/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 
  filter : [o * o * o] --> o 
  mark : [o] --> o 
  nats : [o] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  s : [o] --> o 
  sieve : [o] --> o 
  top : [o] --> o 
  zprimes : [] --> o 

  active(filter(cons(X, Y), 0, Z)) => mark(cons(0, filter(Y, Z, Z))) 
  active(filter(cons(X, Y), s(Z), U)) => mark(cons(X, filter(Y, Z, U))) 
  active(sieve(cons(0, X))) => mark(cons(0, sieve(X))) 
  active(sieve(cons(s(X), Y))) => mark(cons(s(X), sieve(filter(Y, X, X)))) 
  active(nats(X)) => mark(cons(X, nats(s(X)))) 
  active(zprimes) => mark(sieve(nats(s(s(0))))) 
  active(filter(X, Y, Z)) => filter(active(X), Y, Z) 
  active(filter(X, Y, Z)) => filter(X, active(Y), Z) 
  active(filter(X, Y, Z)) => filter(X, Y, active(Z)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(s(X)) => s(active(X)) 
  active(sieve(X)) => sieve(active(X)) 
  active(nats(X)) => nats(active(X)) 
  filter(mark(X), Y, Z) => mark(filter(X, Y, Z)) 
  filter(X, mark(Y), Z) => mark(filter(X, Y, Z)) 
  filter(X, Y, mark(Z)) => mark(filter(X, Y, Z)) 
  cons(mark(X), Y) => mark(cons(X, Y)) 
  s(mark(X)) => mark(s(X)) 
  sieve(mark(X)) => mark(sieve(X)) 
  nats(mark(X)) => mark(nats(X)) 
  proper(filter(X, Y, Z)) => filter(proper(X), proper(Y), proper(Z)) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(0) => ok(0) 
  proper(s(X)) => s(proper(X)) 
  proper(sieve(X)) => sieve(proper(X)) 
  proper(nats(X)) => nats(proper(X)) 
  proper(zprimes) => ok(zprimes) 
  filter(ok(X), ok(Y), ok(Z)) => ok(filter(X, Y, Z)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  s(ok(X)) => ok(s(X)) 
  sieve(ok(X)) => ok(sieve(X)) 
  nats(ok(X)) => ok(nats(X)) 
  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#(filter(cons(X, Y), 0, Z)) =#> cons#(0, filter(Y, Z, Z))   
    1] active#(filter(cons(X, Y), 0, Z)) =#> filter#(Y, Z, Z)   
    2] active#(filter(cons(X, Y), s(Z), U)) =#> cons#(X, filter(Y, Z, U))   
    3] active#(filter(cons(X, Y), s(Z), U)) =#> filter#(Y, Z, U)   
    4] active#(sieve(cons(0, X))) =#> cons#(0, sieve(X))   
    5] active#(sieve(cons(0, X))) =#> sieve#(X)   
    6] active#(sieve(cons(s(X), Y))) =#> cons#(s(X), sieve(filter(Y, X, X)))   
    7] active#(sieve(cons(s(X), Y))) =#> s#(X)   
    8] active#(sieve(cons(s(X), Y))) =#> sieve#(filter(Y, X, X))   
    9] active#(sieve(cons(s(X), Y))) =#> filter#(Y, X, X)   
