/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 
  from : [o] --> o 
  fst : [o * o] --> o 
  len : [o] --> o 
  mark : [o] --> o 
  nil : [] --> o 
  s : [o] --> o 

  active(fst(0, X)) => mark(nil) 
  active(fst(s(X), cons(Y, Z))) => mark(cons(Y, fst(X, Z))) 
  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(len(nil)) => mark(0) 
  active(len(cons(X, Y))) => mark(s(len(Y))) 
  mark(fst(X, Y)) => active(fst(mark(X), mark(Y))) 
  mark(0) => active(0) 
  mark(nil) => active(nil) 
  mark(s(X)) => active(s(X)) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(from(X)) => active(from(mark(X))) 
  mark(add(X, Y)) => active(add(mark(X), mark(Y))) 
  mark(len(X)) => active(len(mark(X))) 
  fst(mark(X), Y) => fst(X, Y) 
  fst(X, mark(Y)) => fst(X, Y) 
  fst(active(X), Y) => fst(X, Y) 
  fst(X, active(Y)) => fst(X, Y) 
  s(mark(X)) => s(X) 
  s(active(X)) => s(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) 
  from(mark(X)) => from(X) 
  from(active(X)) => from(X) 
  add(mark(X), Y) => add(X, Y) 
  add(X, mark(Y)) => add(X, Y) 
  add(active(X), Y) => add(X, Y) 
  add(X, active(Y)) => add(X, Y) 
  len(mark(X)) => len(X) 
  len(active(X)) => len(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#(fst(0, X)) =#> mark#(nil)   
    1] active#(fst(s(X), cons(Y, Z))) =#> mark#(cons(Y, fst(X, Z)))   
    2] active#(fst(s(X), cons(Y, Z))) =#> cons#(Y, fst(X, Z))   
    3] active#(fst(s(X), cons(Y, Z))) =#> fst#(X, Z)   
    4] active#(from(X)) =#> mark#(cons(X, from(s(X))))   
    5] active#(from(X)) =#> cons#(X, from(s(X)))   
    6] active#(from(X)) =#> from#(s(X))   
    7] active#(from(X)) =#> s#(X)   
    8] active#(add(0, X)) =#> mark#(X)   
    9] active#(add(s(X), Y)) =#> mark#(s(add(X, Y)))   
    10] active#(add(s(X), Y)) =#> s#(add(X, Y))   
    11] active#(add(s(X), Y)) =#> add#(X, Y)   
    12] active#(len(nil)) =#> mark#(0)   
    13] active#(len(cons(X, Y))) =#> mark#(s(len(Y)))   
    14] active#(len(cons(X, Y))) =#> s#(len(Y))   
    15] active#(len(cons(X, Y))) =#> len#(Y)   
    16] mark#(fst(X, Y)) =#> active#(fst(mark(X), mark(Y)))   
    17] mark#(fst(X, Y)) =#> fst#(mark(X), mark(Y))   
    18] mark#(fst(X, Y)) =#> mark#(X)   
    19] mark#(fst(X, Y)) =#> mark#(Y)   
    20] mark#(0) =#> active#(0)   
    21] mark#(nil) =#> active#(nil)   
    22] mark#(s(X)) =#> active#(s(X))   
    23] mark#(s(X)) =#> s#(X)   
    24] mark#(cons(X, Y)) =#> active#(cons(mark(X), Y))   
    25] mark#(cons(X, Y)) =#> cons#(mark(X), Y)   
    26] mark#(cons(X, Y)) =#> mark#(X)   
    27] mark#(from(X)) =#> active#(from(mark(X)))   
    28] mark#(from(X)) =#> from#(mark(X))   
    29] mark#(from(X)) =#> mark#(X)   
    30] mark#(add(X, Y)) =#> active#(add(mark(X), mark(Y)))   
    31] mark#(add(X, Y)) =#> add#(mark(X), mark(Y))   
    32] mark#(add(X, Y)) =#> mark#(X)   
    33] mark#(add(X, Y)) =#> mark#(Y)   
    34] mark#(len(X)) =#> active#(len(mark(X)))   
    35] mark#(len(X)) =#> len#(mark(X))   
    36] mark#(len(X)) =#> mark#(X)   
    37] fst#(mark(X), Y) =#> fst#(X, Y)   
    38] fst#(X, mark(Y)) =#> fst#(X, Y)   
    39] fst#(active(X), Y) =#> fst#(X, Y)   
    40] fst#(X, active(Y)) =#> fst#(X, Y)   
    41] s#(mark(X)) =#> s#(X)   
    42] s#(active(X)) =#> s#(X)   
    43] cons#(mark(X), Y) =#> cons#(X, Y)   
    44] cons#(X, mark(Y)) =#> cons#(X, Y)   
    45] cons#(active(X), Y) =#> cons#(X, Y)   
    46] cons#(X, active(Y)) =#> cons#(X, Y)   
    47] from#(mark(X)) =#> from#(X)   
    48] from#(active(X)) =#> from#(X)   
    49] add#(mark(X), Y) =#> add#(X, Y)   
    50] add#(X, mark(Y)) =#> add#(X, Y)   
    51] add#(active(X), Y) =#> add#(X, Y)   
    52] add#(X, active(Y)) =#> add#(X, Y)   
    53] len#(mark(X)) =#> len#(X)   
    54] len#(active(X)) =#> len#(X)   

