu(if#(X, Y, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains:

  if#(X, mark(Y), Z) =#> if#(X, Y, Z)   
  if#(X, Y, mark(Z)) =#> if#(X, Y, Z)   
  if#(X, active(Y), Z) =#> if#(X, Y, Z)   
  if#(X, Y, active(Z)) =#> if#(X, Y, Z)   

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 apply the subterm criterion with the following projection function:

  nu(if#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(if#(X, mark(Y), Z)) = mark(Y) |> Y = nu(if#(X, Y, Z)) 
  nu(if#(X, Y, mark(Z))) = Y = Y = nu(if#(X, Y, Z)) 
  nu(if#(X, active(Y), Z)) = active(Y) |> Y = nu(if#(X, Y, Z)) 
  nu(if#(X, Y, active(Z))) = Y = Y = nu(if#(X, Y, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by (P_5, R_0, minimal, f), where P_5 contains:

  if#(X, Y, mark(Z)) =#> if#(X, Y, Z)   
  if#(X, Y, active(Z)) =#> if#(X, Y, Z)   

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 apply the subterm criterion with the following projection function:

  nu(if#) = 3 

Thus, we can orient the dependency pairs as follows:

  nu(if#(X, Y, mark(Z))) = mark(Z) |> Z = nu(if#(X, Y, Z)) 
  nu(if#(X, Y, active(Z))) = active(Z) |> Z = nu(if#(X, 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.

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(f#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(f#(mark(X))) = mark(X) |> X = nu(f#(X)) 
  nu(f#(active(X))) = active(X) |> X = nu(f#(X)) 

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

  active#(f(X)) >? mark#(if(X, c, f(true))) 
  active#(if(true, X, Y)) >? mark#(X) 
  mark#(f(X)) >? active#(f(mark(X))) 
  mark#(f(X)) >? mark#(X) 
  mark#(if(X, Y, Z)) >? active#(if(mark(X), mark(Y), Z)) 
  mark#(if(X, Y, Z)) >? mark#(X) 
  mark#(if(X, Y, Z)) >? mark#(Y) 
  active(f(X)) >= mark(if(X, c, f(true))) 
  active(if(true, X, Y)) >= mark(X) 
  mark(f(X)) >= active(f(mark(X))) 
  mark(if(X, Y, Z)) >= active(if(mark(X), mark(Y), Z)) 
  mark(c) >= active(c) 
  mark(true) >= active(true) 
  f(mark(X)) >= f(X) 
  f(active(X)) >= f(X) 
  if(mark(X), Y, Z) >= if(X, Y, Z) 
  if(X, mark(Y), Z) >= if(X, Y, Z) 
  if(X, Y, mark(Z)) >= if(X, Y, Z) 
  if(active(X), Y, Z) >= if(X, Y, Z) 
  if(X, active(Y), Z) >= if(X, Y, Z) 
  if(X, Y, active(Z)) >= if(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  active# = \y0.y0 
  c = 0 
  f = \y0.1 + 2y0 
  if = \y0y1y2.y0 + 2y1 
  mark = \y0.2y0 
  mark# = \y0.1 + 2y0 
  true = 2 

Using this interpretation, the requirements translate to:

  [[active#(f(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark#(if(_x0, c, f(true)))]] 
  [[active#(if(true, _x0, _x1))]] = 2 + 2x0 > 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(f(_x0))]] = 3 + 4x0 > 1 + 4x0 = [[active#(f(mark(_x0)))]] 
  [[mark#(f(_x0))]] = 3 + 4x0 > 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(if(_x0, _x1, _x2))]] = 1 + 2x0 + 4x1 > 2x0 + 4x1 = [[active#(if(mark(_x0), mark(_x1), _x2))]] 
  [[mark#(if(_x0, _x1, _x2))]] = 1 + 2x0 + 4x1 >= 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(if(_x0, _x1, _x2))]] = 1 + 2x0 + 4x1 >= 1 + 2x1 = [[mark#(_x1)]] 
  [[active(f(_x0))]] = 1 + 2x0 >= 2x0 = [[mark(if(_x0, c, f(true)))]] 
  [[active(if(true, _x0, _x1))]] = 2 + 2x0 >= 2x0 = [[mark(_x0)]] 
  [[mark(f(_x0))]] = 2 + 4x0 >= 1 + 4x0 = [[active(f(mark(_x0)))]] 
  [[mark(if(_x0, _x1, _x2))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[active(if(mark(_x0), mark(_x1), _x2))]] 
  [[mark(c)]] = 0 >= 0 = [[active(c)]] 
  [[mark(true)]] = 4 >= 2 = [[active(true)]] 
  [[f(mark(_x0))]] = 1 + 4x0 >= 1 + 2x0 = [[f(_x0)]] 
  [[f(active(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[f(_x0)]] 
  [[if(mark(_x0), _x1, _x2)]] = 2x0 + 2x1 >= x0 + 2x1 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, mark(_x1), _x2)]] = x0 + 4x1 >= x0 + 2x1 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, _x1, mark(_x2))]] = x0 + 2x1 >= x0 + 2x1 = [[if(_x0, _x1, _x2)]] 
  [[if(active(_x0), _x1, _x2)]] = x0 + 2x1 >= x0 + 2x1 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, active(_x1), _x2)]] = x0 + 2x1 >= x0 + 2x1 = [[if(_x0, _x1, _x2)]] 
  [[if(_x0, _x1, active(_x2))]] = x0 + 2x1 >= x0 + 2x1 = [[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_1, R_0, minimal, formative) by (P_6, R_0, minimal, formative), where P_6 consists of:

  active#(f(X)) =#> mark#(if(X, c, f(true)))   
  mark#(if(X, Y, Z)) =#> mark#(X)   
  mark#(if(X, Y, Z)) =#> mark#(Y)   

Thus, the original system is terminating if (P_6, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_6, 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, 2 
    * 1 : 1, 2 
    * 2 : 1, 2 

This graph has the following strongly connected components:

  P_7:

    mark#(if(X, Y, Z)) =#> mark#(X)   
    mark#(if(X, Y, Z)) =#> mark#(Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_6, R_0, m, f) by (P_7, R_0, m, f).

Thus, the original system is terminating if (P_7, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_7, 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#(if(X, Y, Z))) = if(X, Y, Z) |> X = nu(mark#(X)) 
  nu(mark#(if(X, Y, Z))) = if(X, Y, Z) |> Y = nu(mark#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_7, 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.