/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  active : [o] --> o 
  add : [o * o] --> o 
  cons : [o * o] --> o 
  dbl : [o] --> o 
  first : [o * o] --> o 
  mark : [o] --> o 
  nil : [] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  recip : [o] --> o 
  s : [o] --> o 
  sqr : [o] --> o 
  terms : [o] --> o 
  top : [o] --> o 

  active(terms(X)) => mark(cons(recip(sqr(X)), terms(s(X)))) 
  active(sqr(0)) => mark(0) 
  active(sqr(s(X))) => mark(s(add(sqr(X), dbl(X)))) 
  active(dbl(0)) => mark(0) 
  active(dbl(s(X))) => mark(s(s(dbl(X)))) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(first(0, X)) => mark(nil) 
  active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) 
  active(terms(X)) => terms(active(X)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(recip(X)) => recip(active(X)) 
  active(sqr(X)) => sqr(active(X)) 
  active(s(X)) => s(active(X)) 
  active(add(X, Y)) => add(active(X), Y) 
  active(add(X, Y)) => add(X, active(Y)) 
  active(dbl(X)) => dbl(active(X)) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  terms(mark(X)) => mark(terms(X)) 
  cons(mark(X), Y) => mark(cons(X, Y)) 
  recip(mark(X)) => mark(recip(X)) 
  sqr(mark(X)) => mark(sqr(X)) 
  s(mark(X)) => mark(s(X)) 
  add(mark(X), Y) => mark(add(X, Y)) 
  add(X, mark(Y)) => mark(add(X, Y)) 
  dbl(mark(X)) => mark(dbl(X)) 
  first(mark(X), Y) => mark(first(X, Y)) 
  first(X, mark(Y)) => mark(first(X, Y)) 
  proper(terms(X)) => terms(proper(X)) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(recip(X)) => recip(proper(X)) 
  proper(sqr(X)) => sqr(proper(X)) 
  proper(s(X)) => s(proper(X)) 
  proper(0) => ok(0) 
  proper(add(X, Y)) => add(proper(X), proper(Y)) 
  proper(dbl(X)) => dbl(proper(X)) 
  proper(first(X, Y)) => first(proper(X), proper(Y)) 
  proper(nil) => ok(nil) 
  terms(ok(X)) => ok(terms(X)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  recip(ok(X)) => ok(recip(X)) 
  sqr(ok(X)) => ok(sqr(X)) 
  s(ok(X)) => ok(s(X)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  dbl(ok(X)) => ok(dbl(X)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 
  top(mark(X)) => top(proper(X)) 
  top(ok(X)) => top(active(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#(terms(X)) =#> cons#(recip(sqr(X)), terms(s(X)))   
    1] active#(terms(X)) =#> recip#(sqr(X))   
    2] active#(terms(X)) =#> sqr#(X)   
    3] active#(terms(X)) =#> terms#(s(X))   
    4] active#(terms(X)) =#> s#(X)   
    5] active#(sqr(s(X))) =#> s#(add(sqr(X), dbl(X)))   
    6] active#(sqr(s(X))) =#> add#(sqr(X), dbl(X))   
    7] active#(sqr(s(X))) =#> sqr#(X)   
    8] active#(sqr(s(X))) =#> dbl#(X)   
    9] active#(dbl(s(X))) =#> s#(s(dbl(X)))   
    10] active#(dbl(s(X))) =#> s#(dbl(X))   
    11] active#(dbl(s(X))) =#> dbl#(X)   
    12] active#(add(s(X), Y)) =#> s#(add(X, Y))   
    13] active#(add(s(X), Y)) =#> add#(X, Y)   
    14] active#(first(s(X), cons(Y, Z))) =#> cons#(Y, first(X, Z))   
    15] active#(first(s(X), cons(Y, Z))) =#> first#(X, Z)   
    16] active#(terms(X)) =#> terms#(active(X))   
    17] active#(terms(X)) =#> active#(X)   
    18] active#(cons(X, Y)) =#> cons#(active(X), Y)   
    19] active#(cons(X, Y)) =#> active#(X)   
    20] active#(recip(X)) =#> recip#(active(X))   
    21] active#(recip(X)) =#> active#(X)   
    22] active#(sqr(X)) =#> sqr#(active(X))   
    23] active#(sqr(X)) =#> active#(X)   
    24] active#(s(X)) =#> s#(active(X))   
    25] active#(s(X)) =#> active#(X)   
    26] active#(add(X, Y)) =#> add#(active(X), Y)   
    27] active#(add(X, Y)) =#> active#(X)   
    28] active#(add(X, Y)) =#> add#(X, active(Y))   
    29] active#(add(X, Y)) =#> active#(Y)   
    30] active#(dbl(X)) =#> dbl#(active(X))   
    31] active#(dbl(X)) =#> active#(X)   
    32] active#(first(X, Y)) =#> first#(active(X), Y)   
    33] active#(first(X, Y)) =#> active#(X)   
    34] active#(first(X, Y)) =#> first#(X, active(Y))   
    35] active#(first(X, Y)) =#> active#(Y)   
    36] terms#(mark(X)) =#> terms#(X)   
    37] cons#(mark(X), Y) =#> cons#(X, Y)   
    38] recip#(mark(X)) =#> recip#(X)   
    39] sqr#(mark(X)) =#> sqr#(X)   
    40] s#(mark(X)) =#> s#(X)   
    41] add#(mark(X), Y) =#> add#(X, Y)   
    42] add#(X, mark(Y)) =#> add#(X, Y)   
    43] dbl#(mark(X)) =#> dbl#(X)   
    44] first#(mark(X), Y) =#> first#(X, Y)   
    45] first#(X, mark(Y)) =#> first#(X, Y)   
    46] proper#(terms(X)) =#> terms#(proper(X))   
    47] proper#(terms(X)) =#> proper#(X)   
    48] proper#(cons(X, Y)) =#> cons#(proper(X), proper(Y))   
    49] proper#(cons(X, Y)) =#> proper#(X)   
    50] proper#(cons(X, Y)) =#> proper#(Y)   
    51] proper#(recip(X)) =#> recip#(proper(X))   
    52] proper#(recip(X)) =#> proper#(X)   
    53] proper#(sqr(X)) =#> sqr#(proper(X))   
    54] proper#(sqr(X)) =#> proper#(X)   
    55] proper#(s(X)) =#> s#(proper(X))   
    56] proper#(s(X)) =#> proper#(X)   
    57] proper#(add(X, Y)) =#> add#(proper(X), proper(Y))   
    58] proper#(add(X, Y)) =#> proper#(X)   
    59] proper#(add(X, Y)) =#> proper#(Y)   
    60] proper#(dbl(X)) =#> dbl#(proper(X))   
    61] proper#(dbl(X)) =#> proper#(X)   
    62] proper#(first(X, Y)) =#> first#(proper(X), proper(Y))   
    63] proper#(first(X, Y)) =#> proper#(X)   
    64] proper#(first(X, Y)) =#> proper#(Y)   
    65] terms#(ok(X)) =#> terms#(X)   
    66] cons#(ok(X), ok(Y)) =#> cons#(X, Y)   
    67] recip#(ok(X)) =#> recip#(X)   
    68] sqr#(ok(X)) =#> sqr#(X)   
    69] s#(ok(X)) =#> s#(X)   
    70] add#(ok(X), ok(Y)) =#> add#(X, Y)   
    71] dbl#(ok(X)) =#> dbl#(X)   
    72] first#(ok(X), ok(Y)) =#> first#(X, Y)   
    73] top#(mark(X)) =#> top#(proper(X))   
    74] top#(mark(X)) =#> proper#(X)   
    75] top#(ok(X)) =#> top#(active(X))   
    76] top#(ok(X)) =#> active#(X)   