  Rules R_0:

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

This graph has the following strongly connected components:

  P_1:

    active#(fst(s(X), cons(Y, Z))) =#> mark#(cons(Y, fst(X, Z)))   
    active#(from(X)) =#> mark#(cons(X, from(s(X))))   
    active#(add(0, X)) =#> mark#(X)   
    mark#(fst(X, Y)) =#> active#(fst(mark(X), mark(Y)))   
    mark#(fst(X, Y)) =#> mark#(X)   
    mark#(fst(X, Y)) =#> mark#(Y)   
    mark#(cons(X, Y)) =#> mark#(X)   
    mark#(from(X)) =#> active#(from(mark(X)))   
    mark#(from(X)) =#> mark#(X)   
    mark#(add(X, Y)) =#> active#(add(mark(X), mark(Y)))   
    mark#(add(X, Y)) =#> mark#(X)   
    mark#(add(X, Y)) =#> mark#(Y)   
    mark#(len(X)) =#> mark#(X)   

  P_2:

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

  P_3:

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

  P_4:

    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_5:

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

  P_6:

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

  P_7:

    len#(mark(X)) =#> len#(X)   
    len#(active(X)) =#> len#(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) and (P_7, 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) 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(len#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, R_0, minimal, f) by (P_8, R_0, minimal, f), where P_8 contains:

  add#(X, mark(Y)) =#> add#(X, Y)   
  add#(X, active(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) 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(add#) = 2 

Thus, we can orient the dependency pairs as follows:

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

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) 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(from#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_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(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_4, R_0, minimal, f) by (P_9, R_0, minimal, f), where P_9 contains:

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

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, 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(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_9, 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(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_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(fst#) = 1 

Thus, we can orient the dependency pairs as follows:

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

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:

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

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(fst#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(fst#(X, mark(Y))) = mark(Y) |> Y = nu(fst#(X, Y)) 
  nu(fst#(X, active(Y))) = active(Y) |> Y = nu(fst#(X, 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 (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#(fst(s(X), cons(Y, Z))) >? mark#(cons(Y, fst(X, Z))) 
  active#(from(X)) >? mark#(cons(X, from(s(X)))) 
  active#(add(0, X)) >? mark#(X) 
  mark#(fst(X, Y)) >? active#(fst(mark(X), mark(Y))) 
  mark#(fst(X, Y)) >? mark#(X) 
  mark#(fst(X, Y)) >? mark#(Y) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(from(X)) >? active#(from(mark(X))) 
  mark#(from(X)) >? mark#(X) 
  mark#(add(X, Y)) >? active#(add(mark(X), mark(Y))) 
  mark#(add(X, Y)) >? mark#(X) 
  mark#(add(X, Y)) >? mark#(Y) 
  mark#(len(X)) >? mark#(X) 
  active(fst(0, X)) >= mark(nil) 
  active(fst(s(X), cons(Y, Z))) >= mark(cons(Y, fst(X, Z))) 
  active(from(X)) >= mark(cons(X, from(s(X)))) 
  active(add(0, X)) >= mark(X) 
  active(add(s(X), Y)) >= mark(s(add(X, Y))) 
  active(len(nil)) >= mark(0) 
  active(len(cons(X, Y))) >= mark(s(len(Y))) 
  mark(fst(X, Y)) >= active(fst(mark(X), mark(Y))) 
  mark(0) >= active(0) 
  mark(nil) >= active(nil) 
  mark(s(X)) >= active(s(X)) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(from(X)) >= active(from(mark(X))) 
  mark(add(X, Y)) >= active(add(mark(X), mark(Y))) 
  mark(len(X)) >= active(len(mark(X))) 
  fst(mark(X), Y) >= fst(X, Y) 
  fst(X, mark(Y)) >= fst(X, Y) 
  fst(active(X), Y) >= fst(X, Y) 
  fst(X, active(Y)) >= fst(X, Y) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(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) 
  from(mark(X)) >= from(X) 
  from(active(X)) >= from(X) 
  add(mark(X), Y) >= add(X, Y) 
  add(X, mark(Y)) >= add(X, Y) 
  add(active(X), Y) >= add(X, Y) 
  add(X, active(Y)) >= add(X, Y) 
  len(mark(X)) >= len(X) 
  len(active(X)) >= len(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 
  add = \y0y1.2y0 + 2y1 
  cons = \y0y1.y0 
  from = \y0.2y0 
  fst = \y0y1.y0 + y1 
  len = \y0.1 + y0 
  mark = \y0.y0 
  mark# = \y0.2y0 
  nil = 0 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[active#(fst(s(_x0), cons(_x1, _x2)))]] = 2x1 >= 2x1 = [[mark#(cons(_x1, fst(_x0, _x2)))]] 
  [[active#(from(_x0))]] = 4x0 >= 2x0 = [[mark#(cons(_x0, from(s(_x0))))]] 
  [[active#(add(0, _x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(fst(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[active#(fst(mark(_x0), mark(_x1)))]] 
  [[mark#(fst(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(fst(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[mark#(_x1)]] 
  [[mark#(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(from(_x0))]] = 4x0 >= 4x0 = [[active#(from(mark(_x0)))]] 
  [[mark#(from(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[active#(add(mark(_x0), mark(_x1)))]] 
  [[mark#(add(_x0, _x1))]] = 4x0 + 4x1 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = 4x0 + 4x1 >= 2x1 = [[mark#(_x1)]] 
  [[mark#(len(_x0))]] = 2 + 2x0 > 2x0 = [[mark#(_x0)]] 
  [[active(fst(0, _x0))]] = x0 >= 0 = [[mark(nil)]] 
  [[active(fst(s(_x0), cons(_x1, _x2)))]] = x1 >= x1 = [[mark(cons(_x1, fst(_x0, _x2)))]] 
  [[active(from(_x0))]] = 2x0 >= x0 = [[mark(cons(_x0, from(s(_x0))))]] 
  [[active(add(0, _x0))]] = 2x0 >= x0 = [[mark(_x0)]] 
  [[active(add(s(_x0), _x1))]] = 2x1 >= 0 = [[mark(s(add(_x0, _x1)))]] 
  [[active(len(nil))]] = 1 >= 0 = [[mark(0)]] 
  [[active(len(cons(_x0, _x1)))]] = 1 + x0 >= 0 = [[mark(s(len(_x1)))]] 
  [[mark(fst(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[active(fst(mark(_x0), mark(_x1)))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(nil)]] = 0 >= 0 = [[active(nil)]] 
  [[mark(s(_x0))]] = 0 >= 0 = [[active(s(_x0))]] 
  [[mark(cons(_x0, _x1))]] = x0 >= x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(from(_x0))]] = 2x0 >= 2x0 = [[active(from(mark(_x0)))]] 
  [[mark(add(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[active(add(mark(_x0), mark(_x1)))]] 
  [[mark(len(_x0))]] = 1 + x0 >= 1 + x0 = [[active(len(mark(_x0)))]] 
  [[fst(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[fst(_x0, _x1)]] 
  [[fst(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[fst(_x0, _x1)]] 
  [[fst(active(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[fst(_x0, _x1)]] 
  [[fst(_x0, active(_x1))]] = x0 + x1 >= x0 + x1 = [[fst(_x0, _x1)]] 
  [[s(mark(_x0))]] = 0 >= 0 = [[s(_x0)]] 
  [[s(active(_x0))]] = 0 >= 0 = [[s(_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)]] 
  [[from(mark(_x0))]] = 2x0 >= 2x0 = [[from(_x0)]] 
  [[from(active(_x0))]] = 2x0 >= 2x0 = [[from(_x0)]] 
  [[add(mark(_x0), _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[add(_x0, _x1)]] 
  [[add(_x0, mark(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[add(_x0, _x1)]] 
  [[add(active(_x0), _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[add(_x0, _x1)]] 
  [[add(_x0, active(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[add(_x0, _x1)]] 
  [[len(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[len(_x0)]] 
  [[len(active(_x0))]] = 1 + x0 >= 1 + x0 = [[len(_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_11, R_0, minimal, formative), where P_11 consists of:

