/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 
  active : [o] --> o 
  cons : [o * o] --> o 
  from : [o] --> o 
  length : [o] --> o 
  length1 : [o] --> o 
  mark : [o] --> o 
  nil : [] --> o 
  s : [o] --> o 

  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(length(nil)) => mark(0) 
  active(length(cons(X, Y))) => mark(s(length1(Y))) 
  active(length1(X)) => mark(length(X)) 
  mark(from(X)) => active(from(mark(X))) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(s(X)) => active(s(mark(X))) 
  mark(length(X)) => active(length(X)) 
  mark(nil) => active(nil) 
  mark(0) => active(0) 
  mark(length1(X)) => active(length1(X)) 
  from(mark(X)) => from(X) 
  from(active(X)) => from(X) 
  cons(mark(X), Y) => cons(X, Y) 
  cons(X, mark(Y)) => cons(X, Y) 
  cons(active(X), Y) => cons(X, Y) 
  cons(X, active(Y)) => cons(X, Y) 
  s(mark(X)) => s(X) 
  s(active(X)) => s(X) 
  length(mark(X)) => length(X) 
  length(active(X)) => length(X) 
  length1(mark(X)) => length1(X) 
  length1(active(X)) => length1(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] active#(from(X)) =#> mark#(cons(X, from(s(X))))   
    1] active#(from(X)) =#> cons#(X, from(s(X)))   
    2] active#(from(X)) =#> from#(s(X))   
    3] active#(from(X)) =#> s#(X)   
    4] active#(length(nil)) =#> mark#(0)   
    5] active#(length(cons(X, Y))) =#> mark#(s(length1(Y)))   
    6] active#(length(cons(X, Y))) =#> s#(length1(Y))   
    7] active#(length(cons(X, Y))) =#> length1#(Y)   
    8] active#(length1(X)) =#> mark#(length(X))   
    9] active#(length1(X)) =#> length#(X)   
    10] mark#(from(X)) =#> active#(from(mark(X)))   
    11] mark#(from(X)) =#> from#(mark(X))   
    12] mark#(from(X)) =#> mark#(X)   
    13] mark#(cons(X, Y)) =#> active#(cons(mark(X), Y))   
    14] mark#(cons(X, Y)) =#> cons#(mark(X), Y)   
    15] mark#(cons(X, Y)) =#> mark#(X)   
    16] mark#(s(X)) =#> active#(s(mark(X)))   
    17] mark#(s(X)) =#> s#(mark(X))   
    18] mark#(s(X)) =#> mark#(X)   
    19] mark#(length(X)) =#> active#(length(X))   
    20] mark#(length(X)) =#> length#(X)   
    21] mark#(nil) =#> active#(nil)   
    22] mark#(0) =#> active#(0)   
    23] mark#(length1(X)) =#> active#(length1(X))   
    24] mark#(length1(X)) =#> length1#(X)   
    25] from#(mark(X)) =#> from#(X)   
    26] from#(active(X)) =#> from#(X)   
    27] cons#(mark(X), Y) =#> cons#(X, Y)   
    28] cons#(X, mark(Y)) =#> cons#(X, Y)   
    29] cons#(active(X), Y) =#> cons#(X, Y)   
    30] cons#(X, active(Y)) =#> cons#(X, Y)   
    31] s#(mark(X)) =#> s#(X)   
    32] s#(active(X)) =#> s#(X)   
    33] length#(mark(X)) =#> length#(X)   
    34] length#(active(X)) =#> length#(X)   
    35] length1#(mark(X)) =#> length1#(X)   
    36] length1#(active(X)) =#> length1#(X)   

  Rules R_0:

    active(from(X)) => mark(cons(X, from(s(X)))) 
    active(length(nil)) => mark(0) 
    active(length(cons(X, Y))) => mark(s(length1(Y))) 
    active(length1(X)) => mark(length(X)) 
    mark(from(X)) => active(from(mark(X))) 
    mark(cons(X, Y)) => active(cons(mark(X), Y)) 
    mark(s(X)) => active(s(mark(X))) 
    mark(length(X)) => active(length(X)) 
    mark(nil) => active(nil) 
    mark(0) => active(0) 
    mark(length1(X)) => active(length1(X)) 
    from(mark(X)) => from(X) 
    from(active(X)) => from(X) 
    cons(mark(X), Y) => cons(X, Y) 
    cons(X, mark(Y)) => cons(X, Y) 
    cons(active(X), Y) => cons(X, Y) 
    cons(X, active(Y)) => cons(X, Y) 
    s(mark(X)) => s(X) 
    s(active(X)) => s(X) 
    length(mark(X)) => length(X) 
    length(active(X)) => length(X) 
    length1(mark(X)) => length1(X) 
    length1(active(X)) => length1(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 : 13, 14, 15 
    * 1 : 27, 29 
    * 2 :  
    * 3 : 31, 32 
    * 4 : 22 
    * 5 : 16, 17, 18 
    * 6 :  
    * 7 : 35, 36 
    * 8 : 19, 20 
    * 9 : 33, 34 
    * 10 : 0, 1, 2, 3 
    * 11 : 25, 26 
    * 12 : 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24 
    * 13 :  
    * 14 : 27, 28, 29, 30 
    * 15 : 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24 
    * 16 :  
    * 17 : 31, 32 
    * 18 : 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24 
    * 19 : 4, 5, 6, 7 
    * 20 : 33, 34 
    * 21 :  
    * 22 :  
    * 23 : 8, 9 
    * 24 : 35, 36 
    * 25 : 25, 26 
    * 26 : 25, 26 
    * 27 : 27, 28, 29, 30 
    * 28 : 27, 28, 29, 30 
    * 29 : 27, 28, 29, 30 
    * 30 : 27, 28, 29, 30 
    * 31 : 31, 32 
    * 32 : 31, 32 
    * 33 : 33, 34 
    * 34 : 33, 34 
    * 35 : 35, 36 
    * 36 : 35, 36 

This graph has the following strongly connected components:

  P_1:

    active#(from(X)) =#> mark#(cons(X, from(s(X))))   
    active#(length(cons(X, Y))) =#> mark#(s(length1(Y)))   
    active#(length1(X)) =#> mark#(length(X))   
    mark#(from(X)) =#> active#(from(mark(X)))   
    mark#(from(X)) =#> mark#(X)   
    mark#(cons(X, Y)) =#> mark#(X)   
    mark#(s(X)) =#> mark#(X)   
    mark#(length(X)) =#> active#(length(X))   
    mark#(length1(X)) =#> active#(length1(X))   

  P_2:

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

  P_3:

    cons#(mark(X), Y) =#> cons#(X, Y)   
    cons#(X, mark(Y)) =#> cons#(X, Y)   
    cons#(active(X), Y) =#> cons#(X, Y)   
    cons#(X, active(Y)) =#> cons#(X, Y)   

  P_4:

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

  P_5:

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

  P_6:

    length1#(mark(X)) =#> length1#(X)   
    length1#(active(X)) =#> length1#(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), (P_5, R_0, m, f) and (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), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, 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(length1#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(length1#(mark(X))) = mark(X) |> X = nu(length1#(X)) 
  nu(length1#(active(X))) = active(X) |> X = nu(length1#(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), (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(length#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(length#(mark(X))) = mark(X) |> X = nu(length#(X)) 
  nu(length#(active(X))) = active(X) |> X = nu(length#(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).

We apply the subterm criterion with the following projection function:

  nu(s#) = 1 

Thus, we can orient the dependency pairs as follows:

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

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(cons#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_7, R_0, minimal, f), where P_7 contains:

  cons#(X, mark(Y)) =#> cons#(X, Y)   
  cons#(X, active(Y)) =#> cons#(X, Y)   

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (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(cons#) = 2 

Thus, we can orient the dependency pairs as follows:

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

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(from#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(from#(mark(X))) = mark(X) |> X = nu(from#(X)) 
  nu(from#(active(X))) = active(X) |> X = nu(from#(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#(from(X)) >? mark#(cons(X, from(s(X)))) 
  active#(length(cons(X, Y))) >? mark#(s(length1(Y))) 
  active#(length1(X)) >? mark#(length(X)) 
  mark#(from(X)) >? active#(from(mark(X))) 
  mark#(from(X)) >? mark#(X) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  mark#(length(X)) >? active#(length(X)) 
  mark#(length1(X)) >? active#(length1(X)) 
  active(from(X)) >= mark(cons(X, from(s(X)))) 
  active(length(nil)) >= mark(0) 
  active(length(cons(X, Y))) >= mark(s(length1(Y))) 
  active(length1(X)) >= mark(length(X)) 
  mark(from(X)) >= active(from(mark(X))) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(s(X)) >= active(s(mark(X))) 
  mark(length(X)) >= active(length(X)) 
  mark(nil) >= active(nil) 
  mark(0) >= active(0) 
  mark(length1(X)) >= active(length1(X)) 
  from(mark(X)) >= from(X) 
  from(active(X)) >= from(X) 
  cons(mark(X), Y) >= cons(X, Y) 
  cons(X, mark(Y)) >= cons(X, Y) 
  cons(active(X), Y) >= cons(X, Y) 
  cons(X, active(Y)) >= cons(X, Y) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(X) 
  length(mark(X)) >= length(X) 
  length(active(X)) >= length(X) 
  length1(mark(X)) >= length1(X) 
  length1(active(X)) >= length1(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  active = \y0.y0 
  active# = \y0.2y0 
  cons = \y0y1.y0 
  from = \y0.1 + y0 
  length = \y0.0 
  length1 = \y0.0 
  mark = \y0.y0 
  mark# = \y0.2y0 
  nil = 0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active#(from(_x0))]] = 2 + 2x0 > 2x0 = [[mark#(cons(_x0, from(s(_x0))))]] 
  [[active#(length(cons(_x0, _x1)))]] = 0 >= 0 = [[mark#(s(length1(_x1)))]] 
  [[active#(length1(_x0))]] = 0 >= 0 = [[mark#(length(_x0))]] 
  [[mark#(from(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[active#(from(mark(_x0)))]] 
  [[mark#(from(_x0))]] = 2 + 2x0 > 2x0 = [[mark#(_x0)]] 
  [[mark#(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(length(_x0))]] = 0 >= 0 = [[active#(length(_x0))]] 
  [[mark#(length1(_x0))]] = 0 >= 0 = [[active#(length1(_x0))]] 
  [[active(from(_x0))]] = 1 + x0 >= x0 = [[mark(cons(_x0, from(s(_x0))))]] 
  [[active(length(nil))]] = 0 >= 0 = [[mark(0)]] 
  [[active(length(cons(_x0, _x1)))]] = 0 >= 0 = [[mark(s(length1(_x1)))]] 
  [[active(length1(_x0))]] = 0 >= 0 = [[mark(length(_x0))]] 
  [[mark(from(_x0))]] = 1 + x0 >= 1 + x0 = [[active(from(mark(_x0)))]] 
  [[mark(cons(_x0, _x1))]] = x0 >= x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[active(s(mark(_x0)))]] 
  [[mark(length(_x0))]] = 0 >= 0 = [[active(length(_x0))]] 
  [[mark(nil)]] = 0 >= 0 = [[active(nil)]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(length1(_x0))]] = 0 >= 0 = [[active(length1(_x0))]] 
  [[from(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[from(_x0)]] 
  [[from(active(_x0))]] = 1 + x0 >= 1 + x0 = [[from(_x0)]] 
  [[cons(mark(_x0), _x1)]] = x0 >= x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = x0 >= x0 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = x0 >= x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = x0 >= x0 = [[cons(_x0, _x1)]] 
  [[s(mark(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[length(mark(_x0))]] = 0 >= 0 = [[length(_x0)]] 
  [[length(active(_x0))]] = 0 >= 0 = [[length(_x0)]] 
  [[length1(mark(_x0))]] = 0 >= 0 = [[length1(_x0)]] 
  [[length1(active(_x0))]] = 0 >= 0 = [[length1(_x0)]] 

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_8, R_0, minimal, formative), where P_8 consists of:

  active#(length(cons(X, Y))) =#> mark#(s(length1(Y)))   
  active#(length1(X)) =#> mark#(length(X))   
  mark#(from(X)) =#> active#(from(mark(X)))   
  mark#(cons(X, Y)) =#> mark#(X)   
  mark#(s(X)) =#> mark#(X)   
  mark#(length(X)) =#> active#(length(X))   
  mark#(length1(X)) =#> active#(length1(X))   

Thus, the original system is terminating if (P_8, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_8, 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 : 4 
    * 1 : 5 
    * 2 :  
    * 3 : 2, 3, 4, 5, 6 
    * 4 : 2, 3, 4, 5, 6 
    * 5 : 0 
    * 6 : 1 

This graph has the following strongly connected components:

  P_9:

    active#(length(cons(X, Y))) =#> mark#(s(length1(Y)))   
    active#(length1(X)) =#> mark#(length(X))   
    mark#(cons(X, Y)) =#> mark#(X)   
    mark#(s(X)) =#> mark#(X)   
    mark#(length(X)) =#> active#(length(X))   
    mark#(length1(X)) =#> active#(length1(X))   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_8, R_0, m, f) by (P_9, R_0, m, f).