    10] active#(nats(X)) =#> cons#(X, nats(s(X)))   
    11] active#(nats(X)) =#> nats#(s(X))   
    12] active#(nats(X)) =#> s#(X)   
    13] active#(zprimes) =#> sieve#(nats(s(s(0))))   
    14] active#(zprimes) =#> nats#(s(s(0)))   
    15] active#(zprimes) =#> s#(s(0))   
    16] active#(zprimes) =#> s#(0)   
    17] active#(filter(X, Y, Z)) =#> filter#(active(X), Y, Z)   
    18] active#(filter(X, Y, Z)) =#> active#(X)   
    19] active#(filter(X, Y, Z)) =#> filter#(X, active(Y), Z)   
    20] active#(filter(X, Y, Z)) =#> active#(Y)   
    21] active#(filter(X, Y, Z)) =#> filter#(X, Y, active(Z))   
    22] active#(filter(X, Y, Z)) =#> active#(Z)   
    23] active#(cons(X, Y)) =#> cons#(active(X), Y)   
    24] active#(cons(X, Y)) =#> active#(X)   
    25] active#(s(X)) =#> s#(active(X))   
    26] active#(s(X)) =#> active#(X)   
    27] active#(sieve(X)) =#> sieve#(active(X))   
    28] active#(sieve(X)) =#> active#(X)   
    29] active#(nats(X)) =#> nats#(active(X))   
    30] active#(nats(X)) =#> active#(X)   
    31] filter#(mark(X), Y, Z) =#> filter#(X, Y, Z)   
    32] filter#(X, mark(Y), Z) =#> filter#(X, Y, Z)   
    33] filter#(X, Y, mark(Z)) =#> filter#(X, Y, Z)   
    34] cons#(mark(X), Y) =#> cons#(X, Y)   
    35] s#(mark(X)) =#> s#(X)   
    36] sieve#(mark(X)) =#> sieve#(X)   
    37] nats#(mark(X)) =#> nats#(X)   
    38] proper#(filter(X, Y, Z)) =#> filter#(proper(X), proper(Y), proper(Z))   
    39] proper#(filter(X, Y, Z)) =#> proper#(X)   
    40] proper#(filter(X, Y, Z)) =#> proper#(Y)   
    41] proper#(filter(X, Y, Z)) =#> proper#(Z)   
    42] proper#(cons(X, Y)) =#> cons#(proper(X), proper(Y))   
    43] proper#(cons(X, Y)) =#> proper#(X)   
    44] proper#(cons(X, Y)) =#> proper#(Y)   
    45] proper#(s(X)) =#> s#(proper(X))   
    46] proper#(s(X)) =#> proper#(X)   
    47] proper#(sieve(X)) =#> sieve#(proper(X))   
    48] proper#(sieve(X)) =#> proper#(X)   
    49] proper#(nats(X)) =#> nats#(proper(X))   
    50] proper#(nats(X)) =#> proper#(X)   
    51] filter#(ok(X), ok(Y), ok(Z)) =#> filter#(X, Y, Z)   
    52] cons#(ok(X), ok(Y)) =#> cons#(X, Y)   
    53] s#(ok(X)) =#> s#(X)   
    54] sieve#(ok(X)) =#> sieve#(X)   
    55] nats#(ok(X)) =#> nats#(X)   
    56] top#(mark(X)) =#> top#(proper(X))   
    57] top#(mark(X)) =#> proper#(X)   
    58] top#(ok(X)) =#> top#(active(X))   
    59] top#(ok(X)) =#> active#(X)   