  Rules R_0:

    active(terms(X)) => mark(cons(recip(sqr(X)), terms(s(X)))) 
    active(sqr(0)) => mark(0) 
    active(sqr(s(X))) => mark(s(add(sqr(X), dbl(X)))) 
    active(dbl(0)) => mark(0) 
    active(dbl(s(X))) => mark(s(s(dbl(X)))) 
    active(add(0, X)) => mark(X) 
    active(add(s(X), Y)) => mark(s(add(X, Y))) 
    active(first(0, X)) => mark(nil) 
    active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) 
    active(terms(X)) => terms(active(X)) 
    active(cons(X, Y)) => cons(active(X), Y) 
    active(recip(X)) => recip(active(X)) 
    active(sqr(X)) => sqr(active(X)) 
    active(s(X)) => s(active(X)) 
    active(add(X, Y)) => add(active(X), Y) 
    active(add(X, Y)) => add(X, active(Y)) 
    active(dbl(X)) => dbl(active(X)) 
    active(first(X, Y)) => first(active(X), Y) 
    active(first(X, Y)) => first(X, active(Y)) 
    terms(mark(X)) => mark(terms(X)) 
    cons(mark(X), Y) => mark(cons(X, Y)) 
    recip(mark(X)) => mark(recip(X)) 
    sqr(mark(X)) => mark(sqr(X)) 
    s(mark(X)) => mark(s(X)) 
    add(mark(X), Y) => mark(add(X, Y)) 
    add(X, mark(Y)) => mark(add(X, Y)) 
    dbl(mark(X)) => mark(dbl(X)) 
    first(mark(X), Y) => mark(first(X, Y)) 
    first(X, mark(Y)) => mark(first(X, Y)) 
    proper(terms(X)) => terms(proper(X)) 
    proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
    proper(recip(X)) => recip(proper(X)) 
    proper(sqr(X)) => sqr(proper(X)) 
    proper(s(X)) => s(proper(X)) 
    proper(0) => ok(0) 
    proper(add(X, Y)) => add(proper(X), proper(Y)) 
    proper(dbl(X)) => dbl(proper(X)) 
    proper(first(X, Y)) => first(proper(X), proper(Y)) 
    proper(nil) => ok(nil) 
    terms(ok(X)) => ok(terms(X)) 
    cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
    recip(ok(X)) => ok(recip(X)) 
    sqr(ok(X)) => ok(sqr(X)) 
    s(ok(X)) => ok(s(X)) 
    add(ok(X), ok(Y)) => ok(add(X, Y)) 
    dbl(ok(X)) => ok(dbl(X)) 
    first(ok(X), ok(Y)) => ok(first(X, Y)) 
    top(mark(X)) => top(proper(X)) 
    top(ok(X)) => top(active(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 : 37, 66 
    * 1 : 38, 67 
    * 2 : 39, 68 
    * 3 : 36, 65 
    * 4 : 40, 69 
    * 5 : 40, 69 
    * 6 : 41, 42, 70 
    * 7 : 39, 68 
    * 8 : 43, 71 
    * 9 : 40, 69 
    * 10 : 40, 69 
    * 11 : 43, 71 
    * 12 : 40, 69 
    * 13 : 41, 42, 70 
    * 14 : 37, 66 
    * 15 : 44, 45, 72 
    * 16 : 36, 65 
    * 17 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 18 : 37, 66 
    * 19 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 20 : 38, 67 
    * 21 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 22 : 39, 68 
    * 23 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 24 : 40, 69 
    * 25 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 26 : 41, 42, 70 
    * 27 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 28 : 41, 42, 70 
    * 29 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 30 : 43, 71 
    * 31 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 32 : 44, 45, 72 
    * 33 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 34 : 44, 45, 72 
    * 35 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 
    * 36 : 36, 65 
    * 37 : 37, 66 
    * 38 : 38, 67 
    * 39 : 39, 68 
    * 40 : 40, 69 
    * 41 : 41, 42, 70 
    * 42 : 41, 42, 70 
    * 43 : 43, 71 
    * 44 : 44, 45, 72 
    * 45 : 44, 45, 72 
    * 46 : 36, 65 
    * 47 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 48 : 37, 66 
    * 49 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 50 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 51 : 38, 67 
    * 52 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 53 : 39, 68 
    * 54 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 55 : 40, 69 
    * 56 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 57 : 41, 42, 70 
    * 58 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 59 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 60 : 43, 71 
    * 61 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 62 : 44, 45, 72 
    * 63 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 64 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 65 : 36, 65 
    * 66 : 37, 66 
    * 67 : 38, 67 
    * 68 : 39, 68 
    * 69 : 40, 69 
    * 70 : 41, 42, 70 
    * 71 : 43, 71 
    * 72 : 44, 45, 72 
    * 73 : 73, 74, 75, 76 
    * 74 : 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64 
    * 75 : 73, 74, 75, 76 
    * 76 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 

