/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 
  head : [o] --> o 
  incr : [o] --> o 
  mark : [o] --> o 
  nats : [] --> o 
  odds : [] --> o 
  pairs : [] --> o 
  s : [o] --> o 
  tail : [o] --> o 

  active(nats) => mark(cons(0, incr(nats))) 
  active(pairs) => mark(cons(0, incr(odds))) 
  active(odds) => mark(incr(pairs)) 
  active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) 
  active(head(cons(X, Y))) => mark(X) 
  active(tail(cons(X, Y))) => mark(Y) 
  mark(nats) => active(nats) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(0) => active(0) 
  mark(incr(X)) => active(incr(mark(X))) 
  mark(pairs) => active(pairs) 
  mark(odds) => active(odds) 
  mark(s(X)) => active(s(mark(X))) 
  mark(head(X)) => active(head(mark(X))) 
  mark(tail(X)) => active(tail(mark(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) 
  incr(mark(X)) => incr(X) 
  incr(active(X)) => incr(X) 
  s(mark(X)) => s(X) 
  s(active(X)) => s(X) 
  head(mark(X)) => head(X) 
  head(active(X)) => head(X) 
  tail(mark(X)) => tail(X) 
  tail(active(X)) => tail(X) 

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(nats) >? mark(cons(0, incr(nats))) 
  active(pairs) >? mark(cons(0, incr(odds))) 
  active(odds) >? mark(incr(pairs)) 
  active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) 
  active(head(cons(X, Y))) >? mark(X) 
  active(tail(cons(X, Y))) >? mark(Y) 
  mark(nats) >? active(nats) 
  mark(cons(X, Y)) >? active(cons(mark(X), Y)) 
  mark(0) >? active(0) 
  mark(incr(X)) >? active(incr(mark(X))) 
  mark(pairs) >? active(pairs) 
  mark(odds) >? active(odds) 
  mark(s(X)) >? active(s(mark(X))) 
  mark(head(X)) >? active(head(mark(X))) 
  mark(tail(X)) >? active(tail(mark(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) 
  incr(mark(X)) >? incr(X) 
  incr(active(X)) >? incr(X) 
  s(mark(X)) >? s(X) 
  s(active(X)) >? s(X) 
  head(mark(X)) >? head(X) 
  head(active(X)) >? head(X) 
  tail(mark(X)) >? tail(X) 
  tail(active(X)) >? tail(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  active = \y0.y0 
  cons = \y0y1.y0 + y1 
  head = \y0.y0 
  incr = \y0.y0 
  mark = \y0.y0 
  nats = 0 
  odds = 0 
  pairs = 0 
  s = \y0.y0 
  tail = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[active(nats)]] = 0 >= 0 = [[mark(cons(0, incr(nats)))]] 
  [[active(pairs)]] = 0 >= 0 = [[mark(cons(0, incr(odds)))]] 
  [[active(odds)]] = 0 >= 0 = [[mark(incr(pairs))]] 
  [[active(incr(cons(_x0, _x1)))]] = x0 + x1 >= x0 + x1 = [[mark(cons(s(_x0), incr(_x1)))]] 
  [[active(head(cons(_x0, _x1)))]] = x0 + x1 >= x0 = [[mark(_x0)]] 
  [[active(tail(cons(_x0, _x1)))]] = 1 + x0 + x1 > x1 = [[mark(_x1)]] 
  [[mark(nats)]] = 0 >= 0 = [[active(nats)]] 
  [[mark(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[active(incr(mark(_x0)))]] 
  [[mark(pairs)]] = 0 >= 0 = [[active(pairs)]] 
  [[mark(odds)]] = 0 >= 0 = [[active(odds)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[active(s(mark(_x0)))]] 
  [[mark(head(_x0))]] = x0 >= x0 = [[active(head(mark(_x0)))]] 
  [[mark(tail(_x0))]] = 1 + x0 >= 1 + x0 = [[active(tail(mark(_x0)))]] 
  [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] 
  [[incr(mark(_x0))]] = x0 >= x0 = [[incr(_x0)]] 
  [[incr(active(_x0))]] = x0 >= x0 = [[incr(_x0)]] 
  [[s(mark(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[head(mark(_x0))]] = x0 >= x0 = [[head(_x0)]] 
  [[head(active(_x0))]] = x0 >= x0 = [[head(_x0)]] 
  [[tail(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[tail(_x0)]] 
  [[tail(active(_x0))]] = 1 + x0 >= 1 + x0 = [[tail(_x0)]] 