  active#(fst(s(X), cons(Y, Z))) =#> mark#(cons(Y, fst(X, Z)))   
  active#(from(X)) =#> mark#(cons(X, from(s(X))))   
  active#(add(0, X)) =#> mark#(X)   
  mark#(fst(X, Y)) =#> active#(fst(mark(X), mark(Y)))   
  mark#(fst(X, Y)) =#> mark#(X)   
  mark#(fst(X, Y)) =#> mark#(Y)   
  mark#(cons(X, Y)) =#> mark#(X)   
  mark#(from(X)) =#> active#(from(mark(X)))   
  mark#(from(X)) =#> mark#(X)   
  mark#(add(X, Y)) =#> active#(add(mark(X), mark(Y)))   
  mark#(add(X, Y)) =#> mark#(X)   
  mark#(add(X, Y)) =#> mark#(Y)   

Thus, the original system is terminating if (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(fst(0, X)) => mark(nil) 
  active(fst(s(X), cons(Y, Z))) => mark(cons(Y, fst(X, Z))) 
  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(len(nil)) => mark(0) 
  active(len(cons(X, Y))) => mark(s(len(Y))) 
  mark(fst(X, Y)) => active(fst(mark(X), mark(Y))) 
  mark(0) => active(0) 
  mark(nil) => active(nil) 
  mark(s(X)) => active(s(X)) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(from(X)) => active(from(mark(X))) 
  mark(add(X, Y)) => active(add(mark(X), mark(Y))) 
  mark(len(X)) => active(len(mark(X))) 
  fst(mark(X), Y) => fst(X, Y) 
  fst(X, mark(Y)) => fst(X, Y) 
  fst(active(X), Y) => fst(X, Y) 
  fst(X, active(Y)) => fst(X, Y) 
  s(mark(X)) => s(X) 
  s(active(X)) => s(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) 
  from(mark(X)) => from(X) 
  from(active(X)) => from(X) 
  add(mark(X), Y) => add(X, Y) 
  add(X, mark(Y)) => add(X, Y) 
  add(active(X), Y) => add(X, Y) 
  add(X, active(Y)) => add(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 (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#(fst(s(X), cons(Y, Z))) >? mark#(cons(Y, fst(X, Z))) 
  active#(from(X)) >? mark#(cons(X, from(s(X)))) 
  active#(add(0, X)) >? mark#(X) 
  mark#(fst(X, Y)) >? active#(fst(mark(X), mark(Y))) 
  mark#(fst(X, Y)) >? mark#(X) 
  mark#(fst(X, Y)) >? mark#(Y) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(from(X)) >? active#(from(mark(X))) 
  mark#(from(X)) >? mark#(X) 
  mark#(add(X, Y)) >? active#(add(mark(X), mark(Y))) 
  mark#(add(X, Y)) >? mark#(X) 
  mark#(add(X, Y)) >? mark#(Y) 
  active(fst(0, X)) >= mark(nil) 
  active(fst(s(X), cons(Y, Z))) >= mark(cons(Y, fst(X, Z))) 
  active(from(X)) >= mark(cons(X, from(s(X)))) 
  active(add(0, X)) >= mark(X) 
  active(add(s(X), Y)) >= mark(s(add(X, Y))) 
  active(len(nil)) >= mark(0) 
  active(len(cons(X, Y))) >= mark(s(len(Y))) 
  mark(fst(X, Y)) >= active(fst(mark(X), mark(Y))) 
  mark(0) >= active(0) 
  mark(nil) >= active(nil) 
  mark(s(X)) >= active(s(X)) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(from(X)) >= active(from(mark(X))) 
  mark(add(X, Y)) >= active(add(mark(X), mark(Y))) 
  mark(len(X)) >= active(len(mark(X))) 
  fst(mark(X), Y) >= fst(X, Y) 
  fst(X, mark(Y)) >= fst(X, Y) 
  fst(active(X), Y) >= fst(X, Y) 
  fst(X, active(Y)) >= fst(X, Y) 
  s(mark(X)) >= s(X) 
  s(active(X)) >= s(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) 
  from(mark(X)) >= from(X) 
  from(active(X)) >= from(X) 
  add(mark(X), Y) >= add(X, Y) 
  add(X, mark(Y)) >= add(X, Y) 
  add(active(X), Y) >= add(X, Y) 
  add(X, active(Y)) >= add(X, Y) 