This graph has the following strongly connected components:

  P_1:

    active#(terms(X)) =#> active#(X)   
    active#(cons(X, Y)) =#> active#(X)   
    active#(recip(X)) =#> active#(X)   
    active#(sqr(X)) =#> active#(X)   
    active#(s(X)) =#> active#(X)   
    active#(add(X, Y)) =#> active#(X)   
    active#(add(X, Y)) =#> active#(Y)   
    active#(dbl(X)) =#> active#(X)   
    active#(first(X, Y)) =#> active#(X)   
    active#(first(X, Y)) =#> active#(Y)   

  P_2:

    terms#(mark(X)) =#> terms#(X)   
    terms#(ok(X)) =#> terms#(X)   

  P_3:

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

  P_4:

    recip#(mark(X)) =#> recip#(X)   
    recip#(ok(X)) =#> recip#(X)   

  P_5:

    sqr#(mark(X)) =#> sqr#(X)   
    sqr#(ok(X)) =#> sqr#(X)   

  P_6:

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

  P_7:

    add#(mark(X), Y) =#> add#(X, Y)   
    add#(X, mark(Y)) =#> add#(X, Y)   
    add#(ok(X), ok(Y)) =#> add#(X, Y)   

  P_8:

    dbl#(mark(X)) =#> dbl#(X)   
    dbl#(ok(X)) =#> dbl#(X)   

  P_9:

    first#(mark(X), Y) =#> first#(X, Y)   
    first#(X, mark(Y)) =#> first#(X, Y)   
    first#(ok(X), ok(Y)) =#> first#(X, Y)   

  P_10:

    proper#(terms(X)) =#> proper#(X)   
    proper#(cons(X, Y)) =#> proper#(X)   
    proper#(cons(X, Y)) =#> proper#(Y)   
    proper#(recip(X)) =#> proper#(X)   
    proper#(sqr(X)) =#> proper#(X)   
    proper#(s(X)) =#> proper#(X)   
    proper#(add(X, Y)) =#> proper#(X)   
    proper#(add(X, Y)) =#> proper#(Y)   
    proper#(dbl(X)) =#> proper#(X)   
    proper#(first(X, Y)) =#> proper#(X)   
    proper#(first(X, Y)) =#> proper#(Y)   

  P_11:

    top#(mark(X)) =#> top#(proper(X))   
    top#(ok(X)) =#> top#(active(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), (P_6, R_0, m, f), (P_7, R_0, m, f), (P_8, R_0, m, f), (P_9, R_0, m, f), (P_10, R_0, m, f) and (P_11, 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), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative), (P_10, R_0, minimal, formative) and (P_11, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_11, R_0, minimal, formative).

The formative rules of (P_11, R_0) are R_1 ::=

  active(terms(X)) => mark(cons(recip(sqr(X)), terms(s(X)))) 
  active(sqr(0)) => mark(0) 
  active(sqr(s(X))) => mark(s(add(sqr(X), dbl(X)))) 
  active(dbl(0)) => mark(0) 
  active(dbl(s(X))) => mark(s(s(dbl(X)))) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(first(0, X)) => mark(nil) 
  active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) 
  active(terms(X)) => terms(active(X)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(recip(X)) => recip(active(X)) 
  active(sqr(X)) => sqr(active(X)) 
  active(s(X)) => s(active(X)) 
  active(add(X, Y)) => add(active(X), Y) 
  active(add(X, Y)) => add(X, active(Y)) 
  active(dbl(X)) => dbl(active(X)) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  terms(mark(X)) => mark(terms(X)) 
  cons(mark(X), Y) => mark(cons(X, Y)) 
  recip(mark(X)) => mark(recip(X)) 
  sqr(mark(X)) => mark(sqr(X)) 
  s(mark(X)) => mark(s(X)) 
  add(mark(X), Y) => mark(add(X, Y)) 
  add(X, mark(Y)) => mark(add(X, Y)) 
  dbl(mark(X)) => mark(dbl(X)) 
  first(mark(X), Y) => mark(first(X, Y)) 
  first(X, mark(Y)) => mark(first(X, Y)) 
  proper(terms(X)) => terms(proper(X)) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(recip(X)) => recip(proper(X)) 
  proper(sqr(X)) => sqr(proper(X)) 
  proper(s(X)) => s(proper(X)) 
  proper(0) => ok(0) 
  proper(add(X, Y)) => add(proper(X), proper(Y)) 
  proper(dbl(X)) => dbl(proper(X)) 
  proper(first(X, Y)) => first(proper(X), proper(Y)) 
  proper(nil) => ok(nil) 
  terms(ok(X)) => ok(terms(X)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  recip(ok(X)) => ok(recip(X)) 
  sqr(ok(X)) => ok(sqr(X)) 
  s(ok(X)) => ok(s(X)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  dbl(ok(X)) => ok(dbl(X)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_11, R_0, minimal, formative) by (P_11, R_1, minimal, formative).