  Rules R_0:

    active(filter(cons(X, Y), 0, Z)) => mark(cons(0, filter(Y, Z, Z))) 
    active(filter(cons(X, Y), s(Z), U)) => mark(cons(X, filter(Y, Z, U))) 
    active(sieve(cons(0, X))) => mark(cons(0, sieve(X))) 
    active(sieve(cons(s(X), Y))) => mark(cons(s(X), sieve(filter(Y, X, X)))) 
    active(nats(X)) => mark(cons(X, nats(s(X)))) 
    active(zprimes) => mark(sieve(nats(s(s(0))))) 
    active(filter(X, Y, Z)) => filter(active(X), Y, Z) 
    active(filter(X, Y, Z)) => filter(X, active(Y), Z) 
    active(filter(X, Y, Z)) => filter(X, Y, active(Z)) 
    active(cons(X, Y)) => cons(active(X), Y) 
    active(s(X)) => s(active(X)) 
    active(sieve(X)) => sieve(active(X)) 
    active(nats(X)) => nats(active(X)) 
    filter(mark(X), Y, Z) => mark(filter(X, Y, Z)) 
    filter(X, mark(Y), Z) => mark(filter(X, Y, Z)) 
    filter(X, Y, mark(Z)) => mark(filter(X, Y, Z)) 
    cons(mark(X), Y) => mark(cons(X, Y)) 
    s(mark(X)) => mark(s(X)) 
    sieve(mark(X)) => mark(sieve(X)) 
    nats(mark(X)) => mark(nats(X)) 
    proper(filter(X, Y, Z)) => filter(proper(X), proper(Y), proper(Z)) 
    proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
    proper(0) => ok(0) 
    proper(s(X)) => s(proper(X)) 
    proper(sieve(X)) => sieve(proper(X)) 
    proper(nats(X)) => nats(proper(X)) 
    proper(zprimes) => ok(zprimes) 
    filter(ok(X), ok(Y), ok(Z)) => ok(filter(X, Y, Z)) 
    cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
    s(ok(X)) => ok(s(X)) 
    sieve(ok(X)) => ok(sieve(X)) 
    nats(ok(X)) => ok(nats(X)) 
    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 :  
    * 1 : 31, 32, 33, 51 
    * 2 : 34, 52 
    * 3 : 31, 32, 33, 51 
    * 4 :  
    * 5 : 36, 54 
    * 6 : 34, 52 
    * 7 : 35, 53 
    * 8 : 36, 54 
    * 9 : 31, 32, 33, 51 
    * 10 : 34, 52 
    * 11 : 37, 55 
    * 12 : 35, 53 
    * 13 : 36, 54 
    * 14 : 37, 55 
    * 15 : 35, 53 
    * 16 :  
    * 17 : 31, 32, 33, 51 
    * 18 : 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 
    * 19 : 31, 32, 33, 51 
    * 20 : 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 
    * 21 : 31, 32, 33, 51 
    * 22 : 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 
    * 23 : 34, 52 
    * 24 : 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 
    * 25 : 35, 53 
    * 26 : 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 
    * 27 : 36, 54 
    * 28 : 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 
    * 29 : 37, 55 
    * 30 : 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 : 31, 32, 33, 51 
    * 32 : 31, 32, 33, 51 
    * 33 : 31, 32, 33, 51 
    * 34 : 34, 52 
    * 35 : 35, 53 
    * 36 : 36, 54 
    * 37 : 37, 55 
    * 38 : 31, 32, 33, 51 
    * 39 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 40 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 41 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 42 : 34, 52 
    * 43 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 44 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 45 : 35, 53 
    * 46 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 47 : 36, 54 
    * 48 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 49 : 37, 55 
    * 50 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 51 : 31, 32, 33, 51 
    * 52 : 34, 52 
    * 53 : 35, 53 
    * 54 : 36, 54 
    * 55 : 37, 55 
    * 56 : 56, 57, 58, 59 
    * 57 : 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 
    * 58 : 56, 57, 58, 59 
    * 59 : 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 

This graph has the following strongly connected components:

  P_1:

    active#(filter(X, Y, Z)) =#> active#(X)   
    active#(filter(X, Y, Z)) =#> active#(Y)   
    active#(filter(X, Y, Z)) =#> active#(Z)   
    active#(cons(X, Y)) =#> active#(X)   
    active#(s(X)) =#> active#(X)   
    active#(sieve(X)) =#> active#(X)   
    active#(nats(X)) =#> active#(X)   

  P_2:

    filter#(mark(X), Y, Z) =#> filter#(X, Y, Z)   
    filter#(X, mark(Y), Z) =#> filter#(X, Y, Z)   
    filter#(X, Y, mark(Z)) =#> filter#(X, Y, Z)   
    filter#(ok(X), ok(Y), ok(Z)) =#> filter#(X, Y, Z)   

  P_3:

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

  P_4:

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

  P_5:

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

  P_6:

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

  P_7:

    proper#(filter(X, Y, Z)) =#> proper#(X)   
    proper#(filter(X, Y, Z)) =#> proper#(Y)   
    proper#(filter(X, Y, Z)) =#> proper#(Z)   
    proper#(cons(X, Y)) =#> proper#(X)   
    proper#(cons(X, Y)) =#> proper#(Y)   
    proper#(s(X)) =#> proper#(X)   
    proper#(sieve(X)) =#> proper#(X)   
    proper#(nats(X)) =#> proper#(X)   

  P_8:

    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) and (P_8, 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) and (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(filter(cons(X, Y), 0, Z)) => mark(cons(0, filter(Y, Z, Z))) 
  active(filter(cons(X, Y), s(Z), U)) => mark(cons(X, filter(Y, Z, U))) 
  active(sieve(cons(0, X))) => mark(cons(0, sieve(X))) 
  active(sieve(cons(s(X), Y))) => mark(cons(s(X), sieve(filter(Y, X, X)))) 
  active(nats(X)) => mark(cons(X, nats(s(X)))) 
  active(zprimes) => mark(sieve(nats(s(s(0))))) 
  active(filter(X, Y, Z)) => filter(active(X), Y, Z) 
  active(filter(X, Y, Z)) => filter(X, active(Y), Z) 
  active(filter(X, Y, Z)) => filter(X, Y, active(Z)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(s(X)) => s(active(X)) 
  active(sieve(X)) => sieve(active(X)) 
  active(nats(X)) => nats(active(X)) 
  filter(mark(X), Y, Z) => mark(filter(X, Y, Z)) 
  filter(X, mark(Y), Z) => mark(filter(X, Y, Z)) 
  filter(X, Y, mark(Z)) => mark(filter(X, Y, Z)) 
  cons(mark(X), Y) => mark(cons(X, Y)) 
  s(mark(X)) => mark(s(X)) 
  sieve(mark(X)) => mark(sieve(X)) 
  nats(mark(X)) => mark(nats(X)) 
  proper(filter(X, Y, Z)) => filter(proper(X), proper(Y), proper(Z)) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(0) => ok(0) 
  proper(s(X)) => s(proper(X)) 
  proper(sieve(X)) => sieve(proper(X)) 
  proper(nats(X)) => nats(proper(X)) 
  proper(zprimes) => ok(zprimes) 
  filter(ok(X), ok(Y), ok(Z)) => ok(filter(X, Y, Z)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  s(ok(X)) => ok(s(X)) 
  sieve(ok(X)) => ok(sieve(X)) 
  nats(ok(X)) => ok(nats(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 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_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:

  top#(mark(X)) >? top#(proper(X)) 
  top#(ok(X)) >? top#(active(X)) 
  active(filter(cons(X, Y), 0, Z)) >= mark(cons(0, filter(Y, Z, Z))) 
  active(filter(cons(X, Y), s(Z), U)) >= mark(cons(X, filter(Y, Z, U))) 
  active(sieve(cons(0, X))) >= mark(cons(0, sieve(X))) 
  active(sieve(cons(s(X), Y))) >= mark(cons(s(X), sieve(filter(Y, X, X)))) 
  active(nats(X)) >= mark(cons(X, nats(s(X)))) 
  active(zprimes) >= mark(sieve(nats(s(s(0))))) 
  active(filter(X, Y, Z)) >= filter(active(X), Y, Z) 
  active(filter(X, Y, Z)) >= filter(X, active(Y), Z) 
  active(filter(X, Y, Z)) >= filter(X, Y, active(Z)) 
  active(cons(X, Y)) >= cons(active(X), Y) 
  active(s(X)) >= s(active(X)) 
  active(sieve(X)) >= sieve(active(X)) 
  active(nats(X)) >= nats(active(X)) 
  filter(mark(X), Y, Z) >= mark(filter(X, Y, Z)) 
  filter(X, mark(Y), Z) >= mark(filter(X, Y, Z)) 
  filter(X, Y, mark(Z)) >= mark(filter(X, Y, Z)) 
  cons(mark(X), Y) >= mark(cons(X, Y)) 
  s(mark(X)) >= mark(s(X)) 
  sieve(mark(X)) >= mark(sieve(X)) 
  nats(mark(X)) >= mark(nats(X)) 
  proper(filter(X, Y, Z)) >= filter(proper(X), proper(Y), proper(Z)) 
  proper(cons(X, Y)) >= cons(proper(X), proper(Y)) 
  proper(0) >= ok(0) 
  proper(s(X)) >= s(proper(X)) 
  proper(sieve(X)) >= sieve(proper(X)) 
  proper(nats(X)) >= nats(proper(X)) 
  proper(zprimes) >= ok(zprimes) 
  filter(ok(X), ok(Y), ok(Z)) >= ok(filter(X, Y, Z)) 
  cons(ok(X), ok(Y)) >= ok(cons(X, Y)) 
  s(ok(X)) >= ok(s(X)) 
  sieve(ok(X)) >= ok(sieve(X)) 
  nats(ok(X)) >= ok(nats(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 = \y0y1.2y0 
  filter = \y0y1y2.1 + y0 + y2 + 2y1 
  mark = \y0.1 + y0 
  nats = \y0.1 + 2y0 
  ok = \y0.y0 
  proper = \y0.y0 
  s = \y0.2y0 
  sieve = \y0.1 + y0 
  top# = \y0.y0 
  zprimes = 3 