We can thus remove the following rules:

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

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(nats) >? mark(cons(0, incr(nats))) 
  active(pairs) >? mark(cons(0, incr(odds))) 
  active(odds) >? mark(incr(pairs)) 
  active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) 
  active(head(cons(X, Y))) >? mark(X) 
  mark(nats) >? active(nats) 
  mark(cons(X, Y)) >? active(cons(mark(X), Y)) 
  mark(0) >? active(0) 
  mark(incr(X)) >? active(incr(mark(X))) 
  mark(pairs) >? active(pairs) 
  mark(odds) >? active(odds) 
  mark(s(X)) >? active(s(mark(X))) 
  mark(head(X)) >? active(head(mark(X))) 
  mark(tail(X)) >? active(tail(mark(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) 
  incr(mark(X)) >? incr(X) 
  incr(active(X)) >? incr(X) 
  s(mark(X)) >? s(X) 
  s(active(X)) >? s(X) 
  head(mark(X)) >? head(X) 
  head(active(X)) >? head(X) 
  tail(mark(X)) >? tail(X) 
  tail(active(X)) >? tail(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  active = \y0.y0 
  cons = \y0y1.y1 + 2y0 
  head = \y0.2 + y0 
  incr = \y0.2y0 
  mark = \y0.y0 
  nats = 0 
  odds = 0 
  pairs = 0 
  s = \y0.2y0 
  tail = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[active(nats)]] = 0 >= 0 = [[mark(cons(0, incr(nats)))]] 
  [[active(pairs)]] = 0 >= 0 = [[mark(cons(0, incr(odds)))]] 
  [[active(odds)]] = 0 >= 0 = [[mark(incr(pairs))]] 
  [[active(incr(cons(_x0, _x1)))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[mark(cons(s(_x0), incr(_x1)))]] 
  [[active(head(cons(_x0, _x1)))]] = 2 + x1 + 2x0 > x0 = [[mark(_x0)]] 
  [[mark(nats)]] = 0 >= 0 = [[active(nats)]] 
  [[mark(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(incr(_x0))]] = 2x0 >= 2x0 = [[active(incr(mark(_x0)))]] 
  [[mark(pairs)]] = 0 >= 0 = [[active(pairs)]] 
  [[mark(odds)]] = 0 >= 0 = [[active(odds)]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[active(s(mark(_x0)))]] 
  [[mark(head(_x0))]] = 2 + x0 >= 2 + x0 = [[active(head(mark(_x0)))]] 
  [[mark(tail(_x0))]] = 2x0 >= 2x0 = [[active(tail(mark(_x0)))]] 
  [[cons(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(_x0, _x1)]] 
  [[incr(mark(_x0))]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[incr(active(_x0))]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[s(mark(_x0))]] = 2x0 >= 2x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = 2x0 >= 2x0 = [[s(_x0)]] 
  [[head(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[head(_x0)]] 
  [[head(active(_x0))]] = 2 + x0 >= 2 + x0 = [[head(_x0)]] 
  [[tail(mark(_x0))]] = 2x0 >= 2x0 = [[tail(_x0)]] 
  [[tail(active(_x0))]] = 2x0 >= 2x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  active(head(cons(X, Y))) => mark(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#(nats) =#> mark#(cons(0, incr(nats)))   
    1] active#(nats) =#> cons#(0, incr(nats))   
    2] active#(nats) =#> incr#(nats)   
    3] active#(pairs) =#> mark#(cons(0, incr(odds)))   
    4] active#(pairs) =#> cons#(0, incr(odds))   
    5] active#(pairs) =#> incr#(odds)   
    6] active#(odds) =#> mark#(incr(pairs))   
    7] active#(odds) =#> incr#(pairs)   
    8] active#(incr(cons(X, Y))) =#> mark#(cons(s(X), incr(Y)))   
    9] active#(incr(cons(X, Y))) =#> cons#(s(X), incr(Y))   
    10] active#(incr(cons(X, Y))) =#> s#(X)   
    11] active#(incr(cons(X, Y))) =#> incr#(Y)   
    12] mark#(nats) =#> active#(nats)   
    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#(0) =#> active#(0)   
    17] mark#(incr(X)) =#> active#(incr(mark(X)))   
    18] mark#(incr(X)) =#> incr#(mark(X))   
    19] mark#(incr(X)) =#> mark#(X)   
    20] mark#(pairs) =#> active#(pairs)   
    21] mark#(odds) =#> active#(odds)   
    22] mark#(s(X)) =#> active#(s(mark(X)))   
    23] mark#(s(X)) =#> s#(mark(X))   
    24] mark#(s(X)) =#> mark#(X)   
    25] mark#(head(X)) =#> active#(head(mark(X)))   
    26] mark#(head(X)) =#> head#(mark(X))   
    27] mark#(head(X)) =#> mark#(X)   
    28] mark#(tail(X)) =#> active#(tail(mark(X)))   
    29] mark#(tail(X)) =#> tail#(mark(X))   
    30] mark#(tail(X)) =#> mark#(X)   
    31] cons#(mark(X), Y) =#> cons#(X, Y)   
    32] cons#(X, mark(Y)) =#> cons#(X, Y)   
    33] cons#(active(X), Y) =#> cons#(X, Y)   
    34] cons#(X, active(Y)) =#> cons#(X, Y)   
    35] incr#(mark(X)) =#> incr#(X)   
    36] incr#(active(X)) =#> incr#(X)   
    37] s#(mark(X)) =#> s#(X)   
    38] s#(active(X)) =#> s#(X)   
    39] head#(mark(X)) =#> head#(X)   
    40] head#(active(X)) =#> head#(X)   
    41] tail#(mark(X)) =#> tail#(X)   
    42] tail#(active(X)) =#> tail#(X)   