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), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative), (P_10, R_0, minimal, formative) and (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:

  top#(mark(X)) >? top#(proper(X)) 
  top#(ok(X)) >? top#(active(X)) 
  active(terms(X)) >= mark(cons(recip(sqr(X)), terms(s(X)))) 
  active(sqr(0)) >= mark(0) 
  active(sqr(s(X))) >= mark(s(add(sqr(X), dbl(X)))) 
  active(dbl(0)) >= mark(0) 
  active(dbl(s(X))) >= mark(s(s(dbl(X)))) 
  active(add(0, X)) >= mark(X) 
  active(add(s(X), Y)) >= mark(s(add(X, Y))) 
  active(first(0, X)) >= mark(nil) 
  active(first(s(X), cons(Y, Z))) >= mark(cons(Y, first(X, Z))) 
  active(terms(X)) >= terms(active(X)) 
  active(cons(X, Y)) >= cons(active(X), Y) 
  active(recip(X)) >= recip(active(X)) 
  active(sqr(X)) >= sqr(active(X)) 
  active(s(X)) >= s(active(X)) 
  active(add(X, Y)) >= add(active(X), Y) 
  active(add(X, Y)) >= add(X, active(Y)) 
  active(dbl(X)) >= dbl(active(X)) 
  active(first(X, Y)) >= first(active(X), Y) 
  active(first(X, Y)) >= first(X, active(Y)) 
  terms(mark(X)) >= mark(terms(X)) 
  cons(mark(X), Y) >= mark(cons(X, Y)) 
  recip(mark(X)) >= mark(recip(X)) 
  sqr(mark(X)) >= mark(sqr(X)) 
  s(mark(X)) >= mark(s(X)) 
  add(mark(X), Y) >= mark(add(X, Y)) 
  add(X, mark(Y)) >= mark(add(X, Y)) 
  dbl(mark(X)) >= mark(dbl(X)) 
  first(mark(X), Y) >= mark(first(X, Y)) 
  first(X, mark(Y)) >= mark(first(X, Y)) 
  proper(terms(X)) >= terms(proper(X)) 
  proper(cons(X, Y)) >= cons(proper(X), proper(Y)) 
  proper(recip(X)) >= recip(proper(X)) 
  proper(sqr(X)) >= sqr(proper(X)) 
  proper(s(X)) >= s(proper(X)) 
  proper(0) >= ok(0) 
  proper(add(X, Y)) >= add(proper(X), proper(Y)) 
  proper(dbl(X)) >= dbl(proper(X)) 
  proper(first(X, Y)) >= first(proper(X), proper(Y)) 
  proper(nil) >= ok(nil) 
  terms(ok(X)) >= ok(terms(X)) 
  cons(ok(X), ok(Y)) >= ok(cons(X, Y)) 
  recip(ok(X)) >= ok(recip(X)) 
  sqr(ok(X)) >= ok(sqr(X)) 
  s(ok(X)) >= ok(s(X)) 
  add(ok(X), ok(Y)) >= ok(add(X, Y)) 
  dbl(ok(X)) >= ok(dbl(X)) 
  first(ok(X), ok(Y)) >= ok(first(X, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[active(x_1)]] = x_1 
  [[cons(x_1, x_2)]] = cons(x_1) 
  [[nil]] = _|_ 
  [[ok(x_1)]] = x_1 
  [[proper(x_1)]] = x_1 

We choose Lex = {} and Mul = {0, add, cons, dbl, first, mark, recip, s, sqr, terms, top#}, and the following precedence: terms > recip > sqr > add > dbl > s > first > cons > 0 > mark = top#