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 
  add = \y0y1.y0 + y1 
  cons = \y0y1.y0 
  from = \y0.2y0 
  fst = \y0y1.1 + 2y0 + 2y1 
  len = \y0.0 
  mark = \y0.y0 
  mark# = \y0.y0 
  nil = 0 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[active#(fst(s(_x0), cons(_x1, _x2)))]] = 1 + 2x1 > x1 = [[mark#(cons(_x1, fst(_x0, _x2)))]] 
  [[active#(from(_x0))]] = 2x0 >= x0 = [[mark#(cons(_x0, from(s(_x0))))]] 
  [[active#(add(0, _x0))]] = x0 >= x0 = [[mark#(_x0)]] 
  [[mark#(fst(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[active#(fst(mark(_x0), mark(_x1)))]] 
  [[mark#(fst(_x0, _x1))]] = 1 + 2x0 + 2x1 > x0 = [[mark#(_x0)]] 
  [[mark#(fst(_x0, _x1))]] = 1 + 2x0 + 2x1 > x1 = [[mark#(_x1)]] 
  [[mark#(cons(_x0, _x1))]] = x0 >= x0 = [[mark#(_x0)]] 
  [[mark#(from(_x0))]] = 2x0 >= 2x0 = [[active#(from(mark(_x0)))]] 
  [[mark#(from(_x0))]] = 2x0 >= x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[active#(add(mark(_x0), mark(_x1)))]] 
  [[mark#(add(_x0, _x1))]] = x0 + x1 >= x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = x0 + x1 >= x1 = [[mark#(_x1)]] 
  [[active(fst(0, _x0))]] = 1 + 2x0 >= 0 = [[mark(nil)]] 
  [[active(fst(s(_x0), cons(_x1, _x2)))]] = 1 + 2x1 >= x1 = [[mark(cons(_x1, fst(_x0, _x2)))]] 
  [[active(from(_x0))]] = 2x0 >= x0 = [[mark(cons(_x0, from(s(_x0))))]] 
  [[active(add(0, _x0))]] = x0 >= x0 = [[mark(_x0)]] 
  [[active(add(s(_x0), _x1))]] = x1 >= 0 = [[mark(s(add(_x0, _x1)))]] 
  [[active(len(nil))]] = 0 >= 0 = [[mark(0)]] 
  [[active(len(cons(_x0, _x1)))]] = 0 >= 0 = [[mark(s(len(_x1)))]] 
  [[mark(fst(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[active(fst(mark(_x0), mark(_x1)))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(nil)]] = 0 >= 0 = [[active(nil)]] 
  [[mark(s(_x0))]] = 0 >= 0 = [[active(s(_x0))]] 
  [[mark(cons(_x0, _x1))]] = x0 >= x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(from(_x0))]] = 2x0 >= 2x0 = [[active(from(mark(_x0)))]] 
  [[mark(add(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[active(add(mark(_x0), mark(_x1)))]] 
  [[mark(len(_x0))]] = 0 >= 0 = [[active(len(mark(_x0)))]] 
  [[fst(mark(_x0), _x1)]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[fst(_x0, _x1)]] 
  [[fst(_x0, mark(_x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[fst(_x0, _x1)]] 
  [[fst(active(_x0), _x1)]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[fst(_x0, _x1)]] 
  [[fst(_x0, active(_x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[fst(_x0, _x1)]] 
  [[s(mark(_x0))]] = 0 >= 0 = [[s(_x0)]] 
  [[s(active(_x0))]] = 0 >= 0 = [[s(_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)]] 
  [[from(mark(_x0))]] = 2x0 >= 2x0 = [[from(_x0)]] 
  [[from(active(_x0))]] = 2x0 >= 2x0 = [[from(_x0)]] 
  [[add(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[add(_x0, _x1)]] 
  [[add(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[add(_x0, _x1)]] 
  [[add(active(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[add(_x0, _x1)]] 
  [[add(_x0, active(_x1))]] = x0 + x1 >= x0 + x1 = [[add(_x0, _x1)]] 