  Rules R_0:

    active(nats) => mark(cons(0, incr(nats))) 
    active(pairs) => mark(cons(0, incr(odds))) 
    active(odds) => mark(incr(pairs)) 
    active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) 
    mark(nats) => active(nats) 
    mark(cons(X, Y)) => active(cons(mark(X), Y)) 
    mark(0) => active(0) 
    mark(incr(X)) => active(incr(mark(X))) 
    mark(pairs) => active(pairs) 
    mark(odds) => active(odds) 
    mark(s(X)) => active(s(mark(X))) 
    mark(head(X)) => active(head(mark(X))) 
    mark(tail(X)) => active(tail(mark(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) 
    incr(mark(X)) => incr(X) 
    incr(active(X)) => incr(X) 
    s(mark(X)) => s(X) 
    s(active(X)) => s(X) 
    head(mark(X)) => head(X) 
    head(active(X)) => head(X) 
    tail(mark(X)) => tail(X) 
    tail(active(X)) => tail(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 :  
    * 2 :  
    * 3 : 13, 14, 15 
    * 4 :  
    * 5 :  
    * 6 : 17, 18, 19 
    * 7 :  
    * 8 : 13, 14, 15 
    * 9 :  
    * 10 : 37, 38 
    * 11 : 35, 36 
    * 12 : 0, 1, 2 
    * 13 :  
    * 14 : 31, 32, 33, 34 
    * 15 : 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 
    * 16 :  
    * 17 : 8, 9, 10, 11 
    * 18 : 35, 36 
    * 19 : 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 
    * 20 : 3, 4, 5 
    * 21 : 6, 7 
    * 22 :  
    * 23 : 37, 38 
    * 24 : 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 
    * 25 :  
    * 26 : 39, 40 
    * 27 : 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 
    * 28 :  
    * 29 : 41, 42 
    * 30 : 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 
    * 31 : 31, 32, 33, 34 
    * 32 : 31, 32, 33, 34 
    * 33 : 31, 32, 33, 34 
    * 34 : 31, 32, 33, 34 
    * 35 : 35, 36 
    * 36 : 35, 36 
    * 37 : 37, 38 
    * 38 : 37, 38 
    * 39 : 39, 40 
    * 40 : 39, 40 
    * 41 : 41, 42 
    * 42 : 41, 42 