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  top#(mark(X)) > top#(X) 
  top#(X) >= top#(X) 
  terms(X) >= mark(cons(recip(sqr(X)))) 
  sqr(0) >= mark(0) 
  sqr(s(X)) >= mark(s(add(sqr(X), dbl(X)))) 
  dbl(0) >= mark(0) 
  dbl(s(X)) >= mark(s(s(dbl(X)))) 
  add(0, X) >= mark(X) 
  add(s(X), Y) >= mark(s(add(X, Y))) 
  first(0, X) >= mark(_|_) 
  first(s(X), cons(Y)) >= mark(cons(Y)) 
  terms(X) >= terms(X) 
  cons(X) >= cons(X) 
  recip(X) >= recip(X) 
  sqr(X) >= sqr(X) 
  s(X) >= s(X) 
  add(X, Y) >= add(X, Y) 
  add(X, Y) >= add(X, Y) 
  dbl(X) >= dbl(X) 
  first(X, Y) >= first(X, Y) 
  first(X, Y) >= first(X, Y) 
  terms(mark(X)) >= mark(terms(X)) 
  cons(mark(X)) >= mark(cons(X)) 
  recip(mark(X)) >= mark(recip(X)) 
  sqr(mark(X)) >= mark(sqr(X)) 
  s(mark(X)) >= mark(s(X)) 
  add(mark(X), Y) >= mark(add(X, Y)) 
  add(X, mark(Y)) >= mark(add(X, Y)) 
  dbl(mark(X)) >= mark(dbl(X)) 
  first(mark(X), Y) >= mark(first(X, Y)) 
  first(X, mark(Y)) >= mark(first(X, Y)) 
  terms(X) >= terms(X) 
  cons(X) >= cons(X) 
  recip(X) >= recip(X) 
  sqr(X) >= sqr(X) 
  s(X) >= s(X) 
  0 >= 0 
  add(X, Y) >= add(X, Y) 
  dbl(X) >= dbl(X) 
  first(X, Y) >= first(X, Y) 
  _|_ >= _|_ 
  terms(X) >= terms(X) 
  cons(X) >= cons(X) 
  recip(X) >= recip(X) 
  sqr(X) >= sqr(X) 
  s(X) >= s(X) 
  add(X, Y) >= add(X, Y) 
  dbl(X) >= dbl(X) 
  first(X, Y) >= first(X, Y) 

With these choices, we have:

  1] top#(mark(X)) > top#(X)  because [2], by definition 
  2] top#*(mark(X)) >= top#(X)  because [3], by (Select) 
  3] mark(X) >= top#(X)  because mark = top#, mark in Mul and [4], by (Fun) 
  4] X >= X  by (Meta) 

  5] top#(X) >= top#(X)  because top# in Mul and [6], by (Fun) 
  6] X >= X  by (Meta) 

  7] terms(X) >= mark(cons(recip(sqr(X))))  because [8], by (Star) 
  8] terms*(X) >= mark(cons(recip(sqr(X))))  because terms > mark and [9], by (Copy) 
  9] terms*(X) >= cons(recip(sqr(X)))  because terms > cons and [10], by (Copy) 
  10] terms*(X) >= recip(sqr(X))  because terms > recip and [11], by (Copy) 
  11] terms*(X) >= sqr(X)  because terms > sqr and [12], by (Copy) 
  12] terms*(X) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 

  14] sqr(0) >= mark(0)  because [15], by (Star) 
  15] sqr*(0) >= mark(0)  because sqr > mark and [16], by (Copy) 
  16] sqr*(0) >= 0  because sqr > 0, by (Copy) 

  17] sqr(s(X)) >= mark(s(add(sqr(X), dbl(X))))  because [18], by (Star) 
  18] sqr*(s(X)) >= mark(s(add(sqr(X), dbl(X))))  because sqr > mark and [19], by (Copy) 
  19] sqr*(s(X)) >= s(add(sqr(X), dbl(X)))  because sqr > s and [20], by (Copy) 
  20] sqr*(s(X)) >= add(sqr(X), dbl(X))  because sqr > add, [21] and [24], by (Copy) 
  21] sqr*(s(X)) >= sqr(X)  because sqr in Mul and [22], by (Stat) 
  22] s(X) > X  because [23], by definition 
  23] s*(X) >= X  because [6], by (Select) 
  24] sqr*(s(X)) >= dbl(X)  because sqr > dbl and [25], by (Copy) 
  25] sqr*(s(X)) >= X  because [26], by (Select) 
  26] s(X) >= X  because [23], by (Star) 

  27] dbl(0) >= mark(0)  because [28], by (Star) 
  28] dbl*(0) >= mark(0)  because dbl > mark and [29], by (Copy) 
  29] dbl*(0) >= 0  because dbl > 0, by (Copy) 

  30] dbl(s(X)) >= mark(s(s(dbl(X))))  because [31], by (Star) 
  31] dbl*(s(X)) >= mark(s(s(dbl(X))))  because dbl > mark and [32], by (Copy) 
  32] dbl*(s(X)) >= s(s(dbl(X)))  because dbl > s and [33], by (Copy) 
  33] dbl*(s(X)) >= s(dbl(X))  because dbl > s and [34], by (Copy) 
  34] dbl*(s(X)) >= dbl(X)  because dbl in Mul and [22], by (Stat) 

  35] add(0, X) >= mark(X)  because [36], by (Star) 
  36] add*(0, X) >= mark(X)  because add > mark and [37], by (Copy) 
  37] add*(0, X) >= X  because [6], by (Select) 

  38] add(s(X), Y) >= mark(s(add(X, Y)))  because [39], by (Star) 
  39] add*(s(X), Y) >= mark(s(add(X, Y)))  because add > mark and [40], by (Copy) 
  40] add*(s(X), Y) >= s(add(X, Y))  because add > s and [41], by (Copy) 
  41] add*(s(X), Y) >= add(X, Y)  because add in Mul, [22] and [42], by (Stat) 
  42] Y >= Y  by (Meta) 

  43] first(0, X) >= mark(_|_)  because [44], by (Star) 
  44] first*(0, X) >= mark(_|_)  because [45], by (Select) 
  45] 0 >= mark(_|_)  because [46], by (Star) 
  46] 0* >= mark(_|_)  because 0 > mark and [47], by (Copy) 
  47] 0* >= _|_  by (Bot) 