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:

  active#(from(X)) =#> mark#(cons(X, from(s(X))))   
  active#(add(0, X)) =#> mark#(X)   
  mark#(fst(X, Y)) =#> active#(fst(mark(X), mark(Y)))   
  mark#(cons(X, Y)) =#> mark#(X)   
  mark#(from(X)) =#> active#(from(mark(X)))   
  mark#(from(X)) =#> mark#(X)   
  mark#(add(X, Y)) =#> active#(add(mark(X), mark(Y)))   
  mark#(add(X, Y)) =#> mark#(X)   
  mark#(add(X, Y)) =#> mark#(Y)   

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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 3 
    * 1 : 2, 3, 4, 5, 6, 7, 8 
    * 2 :  
    * 3 : 2, 3, 4, 5, 6, 7, 8 
    * 4 : 0 
    * 5 : 2, 3, 4, 5, 6, 7, 8 
    * 6 : 1 
    * 7 : 2, 3, 4, 5, 6, 7, 8 
    * 8 : 2, 3, 4, 5, 6, 7, 8 

This graph has the following strongly connected components:

  P_13:

    active#(from(X)) =#> mark#(cons(X, from(s(X))))   
    active#(add(0, X)) =#> mark#(X)   
    mark#(cons(X, Y)) =#> mark#(X)   
    mark#(from(X)) =#> active#(from(mark(X)))   
    mark#(from(X)) =#> mark#(X)   
    mark#(add(X, Y)) =#> active#(add(mark(X), mark(Y)))   
    mark#(add(X, Y)) =#> mark#(X)   
    mark#(add(X, Y)) =#> mark#(Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_12, R_1, m, f) by (P_13, R_1, m, f).

Thus, the original system is terminating if (P_13, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_13, R_1, minimal, formative).

The formative rules of (P_13, R_1) are R_2 ::=

  active(fst(0, X)) => mark(nil) 
  active(fst(s(X), cons(Y, Z))) => mark(cons(Y, fst(X, Z))) 
  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(len(nil)) => mark(0) 
  active(len(cons(X, Y))) => mark(s(len(Y))) 
  mark(fst(X, Y)) => active(fst(mark(X), mark(Y))) 
  mark(0) => active(0) 
  mark(nil) => active(nil) 
  mark(s(X)) => active(s(X)) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(from(X)) => active(from(mark(X))) 
  mark(add(X, Y)) => active(add(mark(X), mark(Y))) 
  mark(len(X)) => active(len(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) 
  from(mark(X)) => from(X) 
  from(active(X)) => from(X) 
  add(mark(X), Y) => add(X, Y) 
  add(X, mark(Y)) => add(X, Y) 
  add(active(X), Y) => add(X, Y) 
  add(X, active(Y)) => add(X, Y) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_13, R_1, minimal, formative) by (P_13, R_2, minimal, formative).

Thus, the original system is terminating if (P_13, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_13, R_2, minimal, formative).

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

  active#(from(X)) >? mark#(cons(X, from(s(X)))) 
  active#(add(0, X)) >? mark#(X) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(from(X)) >? active#(from(mark(X))) 
  mark#(from(X)) >? mark#(X) 
  mark#(add(X, Y)) >? active#(add(mark(X), mark(Y))) 
  mark#(add(X, Y)) >? mark#(X) 
  mark#(add(X, Y)) >? mark#(Y) 
  active(fst(0, X)) >= mark(nil) 
  active(fst(s(X), cons(Y, Z))) >= mark(cons(Y, fst(X, Z))) 
  active(from(X)) >= mark(cons(X, from(s(X)))) 
  active(add(0, X)) >= mark(X) 
  active(add(s(X), Y)) >= mark(s(add(X, Y))) 
  active(len(nil)) >= mark(0) 
  active(len(cons(X, Y))) >= mark(s(len(Y))) 
  mark(fst(X, Y)) >= active(fst(mark(X), mark(Y))) 
  mark(0) >= active(0) 
  mark(nil) >= active(nil) 
  mark(s(X)) >= active(s(X)) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(from(X)) >= active(from(mark(X))) 
  mark(add(X, Y)) >= active(add(mark(X), mark(Y))) 
  mark(len(X)) >= active(len(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) 
  from(mark(X)) >= from(X) 
  from(active(X)) >= from(X) 
  add(mark(X), Y) >= add(X, Y) 
  add(X, mark(Y)) >= add(X, Y) 
  add(active(X), Y) >= add(X, Y) 
  add(X, active(Y)) >= add(X, Y) 