This graph has the following strongly connected components:

  P_1:

    active#(nats) =#> mark#(cons(0, incr(nats)))   
    active#(pairs) =#> mark#(cons(0, incr(odds)))   
    active#(odds) =#> mark#(incr(pairs))   
    active#(incr(cons(X, Y))) =#> mark#(cons(s(X), incr(Y)))   
    mark#(nats) =#> active#(nats)   
    mark#(cons(X, Y)) =#> mark#(X)   
    mark#(incr(X)) =#> active#(incr(mark(X)))   
    mark#(incr(X)) =#> mark#(X)   
    mark#(pairs) =#> active#(pairs)   
    mark#(odds) =#> active#(odds)   
    mark#(s(X)) =#> mark#(X)   
    mark#(head(X)) =#> mark#(X)   
    mark#(tail(X)) =#> mark#(X)   

  P_2:

    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_3:

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

  P_4:

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

  P_5:

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

  P_6:

    tail#(mark(X)) =#> tail#(X)   
    tail#(active(X)) =#> tail#(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(tail#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(tail#(mark(X))) = mark(X) |> X = nu(tail#(X)) 
  nu(tail#(active(X))) = active(X) |> X = nu(tail#(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(head#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(head#(mark(X))) = mark(X) |> X = nu(head#(X)) 
  nu(head#(active(X))) = active(X) |> X = nu(head#(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(incr#) = 1 

Thus, we can orient the dependency pairs as follows:

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

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(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_2, 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) 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 (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#(nats) >? mark#(cons(0, incr(nats))) 
  active#(pairs) >? mark#(cons(0, incr(odds))) 
  active#(odds) >? mark#(incr(pairs)) 
  active#(incr(cons(X, Y))) >? mark#(cons(s(X), incr(Y))) 
  mark#(nats) >? active#(nats) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(incr(X)) >? active#(incr(mark(X))) 
  mark#(incr(X)) >? mark#(X) 
  mark#(pairs) >? active#(pairs) 
  mark#(odds) >? active#(odds) 
  mark#(s(X)) >? mark#(X) 
  mark#(head(X)) >? mark#(X) 
  mark#(tail(X)) >? mark#(X) 
  active(nats) >= mark(cons(0, incr(nats))) 
  active(pairs) >= mark(cons(0, incr(odds))) 
  active(odds) >= mark(incr(pairs)) 
  active(incr(cons(X, Y))) >= mark(cons(s(X), incr(Y))) 
  mark(nats) >= active(nats) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(0) >= active(0) 
  mark(incr(X)) >= active(incr(mark(X))) 
  mark(pairs) >= active(pairs) 
  mark(odds) >= active(odds) 
  mark(s(X)) >= active(s(mark(X))) 
  mark(head(X)) >= active(head(mark(X))) 
  mark(tail(X)) >= active(tail(mark(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) 
  incr(mark(X)) >= incr(X) 
  incr(active(X)) >= incr(X) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(X) 
  head(mark(X)) >= head(X) 
  head(active(X)) >= head(X) 
  tail(mark(X)) >= tail(X) 
  tail(active(X)) >= tail(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.y0 
  cons = \y0y1.2y0 
  head = \y0.2 + y0 
  incr = \y0.2y0 
  mark = \y0.2y0 
  mark# = \y0.2y0 
  nats = 1 
  odds = 0 
  pairs = 0 
  s = \y0.y0 
  tail = \y0.2 + 2y0 

Using this interpretation, the requirements translate to:

  [[active#(nats)]] = 1 > 0 = [[mark#(cons(0, incr(nats)))]] 
  [[active#(pairs)]] = 0 >= 0 = [[mark#(cons(0, incr(odds)))]] 
  [[active#(odds)]] = 0 >= 0 = [[mark#(incr(pairs))]] 
  [[active#(incr(cons(_x0, _x1)))]] = 4x0 >= 4x0 = [[mark#(cons(s(_x0), incr(_x1)))]] 
  [[mark#(nats)]] = 2 > 1 = [[active#(nats)]] 
  [[mark#(cons(_x0, _x1))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(incr(_x0))]] = 4x0 >= 4x0 = [[active#(incr(mark(_x0)))]] 
  [[mark#(incr(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(pairs)]] = 0 >= 0 = [[active#(pairs)]] 
  [[mark#(odds)]] = 0 >= 0 = [[active#(odds)]] 
  [[mark#(s(_x0))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(head(_x0))]] = 4 + 2x0 > 2x0 = [[mark#(_x0)]] 
  [[mark#(tail(_x0))]] = 4 + 4x0 > 2x0 = [[mark#(_x0)]] 
  [[active(nats)]] = 1 >= 0 = [[mark(cons(0, incr(nats)))]] 
  [[active(pairs)]] = 0 >= 0 = [[mark(cons(0, incr(odds)))]] 
  [[active(odds)]] = 0 >= 0 = [[mark(incr(pairs))]] 
  [[active(incr(cons(_x0, _x1)))]] = 4x0 >= 4x0 = [[mark(cons(s(_x0), incr(_x1)))]] 
  [[mark(nats)]] = 2 >= 1 = [[active(nats)]] 
  [[mark(cons(_x0, _x1))]] = 4x0 >= 4x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[active(incr(mark(_x0)))]] 
  [[mark(pairs)]] = 0 >= 0 = [[active(pairs)]] 
  [[mark(odds)]] = 0 >= 0 = [[active(odds)]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[active(s(mark(_x0)))]] 
  [[mark(head(_x0))]] = 4 + 2x0 >= 2 + 2x0 = [[active(head(mark(_x0)))]] 
  [[mark(tail(_x0))]] = 4 + 4x0 >= 2 + 4x0 = [[active(tail(mark(_x0)))]] 
  [[cons(mark(_x0), _x1)]] = 4x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[incr(mark(_x0))]] = 4x0 >= 2x0 = [[incr(_x0)]] 
  [[incr(active(_x0))]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[s(mark(_x0))]] = 2x0 >= x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[head(mark(_x0))]] = 2 + 2x0 >= 2 + x0 = [[head(_x0)]] 
  [[head(active(_x0))]] = 2 + x0 >= 2 + x0 = [[head(_x0)]] 
  [[tail(mark(_x0))]] = 2 + 4x0 >= 2 + 2x0 = [[tail(_x0)]] 
  [[tail(active(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[tail(_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#(pairs) =#> mark#(cons(0, incr(odds)))   
  active#(odds) =#> mark#(incr(pairs))   
  active#(incr(cons(X, Y))) =#> mark#(cons(s(X), incr(Y)))   
  mark#(cons(X, Y)) =#> mark#(X)   
  mark#(incr(X)) =#> active#(incr(mark(X)))   
  mark#(incr(X)) =#> mark#(X)   
  mark#(pairs) =#> active#(pairs)   
  mark#(odds) =#> active#(odds)   
  mark#(s(X)) =#> mark#(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).

The formative rules of (P_8, R_0) are R_1 ::=

  active(nats) => mark(cons(0, incr(nats))) 
  active(pairs) => mark(cons(0, incr(odds))) 
  active(odds) => mark(incr(pairs)) 
  active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) 
  mark(nats) => active(nats) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(0) => active(0) 
  mark(incr(X)) => active(incr(mark(X))) 
  mark(pairs) => active(pairs) 
  mark(odds) => active(odds) 
  mark(s(X)) => active(s(mark(X))) 
  mark(head(X)) => active(head(mark(X))) 
  mark(tail(X)) => active(tail(mark(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) 
  incr(mark(X)) => incr(X) 
  incr(active(X)) => incr(X) 
  s(mark(X)) => s(X) 
  s(active(X)) => s(X) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_8, R_0, minimal, formative) by (P_8, R_1, minimal, formative).