Thus, the original system is terminating if (P_9, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_9, R_0, minimal, formative).

The formative rules of (P_9, R_0) are R_1 ::=

  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(length(nil)) => mark(0) 
  active(length(cons(X, Y))) => mark(s(length1(Y))) 
  active(length1(X)) => mark(length(X)) 
  mark(from(X)) => active(from(mark(X))) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(s(X)) => active(s(mark(X))) 
  mark(length(X)) => active(length(X)) 
  mark(nil) => active(nil) 
  mark(0) => active(0) 
  mark(length1(X)) => active(length1(X)) 
  cons(mark(X), Y) => cons(X, Y) 
  cons(X, mark(Y)) => cons(X, Y) 
  cons(active(X), Y) => cons(X, Y) 
  cons(X, active(Y)) => cons(X, Y) 
  s(mark(X)) => s(X) 
  s(active(X)) => s(X) 
  length(mark(X)) => length(X) 
  length(active(X)) => length(X) 
  length1(mark(X)) => length1(X) 
  length1(active(X)) => length1(X) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_9, R_0, minimal, formative) by (P_9, R_1, minimal, formative).

Thus, the original system is terminating if (P_9, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_9, R_1, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_9, R_1) are:

  s(mark(X)) => s(X) 
  s(active(X)) => s(X) 
  length(mark(X)) => length(X) 
  length(active(X)) => length(X) 
  length1(mark(X)) => length1(X) 
  length1(active(X)) => length1(X) 

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

  active#(length(cons(X, Y))) >? mark#(s(length1(Y))) 
  active#(length1(X)) >? mark#(length(X)) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  mark#(length(X)) >? active#(length(X)) 
  mark#(length1(X)) >? active#(length1(X)) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(X) 
  length(mark(X)) >= length(X) 
  length(active(X)) >= length(X) 
  length1(mark(X)) >= length1(X) 
  length1(active(X)) >= length1(X) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  active# = \y0.y0 
  cons = \y0y1.3 + 2y0 + 2y1 
  length = \y0.y0 
  length1 = \y0.2y0 
  mark = \y0.2y0 
  mark# = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active#(length(cons(_x0, _x1)))]] = 3 + 2x0 + 2x1 > 2x1 = [[mark#(s(length1(_x1)))]] 
  [[active#(length1(_x0))]] = 2x0 >= x0 = [[mark#(length(_x0))]] 
  [[mark#(cons(_x0, _x1))]] = 3 + 2x0 + 2x1 > x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = x0 >= x0 = [[mark#(_x0)]] 
  [[mark#(length(_x0))]] = x0 >= x0 = [[active#(length(_x0))]] 
  [[mark#(length1(_x0))]] = 2x0 >= 2x0 = [[active#(length1(_x0))]] 
  [[s(mark(_x0))]] = 2x0 >= x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[length(mark(_x0))]] = 2x0 >= x0 = [[length(_x0)]] 
  [[length(active(_x0))]] = x0 >= x0 = [[length(_x0)]] 
  [[length1(mark(_x0))]] = 4x0 >= 2x0 = [[length1(_x0)]] 
  [[length1(active(_x0))]] = 2x0 >= 2x0 = [[length1(_x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_9, R_1, minimal, formative) by (P_10, R_1, minimal, formative), where P_10 consists of:

  active#(length1(X)) =#> mark#(length(X))   
  mark#(s(X)) =#> mark#(X)   
  mark#(length(X)) =#> active#(length(X))   
  mark#(length1(X)) =#> active#(length1(X))   

Thus, the original system is terminating if (P_10, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_10, R_1, 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 : 2 
    * 1 : 1, 2, 3 
    * 2 :  
    * 3 : 0 

This graph has the following strongly connected components:

  P_11:

    mark#(s(X)) =#> mark#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_10, R_1, m, f) by (P_11, R_1, m, f).

Thus, the original system is terminating if (P_11, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_11, R_1, 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#(s(X))) = s(X) |> X = nu(mark#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_11, R_1, minimal, f) by ({}, R_1, 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.