Using this interpretation, the requirements translate to:

  [[top#(mark(_x0))]] = 1 + x0 > x0 = [[top#(proper(_x0))]] 
  [[top#(ok(_x0))]] = x0 >= x0 = [[top#(active(_x0))]] 
  [[active(filter(cons(_x0, _x1), 0, _x2))]] = 1 + x2 + 2x0 >= 1 = [[mark(cons(0, filter(_x1, _x2, _x2)))]] 
  [[active(filter(cons(_x0, _x1), s(_x2), _x3))]] = 1 + x3 + 2x0 + 4x2 >= 1 + 2x0 = [[mark(cons(_x0, filter(_x1, _x2, _x3)))]] 
  [[active(sieve(cons(0, _x0)))]] = 1 >= 1 = [[mark(cons(0, sieve(_x0)))]] 
  [[active(sieve(cons(s(_x0), _x1)))]] = 1 + 4x0 >= 1 + 4x0 = [[mark(cons(s(_x0), sieve(filter(_x1, _x0, _x0))))]] 
  [[active(nats(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark(cons(_x0, nats(s(_x0))))]] 
  [[active(zprimes)]] = 3 >= 3 = [[mark(sieve(nats(s(s(0)))))]] 
  [[active(filter(_x0, _x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[filter(active(_x0), _x1, _x2)]] 
  [[active(filter(_x0, _x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[filter(_x0, active(_x1), _x2)]] 
  [[active(filter(_x0, _x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[filter(_x0, _x1, active(_x2))]] 
  [[active(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[cons(active(_x0), _x1)]] 
  [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] 
  [[active(sieve(_x0))]] = 1 + x0 >= 1 + x0 = [[sieve(active(_x0))]] 
  [[active(nats(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[nats(active(_x0))]] 
  [[filter(mark(_x0), _x1, _x2)]] = 2 + x0 + x2 + 2x1 >= 2 + x0 + x2 + 2x1 = [[mark(filter(_x0, _x1, _x2))]] 
  [[filter(_x0, mark(_x1), _x2)]] = 3 + x0 + x2 + 2x1 >= 2 + x0 + x2 + 2x1 = [[mark(filter(_x0, _x1, _x2))]] 
  [[filter(_x0, _x1, mark(_x2))]] = 2 + x0 + x2 + 2x1 >= 2 + x0 + x2 + 2x1 = [[mark(filter(_x0, _x1, _x2))]] 
  [[cons(mark(_x0), _x1)]] = 2 + 2x0 >= 1 + 2x0 = [[mark(cons(_x0, _x1))]] 
  [[s(mark(_x0))]] = 2 + 2x0 >= 1 + 2x0 = [[mark(s(_x0))]] 
  [[sieve(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(sieve(_x0))]] 
  [[nats(mark(_x0))]] = 3 + 2x0 >= 2 + 2x0 = [[mark(nats(_x0))]] 
  [[proper(filter(_x0, _x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[filter(proper(_x0), proper(_x1), proper(_x2))]] 
  [[proper(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[cons(proper(_x0), proper(_x1))]] 
  [[proper(0)]] = 0 >= 0 = [[ok(0)]] 
  [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] 
  [[proper(sieve(_x0))]] = 1 + x0 >= 1 + x0 = [[sieve(proper(_x0))]] 
  [[proper(nats(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[nats(proper(_x0))]] 
  [[proper(zprimes)]] = 3 >= 3 = [[ok(zprimes)]] 
  [[filter(ok(_x0), ok(_x1), ok(_x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[ok(filter(_x0, _x1, _x2))]] 
  [[cons(ok(_x0), ok(_x1))]] = 2x0 >= 2x0 = [[ok(cons(_x0, _x1))]] 
  [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] 
  [[sieve(ok(_x0))]] = 1 + x0 >= 1 + x0 = [[ok(sieve(_x0))]] 
  [[nats(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(nats(_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:

  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) and (P_9, R_1, minimal, formative) is finite.

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

The formative rules of (P_9, R_1) are R_2 ::=

  active(filter(X, Y, Z)) => filter(active(X), Y, Z) 
  active(filter(X, Y, Z)) => filter(X, active(Y), Z) 
  active(filter(X, Y, Z)) => filter(X, Y, active(Z)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(s(X)) => s(active(X)) 
  active(sieve(X)) => sieve(active(X)) 
  active(nats(X)) => nats(active(X)) 
  proper(filter(X, Y, Z)) => filter(proper(X), proper(Y), proper(Z)) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(0) => ok(0) 
  proper(s(X)) => s(proper(X)) 
  proper(sieve(X)) => sieve(proper(X)) 
  proper(nats(X)) => nats(proper(X)) 
  proper(zprimes) => ok(zprimes) 
  filter(ok(X), ok(Y), ok(Z)) => ok(filter(X, Y, Z)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  s(ok(X)) => ok(s(X)) 
  sieve(ok(X)) => ok(sieve(X)) 
  nats(ok(X)) => ok(nats(X)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_9, R_1, minimal, formative) by (P_9, 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) and (P_9, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_9, R_2, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_9, R_2) are:

  active(filter(X, Y, Z)) => filter(active(X), Y, Z) 
  active(filter(X, Y, Z)) => filter(X, active(Y), Z) 
  active(filter(X, Y, Z)) => filter(X, Y, active(Z)) 
  active(cons(X, Y)) => cons(active(X), Y) 
  active(s(X)) => s(active(X)) 
  active(sieve(X)) => sieve(active(X)) 
  active(nats(X)) => nats(active(X)) 
  filter(ok(X), ok(Y), ok(Z)) => ok(filter(X, Y, Z)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  s(ok(X)) => ok(s(X)) 
  sieve(ok(X)) => ok(sieve(X)) 
  nats(ok(X)) => ok(nats(X)) 