Thus, the original system is terminating if (P_8, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_8, R_1, 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#(pairs) >? mark#(cons(0, incr(odds))) 
  active#(odds) >? mark#(incr(pairs)) 
  active#(incr(cons(X, Y))) >? mark#(cons(s(X), incr(Y))) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(incr(X)) >? active#(incr(mark(X))) 
  mark#(incr(X)) >? mark#(X) 
  mark#(pairs) >? active#(pairs) 
  mark#(odds) >? active#(odds) 
  mark#(s(X)) >? mark#(X) 
  active(nats) >= mark(cons(0, incr(nats))) 
  active(pairs) >= mark(cons(0, incr(odds))) 
  active(odds) >= mark(incr(pairs)) 
  active(incr(cons(X, Y))) >= mark(cons(s(X), incr(Y))) 
  mark(nats) >= active(nats) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(0) >= active(0) 
  mark(incr(X)) >= active(incr(mark(X))) 
  mark(pairs) >= active(pairs) 
  mark(odds) >= active(odds) 
  mark(s(X)) >= active(s(mark(X))) 
  mark(head(X)) >= active(head(mark(X))) 
  mark(tail(X)) >= active(tail(mark(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) 
  incr(mark(X)) >= incr(X) 
  incr(active(X)) >= incr(X) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(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.y0 
  cons = \y0y1.2y0 
  head = \y0.0 
  incr = \y0.2y0 
  mark = \y0.2y0 
  mark# = \y0.2y0 
  nats = 0 
  odds = 2 
  pairs = 0 
  s = \y0.y0 
  tail = \y0.0 

Using this interpretation, the requirements translate to:

  [[active#(pairs)]] = 0 >= 0 = [[mark#(cons(0, incr(odds)))]] 
  [[active#(odds)]] = 2 > 0 = [[mark#(incr(pairs))]] 
  [[active#(incr(cons(_x0, _x1)))]] = 4x0 >= 4x0 = [[mark#(cons(s(_x0), incr(_x1)))]] 
  [[mark#(cons(_x0, _x1))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(incr(_x0))]] = 4x0 >= 4x0 = [[active#(incr(mark(_x0)))]] 
  [[mark#(incr(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(pairs)]] = 0 >= 0 = [[active#(pairs)]] 
  [[mark#(odds)]] = 4 > 2 = [[active#(odds)]] 
  [[mark#(s(_x0))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[active(nats)]] = 0 >= 0 = [[mark(cons(0, incr(nats)))]] 
  [[active(pairs)]] = 0 >= 0 = [[mark(cons(0, incr(odds)))]] 
  [[active(odds)]] = 2 >= 0 = [[mark(incr(pairs))]] 
  [[active(incr(cons(_x0, _x1)))]] = 4x0 >= 4x0 = [[mark(cons(s(_x0), incr(_x1)))]] 
  [[mark(nats)]] = 0 >= 0 = [[active(nats)]] 
  [[mark(cons(_x0, _x1))]] = 4x0 >= 4x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[active(incr(mark(_x0)))]] 
  [[mark(pairs)]] = 0 >= 0 = [[active(pairs)]] 
  [[mark(odds)]] = 4 >= 2 = [[active(odds)]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[active(s(mark(_x0)))]] 
  [[mark(head(_x0))]] = 0 >= 0 = [[active(head(mark(_x0)))]] 
  [[mark(tail(_x0))]] = 0 >= 0 = [[active(tail(mark(_x0)))]] 
  [[cons(mark(_x0), _x1)]] = 4x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[incr(mark(_x0))]] = 4x0 >= 2x0 = [[incr(_x0)]] 
  [[incr(active(_x0))]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[s(mark(_x0))]] = 2x0 >= x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = x0 >= x0 = [[s(_x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_8, R_1, minimal, formative) by (P_9, R_1, minimal, formative), where P_9 consists of:

  active#(pairs) =#> mark#(cons(0, incr(odds)))   
  active#(incr(cons(X, Y))) =#> mark#(cons(s(X), incr(Y)))   
  mark#(cons(X, Y)) =#> mark#(X)   
  mark#(incr(X)) =#> active#(incr(mark(X)))   
  mark#(incr(X)) =#> mark#(X)   
  mark#(pairs) =#> active#(pairs)   
  mark#(s(X)) =#> mark#(X)   