  48] first(s(X), cons(Y)) >= mark(cons(Y))  because [49], by (Star) 
  49] first*(s(X), cons(Y)) >= mark(cons(Y))  because first > mark and [50], by (Copy) 
  50] first*(s(X), cons(Y)) >= cons(Y)  because first > cons and [51], by (Copy) 
  51] first*(s(X), cons(Y)) >= Y  because [52], by (Select) 
  52] cons(Y) >= Y  because [53], by (Star) 
  53] cons*(Y) >= Y  because [42], by (Select) 

  54] terms(X) >= terms(X)  because terms in Mul and [55], by (Fun) 
  55] X >= X  by (Meta) 

  56] cons(X) >= cons(X)  because cons in Mul and [57], by (Fun) 
  57] X >= X  by (Meta) 

  58] recip(X) >= recip(X)  because recip in Mul and [55], by (Fun) 

  59] sqr(X) >= sqr(X)  because sqr in Mul and [55], by (Fun) 

  60] s(X) >= s(X)  because s in Mul and [55], by (Fun) 

  61] add(X, Y) >= add(X, Y)  because add in Mul, [57] and [62], by (Fun) 
  62] Y >= Y  by (Meta) 

  63] add(X, Y) >= add(X, Y)  because add in Mul, [57] and [64], by (Fun) 
  64] Y >= Y  by (Meta) 

  65] dbl(X) >= dbl(X)  because dbl in Mul and [55], by (Fun) 

  66] first(X, Y) >= first(X, Y)  because first in Mul, [57] and [64], by (Fun) 

  67] first(X, Y) >= first(X, Y)  because first in Mul, [57] and [64], by (Fun) 

  68] terms(mark(X)) >= mark(terms(X))  because [69], by (Star) 
  69] terms*(mark(X)) >= mark(terms(X))  because terms > mark and [70], by (Copy) 
  70] terms*(mark(X)) >= terms(X)  because terms in Mul and [71], by (Stat) 
  71] mark(X) > X  because [72], by definition 
  72] mark*(X) >= X  because [55], by (Select) 

  73] cons(mark(X)) >= mark(cons(X))  because [74], by (Star) 
  74] cons*(mark(X)) >= mark(cons(X))  because cons > mark and [75], by (Copy) 
  75] cons*(mark(X)) >= cons(X)  because cons in Mul and [76], by (Stat) 
  76] mark(X) > X  because [77], by definition 
  77] mark*(X) >= X  because [57], by (Select) 

  78] recip(mark(X)) >= mark(recip(X))  because [79], by (Star) 
  79] recip*(mark(X)) >= mark(recip(X))  because recip > mark and [80], by (Copy) 
  80] recip*(mark(X)) >= recip(X)  because recip in Mul and [71], by (Stat) 

  81] sqr(mark(X)) >= mark(sqr(X))  because [82], by (Star) 
  82] sqr*(mark(X)) >= mark(sqr(X))  because sqr > mark and [83], by (Copy) 
  83] sqr*(mark(X)) >= sqr(X)  because sqr in Mul and [71], by (Stat) 

  84] s(mark(X)) >= mark(s(X))  because [85], by (Star) 
  85] s*(mark(X)) >= mark(s(X))  because s > mark and [86], by (Copy) 
  86] s*(mark(X)) >= s(X)  because s in Mul and [71], by (Stat) 

  87] add(mark(X), Y) >= mark(add(X, Y))  because [88], by (Star) 
  88] add*(mark(X), Y) >= mark(add(X, Y))  because add > mark and [89], by (Copy) 
  89] add*(mark(X), Y) >= add(X, Y)  because add in Mul, [76] and [64], by (Stat) 

  90] add(X, mark(Y)) >= mark(add(X, Y))  because [91], by (Star) 
  91] add*(X, mark(Y)) >= mark(add(X, Y))  because add > mark and [92], by (Copy) 
  92] add*(X, mark(Y)) >= add(X, Y)  because add in Mul, [57] and [93], by (Stat) 
  93] mark(Y) > Y  because [94], by definition 
  94] mark*(Y) >= Y  because [64], by (Select) 

  95] dbl(mark(X)) >= mark(dbl(X))  because [96], by (Star) 
  96] dbl*(mark(X)) >= mark(dbl(X))  because dbl > mark and [97], by (Copy) 
  97] dbl*(mark(X)) >= dbl(X)  because dbl in Mul and [71], by (Stat) 

  98] first(mark(X), Y) >= mark(first(X, Y))  because [99], by (Star) 
  99] first*(mark(X), Y) >= mark(first(X, Y))  because first > mark and [100], by (Copy) 
  100] first*(mark(X), Y) >= first(X, Y)  because first in Mul, [76] and [64], by (Stat) 

  101] first(X, mark(Y)) >= mark(first(X, Y))  because [102], by (Star) 
  102] first*(X, mark(Y)) >= mark(first(X, Y))  because first > mark and [103], by (Copy) 
  103] first*(X, mark(Y)) >= first(X, Y)  because first in Mul, [57] and [93], by (Stat) 

  104] terms(X) >= terms(X)  because terms in Mul and [4], by (Fun) 

  105] cons(X) >= cons(X)  because cons in Mul and [106], by (Fun) 
  106] X >= X  by (Meta) 

  107] recip(X) >= recip(X)  because recip in Mul and [4], by (Fun) 

  108] sqr(X) >= sqr(X)  because sqr in Mul and [4], by (Fun) 

  109] s(X) >= s(X)  because s in Mul and [4], by (Fun) 

  110] 0 >= 0  by (Fun) 

  111] add(X, Y) >= add(X, Y)  because add in Mul, [106] and [112], by (Fun) 
  112] Y >= Y  by (Meta) 

  113] dbl(X) >= dbl(X)  because dbl in Mul and [4], by (Fun) 