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

  top#(ok(X)) >? top#(active(X)) 
  active(filter(X, Y, Z)) >= filter(active(X), Y, Z) 
  active(filter(X, Y, Z)) >= filter(X, active(Y), Z) 
  active(filter(X, Y, Z)) >= filter(X, Y, active(Z)) 
  active(cons(X, Y)) >= cons(active(X), Y) 
  active(s(X)) >= s(active(X)) 
  active(sieve(X)) >= sieve(active(X)) 
  active(nats(X)) >= nats(active(X)) 
  filter(ok(X), ok(Y), ok(Z)) >= ok(filter(X, Y, Z)) 
  cons(ok(X), ok(Y)) >= ok(cons(X, Y)) 
  s(ok(X)) >= ok(s(X)) 
  sieve(ok(X)) >= ok(sieve(X)) 
  nats(ok(X)) >= ok(nats(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.1 + 2y0 
  cons = \y0y1.2y1 
  filter = \y0y1y2.y0 
  nats = \y0.y0 
  ok = \y0.3 + 2y0 
  s = \y0.y0 
  sieve = \y0.y0 
  top# = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[top#(ok(_x0))]] = 9 + 6x0 > 3 + 6x0 = [[top#(active(_x0))]] 
  [[active(filter(_x0, _x1, _x2))]] = 1 + 2x0 >= 1 + 2x0 = [[filter(active(_x0), _x1, _x2)]] 
  [[active(filter(_x0, _x1, _x2))]] = 1 + 2x0 >= x0 = [[filter(_x0, active(_x1), _x2)]] 
  [[active(filter(_x0, _x1, _x2))]] = 1 + 2x0 >= x0 = [[filter(_x0, _x1, active(_x2))]] 
  [[active(cons(_x0, _x1))]] = 1 + 4x1 >= 2x1 = [[cons(active(_x0), _x1)]] 
  [[active(s(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[s(active(_x0))]] 
  [[active(sieve(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[sieve(active(_x0))]] 
  [[active(nats(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[nats(active(_x0))]] 
  [[filter(ok(_x0), ok(_x1), ok(_x2))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(filter(_x0, _x1, _x2))]] 
  [[cons(ok(_x0), ok(_x1))]] = 6 + 4x1 >= 3 + 4x1 = [[ok(cons(_x0, _x1))]] 
  [[s(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(s(_x0))]] 
  [[sieve(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(sieve(_x0))]] 
  [[nats(ok(_x0))]] = 3 + 2x0 >= 3 + 2x0 = [[ok(nats(_x0))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_9, 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) 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(proper#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(proper#(filter(X, Y, Z))) = filter(X, Y, Z) |> X = nu(proper#(X)) 
  nu(proper#(filter(X, Y, Z))) = filter(X, Y, Z) |> Y = nu(proper#(Y)) 
  nu(proper#(filter(X, Y, Z))) = filter(X, Y, Z) |> Z = nu(proper#(Z)) 
  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#(s(X))) = s(X) |> X = nu(proper#(X)) 
  nu(proper#(sieve(X))) = sieve(X) |> X = nu(proper#(X)) 
  nu(proper#(nats(X))) = nats(X) |> X = nu(proper#(X)) 

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), (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(nats#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(nats#(mark(X))) = mark(X) |> X = nu(nats#(X)) 
  nu(nats#(ok(X))) = ok(X) |> X = nu(nats#(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(sieve#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sieve#(mark(X))) = mark(X) |> X = nu(sieve#(X)) 
  nu(sieve#(ok(X))) = ok(X) |> X = nu(sieve#(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#(ok(X))) = ok(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#(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(filter#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_10, R_0, minimal, f), where P_10 contains:

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

Thus, the original system is terminating if each of (P_1, 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(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_10, R_0, minimal, f) by (P_11, R_0, minimal, f), where P_11 contains:

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

Thus, the original system is terminating if each of (P_1, 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).

We apply the subterm criterion with the following projection function:

  nu(filter#) = 3 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_11, 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#(filter(X, Y, Z))) = filter(X, Y, Z) |> X = nu(active#(X)) 
  nu(active#(filter(X, Y, Z))) = filter(X, Y, Z) |> Y = nu(active#(Y)) 
  nu(active#(filter(X, Y, Z))) = filter(X, Y, Z) |> Z = nu(active#(Z)) 
  nu(active#(cons(X, Y))) = cons(X, Y) |> X = nu(active#(X)) 
  nu(active#(s(X))) = s(X) |> X = nu(active#(X)) 
  nu(active#(sieve(X))) = sieve(X) |> X = nu(active#(X)) 
  nu(active#(nats(X))) = nats(X) |> X = nu(active#(X)) 

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.