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 
  add = \y0y1.1 + y0 + y1 
  cons = \y0y1.2y0 
  from = \y0.2y0 
  fst = \y0y1.2y1 
  len = \y0.0 
  mark = \y0.y0 
  mark# = \y0.y0 
  nil = 0 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[active#(from(_x0))]] = 2x0 >= 2x0 = [[mark#(cons(_x0, from(s(_x0))))]] 
  [[active#(add(0, _x0))]] = 1 + x0 > x0 = [[mark#(_x0)]] 
  [[mark#(cons(_x0, _x1))]] = 2x0 >= x0 = [[mark#(_x0)]] 
  [[mark#(from(_x0))]] = 2x0 >= 2x0 = [[active#(from(mark(_x0)))]] 
  [[mark#(from(_x0))]] = 2x0 >= x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[active#(add(mark(_x0), mark(_x1)))]] 
  [[mark#(add(_x0, _x1))]] = 1 + x0 + x1 > x0 = [[mark#(_x0)]] 
  [[mark#(add(_x0, _x1))]] = 1 + x0 + x1 > x1 = [[mark#(_x1)]] 
  [[active(fst(0, _x0))]] = 2x0 >= 0 = [[mark(nil)]] 
  [[active(fst(s(_x0), cons(_x1, _x2)))]] = 4x1 >= 2x1 = [[mark(cons(_x1, fst(_x0, _x2)))]] 
  [[active(from(_x0))]] = 2x0 >= 2x0 = [[mark(cons(_x0, from(s(_x0))))]] 
  [[active(add(0, _x0))]] = 1 + x0 >= x0 = [[mark(_x0)]] 
  [[active(add(s(_x0), _x1))]] = 1 + x1 >= 0 = [[mark(s(add(_x0, _x1)))]] 
  [[active(len(nil))]] = 0 >= 0 = [[mark(0)]] 
  [[active(len(cons(_x0, _x1)))]] = 0 >= 0 = [[mark(s(len(_x1)))]] 
  [[mark(fst(_x0, _x1))]] = 2x1 >= 2x1 = [[active(fst(mark(_x0), mark(_x1)))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(nil)]] = 0 >= 0 = [[active(nil)]] 
  [[mark(s(_x0))]] = 0 >= 0 = [[active(s(_x0))]] 
  [[mark(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(from(_x0))]] = 2x0 >= 2x0 = [[active(from(mark(_x0)))]] 
  [[mark(add(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[active(add(mark(_x0), mark(_x1)))]] 
  [[mark(len(_x0))]] = 0 >= 0 = [[active(len(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)]] 
  [[from(mark(_x0))]] = 2x0 >= 2x0 = [[from(_x0)]] 
  [[from(active(_x0))]] = 2x0 >= 2x0 = [[from(_x0)]] 
  [[add(mark(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[add(_x0, _x1)]] 
  [[add(_x0, mark(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[add(_x0, _x1)]] 
  [[add(active(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[add(_x0, _x1)]] 
  [[add(_x0, active(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[add(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_13, R_2, minimal, formative) by (P_14, R_2, minimal, formative), where P_14 consists of:

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

Thus, the original system is terminating if (P_14, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_14, R_2, 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 
    * 2 : 0 
    * 3 : 1, 2, 3, 4 
    * 4 :  

This graph has the following strongly connected components:

  P_15:

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

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_14, R_2, m, f) by (P_15, R_2, m, f).

Thus, the original system is terminating if (P_15, R_2, minimal, formative) is finite.

We consider the dependency pair problem (P_15, R_2, minimal, formative).