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

  active#(pairs) >? mark#(cons(0, incr(odds))) 
  active#(incr(cons(X, Y))) >? mark#(cons(s(X), incr(Y))) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(incr(X)) >? active#(incr(mark(X))) 
  mark#(incr(X)) >? mark#(X) 
  mark#(pairs) >? active#(pairs) 
  mark#(s(X)) >? mark#(X) 
  active(nats) >= mark(cons(0, incr(nats))) 
  active(pairs) >= mark(cons(0, incr(odds))) 
  active(odds) >= mark(incr(pairs)) 
  active(incr(cons(X, Y))) >= mark(cons(s(X), incr(Y))) 
  mark(nats) >= active(nats) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(0) >= active(0) 
  mark(incr(X)) >= active(incr(mark(X))) 
  mark(pairs) >= active(pairs) 
  mark(odds) >= active(odds) 
  mark(s(X)) >= active(s(mark(X))) 
  mark(head(X)) >= active(head(mark(X))) 
  mark(tail(X)) >= active(tail(mark(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) 
  incr(mark(X)) >= incr(X) 
  incr(active(X)) >= incr(X) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(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 
  head = \y0.0 
  incr = \y0.2y0 
  mark = \y0.y0 
  mark# = \y0.2y0 
  nats = 0 
  odds = 2 
  pairs = 1 
  s = \y0.y0 
  tail = \y0.0 

Using this interpretation, the requirements translate to:

  [[active#(pairs)]] = 2 > 0 = [[mark#(cons(0, incr(odds)))]] 
  [[active#(incr(cons(_x0, _x1)))]] = 4x0 >= 2x0 = [[mark#(cons(s(_x0), incr(_x1)))]] 
  [[mark#(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(incr(_x0))]] = 4x0 >= 4x0 = [[active#(incr(mark(_x0)))]] 
  [[mark#(incr(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(pairs)]] = 2 >= 2 = [[active#(pairs)]] 
  [[mark#(s(_x0))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[active(nats)]] = 0 >= 0 = [[mark(cons(0, incr(nats)))]] 
  [[active(pairs)]] = 1 >= 0 = [[mark(cons(0, incr(odds)))]] 
  [[active(odds)]] = 2 >= 2 = [[mark(incr(pairs))]] 
  [[active(incr(cons(_x0, _x1)))]] = 2x0 >= x0 = [[mark(cons(s(_x0), incr(_x1)))]] 
  [[mark(nats)]] = 0 >= 0 = [[active(nats)]] 
  [[mark(cons(_x0, _x1))]] = x0 >= x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(incr(_x0))]] = 2x0 >= 2x0 = [[active(incr(mark(_x0)))]] 
  [[mark(pairs)]] = 1 >= 1 = [[active(pairs)]] 
  [[mark(odds)]] = 2 >= 2 = [[active(odds)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[active(s(mark(_x0)))]] 
  [[mark(head(_x0))]] = 0 >= 0 = [[active(head(mark(_x0)))]] 
  [[mark(tail(_x0))]] = 0 >= 0 = [[active(tail(mark(_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)]] 
  [[incr(mark(_x0))]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[incr(active(_x0))]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[s(mark(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = x0 >= x0 = [[s(_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#(incr(cons(X, Y))) =#> mark#(cons(s(X), incr(Y)))   
  mark#(cons(X, Y)) =#> mark#(X)   
  mark#(incr(X)) =#> active#(incr(mark(X)))   
  mark#(incr(X)) =#> mark#(X)   
  mark#(pairs) =#> active#(pairs)   
  mark#(s(X)) =#> mark#(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 : 1 
    * 1 : 1, 2, 3, 4, 5 
    * 2 : 0 
    * 3 : 1, 2, 3, 4, 5 
    * 4 :  
    * 5 : 1, 2, 3, 4, 5 

This graph has the following strongly connected components:

  P_11:

    active#(incr(cons(X, Y))) =#> mark#(cons(s(X), incr(Y)))   
    mark#(cons(X, Y)) =#> mark#(X)   
    mark#(incr(X)) =#> active#(incr(mark(X)))   
    mark#(incr(X)) =#> mark#(X)   
    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 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#(incr(cons(X, Y))) >? mark#(cons(s(X), incr(Y))) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(incr(X)) >? active#(incr(mark(X))) 
  mark#(incr(X)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  active(nats) >= mark(cons(0, incr(nats))) 
  active(pairs) >= mark(cons(0, incr(odds))) 
  active(odds) >= mark(incr(pairs)) 
  active(incr(cons(X, Y))) >= mark(cons(s(X), incr(Y))) 
  mark(nats) >= active(nats) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(0) >= active(0) 
  mark(incr(X)) >= active(incr(mark(X))) 
  mark(pairs) >= active(pairs) 
  mark(odds) >= active(odds) 
  mark(s(X)) >= active(s(mark(X))) 
  mark(head(X)) >= active(head(mark(X))) 
  mark(tail(X)) >= active(tail(mark(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) 
  incr(mark(X)) >= incr(X) 
  incr(active(X)) >= incr(X) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(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.y0 
  cons = \y0y1.2y0 
  head = \y0.0 
  incr = \y0.2 + y0 
  mark = \y0.y0 
  mark# = \y0.1 + y0 
  nats = 0 
  odds = 2 
  pairs = 0 
  s = \y0.y0 
  tail = \y0.0 

Using this interpretation, the requirements translate to:

  [[active#(incr(cons(_x0, _x1)))]] = 2 + 2x0 > 1 + 2x0 = [[mark#(cons(s(_x0), incr(_x1)))]] 
  [[mark#(cons(_x0, _x1))]] = 1 + 2x0 >= 1 + x0 = [[mark#(_x0)]] 
  [[mark#(incr(_x0))]] = 3 + x0 > 2 + x0 = [[active#(incr(mark(_x0)))]] 
  [[mark#(incr(_x0))]] = 3 + x0 > 1 + x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = 1 + x0 >= 1 + x0 = [[mark#(_x0)]] 
  [[active(nats)]] = 0 >= 0 = [[mark(cons(0, incr(nats)))]] 
  [[active(pairs)]] = 0 >= 0 = [[mark(cons(0, incr(odds)))]] 
  [[active(odds)]] = 2 >= 2 = [[mark(incr(pairs))]] 
  [[active(incr(cons(_x0, _x1)))]] = 2 + 2x0 >= 2x0 = [[mark(cons(s(_x0), incr(_x1)))]] 
  [[mark(nats)]] = 0 >= 0 = [[active(nats)]] 
  [[mark(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(incr(_x0))]] = 2 + x0 >= 2 + x0 = [[active(incr(mark(_x0)))]] 
  [[mark(pairs)]] = 0 >= 0 = [[active(pairs)]] 
  [[mark(odds)]] = 2 >= 2 = [[active(odds)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[active(s(mark(_x0)))]] 
  [[mark(head(_x0))]] = 0 >= 0 = [[active(head(mark(_x0)))]] 
  [[mark(tail(_x0))]] = 0 >= 0 = [[active(tail(mark(_x0)))]] 
  [[cons(mark(_x0), _x1)]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, mark(_x1))]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(active(_x0), _x1)]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[cons(_x0, active(_x1))]] = 2x0 >= 2x0 = [[cons(_x0, _x1)]] 
  [[incr(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[incr(_x0)]] 
  [[incr(active(_x0))]] = 2 + x0 >= 2 + x0 = [[incr(_x0)]] 
  [[s(mark(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[s(active(_x0))]] = x0 >= x0 = [[s(_x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_11, R_1, minimal, formative) by (P_12, R_1, minimal, formative), where P_12 consists of:

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

Thus, the original system is terminating if (P_12, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_12, 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#(cons(X, Y))) = cons(X, Y) |> X = nu(mark#(X)) 
  nu(mark#(s(X))) = s(X) |> X = nu(mark#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_12, 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.