  114] first(X, Y) >= first(X, Y)  because first in Mul, [106] and [112], by (Fun) 

  115] _|_ >= _|_  by (Bot) 

  116] terms(X) >= terms(X)  because terms in Mul and [6], by (Fun) 

  117] cons(X) >= cons(X)  because cons in Mul and [118], by (Fun) 
  118] X >= X  by (Meta) 

  119] recip(X) >= recip(X)  because recip in Mul and [6], by (Fun) 

  120] sqr(X) >= sqr(X)  because sqr in Mul and [6], by (Fun) 

  121] s(X) >= s(X)  because s in Mul and [6], by (Fun) 

  122] add(X, Y) >= add(X, Y)  because add in Mul, [118] and [123], by (Fun) 
  123] Y >= Y  by (Meta) 

  124] dbl(X) >= dbl(X)  because dbl in Mul and [6], by (Fun) 

  125] first(X, Y) >= first(X, Y)  because first in Mul, [118] and [123], by (Fun) 

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:

  top#(ok(X)) =#> top#(active(X))   

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), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative), (P_10, R_0, minimal, formative) and (P_12, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_12, R_1, minimal, formative).

The formative rules of (P_12, R_1) are R_2 ::=

  active(terms(X)) => terms(active(X)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(recip(X)) => recip(active(X)) 
  active(sqr(X)) => sqr(active(X)) 
  active(s(X)) => s(active(X)) 
  active(add(X, Y)) => add(active(X), Y) 
  active(add(X, Y)) => add(X, active(Y)) 
  active(dbl(X)) => dbl(active(X)) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  proper(terms(X)) => terms(proper(X)) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(recip(X)) => recip(proper(X)) 
  proper(sqr(X)) => sqr(proper(X)) 
  proper(s(X)) => s(proper(X)) 
  proper(0) => ok(0) 
  proper(add(X, Y)) => add(proper(X), proper(Y)) 
  proper(dbl(X)) => dbl(proper(X)) 
  proper(first(X, Y)) => first(proper(X), proper(Y)) 
  proper(nil) => ok(nil) 
  terms(ok(X)) => ok(terms(X)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  recip(ok(X)) => ok(recip(X)) 
  sqr(ok(X)) => ok(sqr(X)) 
  s(ok(X)) => ok(s(X)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  dbl(ok(X)) => ok(dbl(X)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_12, R_1, minimal, formative) by (P_12, R_2, minimal, formative).

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), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative), (P_10, R_0, minimal, formative) and (P_12, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_12, R_2, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_12, R_2) are:

  active(terms(X)) => terms(active(X)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(recip(X)) => recip(active(X)) 
  active(sqr(X)) => sqr(active(X)) 
  active(s(X)) => s(active(X)) 
  active(add(X, Y)) => add(active(X), Y) 
  active(add(X, Y)) => add(X, active(Y)) 
  active(dbl(X)) => dbl(active(X)) 
  active(first(X, Y)) => first(active(X), Y) 
  active(first(X, Y)) => first(X, active(Y)) 
  terms(ok(X)) => ok(terms(X)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  recip(ok(X)) => ok(recip(X)) 
  sqr(ok(X)) => ok(sqr(X)) 
  s(ok(X)) => ok(s(X)) 
  add(ok(X), ok(Y)) => ok(add(X, Y)) 
  dbl(ok(X)) => ok(dbl(X)) 
  first(ok(X), ok(Y)) => ok(first(X, Y)) 

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

  top#(ok(X)) >? top#(active(X)) 
  active(terms(X)) >= terms(active(X)) 
  active(cons(X, Y)) >= cons(active(X), Y) 
  active(recip(X)) >= recip(active(X)) 
  active(sqr(X)) >= sqr(active(X)) 
  active(s(X)) >= s(active(X)) 
  active(add(X, Y)) >= add(active(X), Y) 
  active(add(X, Y)) >= add(X, active(Y)) 
  active(dbl(X)) >= dbl(active(X)) 
  active(first(X, Y)) >= first(active(X), Y) 
  active(first(X, Y)) >= first(X, active(Y)) 
  terms(ok(X)) >= ok(terms(X)) 
  cons(ok(X), ok(Y)) >= ok(cons(X, Y)) 
  recip(ok(X)) >= ok(recip(X)) 
  sqr(ok(X)) >= ok(sqr(X)) 
  s(ok(X)) >= ok(s(X)) 
  add(ok(X), ok(Y)) >= ok(add(X, Y)) 
  dbl(ok(X)) >= ok(dbl(X)) 
  first(ok(X), ok(Y)) >= ok(first(X, Y)) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  add = \y0y1.3y1 
  cons = \y0y1.3y0 
  dbl = \y0.y0 
  first = \y0y1.y1 
  ok = \y0.3 + 2y0 
  recip = \y0.y0 
  s = \y0.y0 
  sqr = \y0.y0 
  terms = \y0.3y0 
  top# = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[top#(ok(_x0))]] = 9 + 6x0 > 3x0 = [[top#(active(_x0))]] 
  [[active(terms(_x0))]] = 3x0 >= 3x0 = [[terms(active(_x0))]] 
  [[active(cons(_x0, _x1))]] = 3x0 >= 3x0 = [[cons(active(_x0), _x1)]] 
  [[active(recip(_x0))]] = x0 >= x0 = [[recip(active(_x0))]] 
  [[active(sqr(_x0))]] = x0 >= x0 = [[sqr(active(_x0))]] 
  [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] 
  [[active(add(_x0, _x1))]] = 3x1 >= 3x1 = [[add(active(_x0), _x1)]] 
  [[active(add(_x0, _x1))]] = 3x1 >= 3x1 = [[add(_x0, active(_x1))]] 
  [[active(dbl(_x0))]] = x0 >= x0 = [[dbl(active(_x0))]] 
  [[active(first(_x0, _x1))]] = x1 >= x1 = [[first(active(_x0), _x1)]] 
  [[active(first(_x0, _x1))]] = x1 >= x1 = [[first(_x0, active(_x1))]] 
  [[terms(ok(_x0))]] = 9 + 6x0 >= 3 + 6x0 = [[ok(terms(_x0))]] 
  [[cons(ok(_x0), ok(_x1))]] = 9 + 6x0 >= 3 + 6x0 = [[ok(cons(_x0, _x1))]] 
  [[recip(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(recip(_x0))]] 
  [[sqr(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(sqr(_x0))]] 
  [[s(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(s(_x0))]] 
  [[add(ok(_x0), ok(_x1))]] = 9 + 6x1 >= 3 + 6x1 = [[ok(add(_x0, _x1))]] 
  [[dbl(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(dbl(_x0))]] 
  [[first(ok(_x0), ok(_x1))]] = 3 + 2x1 >= 3 + 2x1 = [[ok(first(_x0, _x1))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_12, R_2) by ({}, R_2).  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), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative), (P_9, R_0, minimal, formative) and (P_10, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_10, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(proper#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(proper#(terms(X))) = terms(X) |> X = nu(proper#(X)) 
  nu(proper#(cons(X, Y))) = cons(X, Y) |> X = nu(proper#(X)) 
  nu(proper#(cons(X, Y))) = cons(X, Y) |> Y = nu(proper#(Y)) 
  nu(proper#(recip(X))) = recip(X) |> X = nu(proper#(X)) 
  nu(proper#(sqr(X))) = sqr(X) |> X = nu(proper#(X)) 
  nu(proper#(s(X))) = s(X) |> X = nu(proper#(X)) 
  nu(proper#(add(X, Y))) = add(X, Y) |> X = nu(proper#(X)) 
  nu(proper#(add(X, Y))) = add(X, Y) |> Y = nu(proper#(Y)) 
  nu(proper#(dbl(X))) = dbl(X) |> X = nu(proper#(X)) 
  nu(proper#(first(X, Y))) = first(X, Y) |> X = nu(proper#(X)) 
  nu(proper#(first(X, Y))) = first(X, Y) |> Y = nu(proper#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_10, 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), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative) and (P_9, R_0, minimal, formative) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(first#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(first#(mark(X), Y)) = mark(X) |> X = nu(first#(X, Y)) 
  nu(first#(X, mark(Y))) = X = X = nu(first#(X, Y)) 
  nu(first#(ok(X), ok(Y))) = ok(X) |> X = nu(first#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_9, R_0, minimal, f) by (P_13, R_0, minimal, f), where P_13 contains:

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

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), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative) and (P_13, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_13, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(first#) = 2 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_13, 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), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative) and (P_8, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_8, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(dbl#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(dbl#(mark(X))) = mark(X) |> X = nu(dbl#(X)) 
  nu(dbl#(ok(X))) = ok(X) |> X = nu(dbl#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_8, 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), (P_5, R_0, minimal, formative), (P_6, 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(add#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(add#(mark(X), Y)) = mark(X) |> X = nu(add#(X, Y)) 
  nu(add#(X, mark(Y))) = X = X = nu(add#(X, Y)) 
  nu(add#(ok(X), ok(Y))) = ok(X) |> X = nu(add#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_7, R_0, minimal, f) by (P_14, R_0, minimal, f), where P_14 contains:

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

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), (P_6, R_0, minimal, formative) and (P_14, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_14, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(add#) = 2 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_14, 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), (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(s#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(s#(mark(X))) = mark(X) |> X = nu(s#(X)) 
  nu(s#(ok(X))) = ok(X) |> X = nu(s#(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(sqr#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sqr#(mark(X))) = mark(X) |> X = nu(sqr#(X)) 
  nu(sqr#(ok(X))) = ok(X) |> X = nu(sqr#(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(recip#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(recip#(mark(X))) = mark(X) |> X = nu(recip#(X)) 
  nu(recip#(ok(X))) = ok(X) |> X = nu(recip#(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#(ok(X), ok(Y))) = ok(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 ({}, 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(terms#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(terms#(mark(X))) = mark(X) |> X = nu(terms#(X)) 
  nu(terms#(ok(X))) = ok(X) |> X = nu(terms#(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 apply the subterm criterion with the following projection function:

  nu(active#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(active#(terms(X))) = terms(X) |> X = nu(active#(X)) 
  nu(active#(cons(X, Y))) = cons(X, Y) |> X = nu(active#(X)) 
  nu(active#(recip(X))) = recip(X) |> X = nu(active#(X)) 
  nu(active#(sqr(X))) = sqr(X) |> X = nu(active#(X)) 
  nu(active#(s(X))) = s(X) |> X = nu(active#(X)) 
  nu(active#(add(X, Y))) = add(X, Y) |> X = nu(active#(X)) 
  nu(active#(add(X, Y))) = add(X, Y) |> Y = nu(active#(Y)) 
  nu(active#(dbl(X))) = dbl(X) |> X = nu(active#(X)) 
  nu(active#(first(X, Y))) = first(X, Y) |> X = nu(active#(X)) 
  nu(active#(first(X, Y))) = first(X, Y) |> Y = nu(active#(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.