The formative rules of (P_15, R_2) are R_3 ::=

  active(fst(0, X)) => mark(nil) 
  active(fst(s(X), cons(Y, Z))) => mark(cons(Y, fst(X, Z))) 
  active(from(X)) => mark(cons(X, from(s(X)))) 
  active(add(0, X)) => mark(X) 
  active(add(s(X), Y)) => mark(s(add(X, Y))) 
  active(len(nil)) => mark(0) 
  active(len(cons(X, Y))) => mark(s(len(Y))) 
  mark(fst(X, Y)) => active(fst(mark(X), mark(Y))) 
  mark(0) => active(0) 
  mark(nil) => active(nil) 
  mark(s(X)) => active(s(X)) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(from(X)) => active(from(mark(X))) 
  mark(add(X, Y)) => active(add(mark(X), mark(Y))) 
  mark(len(X)) => active(len(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) 
  from(mark(X)) => from(X) 
  from(active(X)) => from(X) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_15, R_2, minimal, formative) by (P_15, R_3, minimal, formative).

Thus, the original system is terminating if (P_15, R_3, minimal, formative) is finite.

We consider the dependency pair problem (P_15, R_3, minimal, formative).

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

  active#(from(X)) >? mark#(cons(X, from(s(X)))) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(from(X)) >? active#(from(mark(X))) 
  mark#(from(X)) >? mark#(X) 
  active(fst(0, X)) >= mark(nil) 
  active(fst(s(X), cons(Y, Z))) >= mark(cons(Y, fst(X, Z))) 
  active(from(X)) >= mark(cons(X, from(s(X)))) 
  active(add(0, X)) >= mark(X) 
  active(add(s(X), Y)) >= mark(s(add(X, Y))) 
  active(len(nil)) >= mark(0) 
  active(len(cons(X, Y))) >= mark(s(len(Y))) 
  mark(fst(X, Y)) >= active(fst(mark(X), mark(Y))) 
  mark(0) >= active(0) 
  mark(nil) >= active(nil) 
  mark(s(X)) >= active(s(X)) 
  mark(cons(X, Y)) >= active(cons(mark(X), Y)) 
  mark(from(X)) >= active(from(mark(X))) 
  mark(add(X, Y)) >= active(add(mark(X), mark(Y))) 
  mark(len(X)) >= active(len(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) 
  from(mark(X)) >= from(X) 
  from(active(X)) >= from(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 
  add = \y0y1.y1 
  cons = \y0y1.y0 
  from = \y0.1 + 2y0 
  fst = \y0y1.2 + y1 
  len = \y0.2 + 3y0 
  mark = \y0.y0 
  mark# = \y0.y0 
  nil = 2 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[active#(from(_x0))]] = 1 + 2x0 > x0 = [[mark#(cons(_x0, from(s(_x0))))]] 
  [[mark#(cons(_x0, _x1))]] = x0 >= x0 = [[mark#(_x0)]] 
  [[mark#(from(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[active#(from(mark(_x0)))]] 
  [[mark#(from(_x0))]] = 1 + 2x0 > x0 = [[mark#(_x0)]] 
  [[active(fst(0, _x0))]] = 2 + x0 >= 2 = [[mark(nil)]] 
  [[active(fst(s(_x0), cons(_x1, _x2)))]] = 2 + x1 >= x1 = [[mark(cons(_x1, fst(_x0, _x2)))]] 
  [[active(from(_x0))]] = 1 + 2x0 >= x0 = [[mark(cons(_x0, from(s(_x0))))]] 
  [[active(add(0, _x0))]] = x0 >= x0 = [[mark(_x0)]] 
  [[active(add(s(_x0), _x1))]] = x1 >= 0 = [[mark(s(add(_x0, _x1)))]] 
  [[active(len(nil))]] = 8 >= 0 = [[mark(0)]] 
  [[active(len(cons(_x0, _x1)))]] = 2 + 3x0 >= 0 = [[mark(s(len(_x1)))]] 
  [[mark(fst(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[active(fst(mark(_x0), mark(_x1)))]] 
  [[mark(0)]] = 0 >= 0 = [[active(0)]] 
  [[mark(nil)]] = 2 >= 2 = [[active(nil)]] 
  [[mark(s(_x0))]] = 0 >= 0 = [[active(s(_x0))]] 
  [[mark(cons(_x0, _x1))]] = x0 >= x0 = [[active(cons(mark(_x0), _x1))]] 
  [[mark(from(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[active(from(mark(_x0)))]] 
  [[mark(add(_x0, _x1))]] = x1 >= x1 = [[active(add(mark(_x0), mark(_x1)))]] 
  [[mark(len(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[active(len(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)]] 
  [[from(mark(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[from(_x0)]] 
  [[from(active(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[from(_x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_15, R_3, minimal, formative) by (P_16, R_3, minimal, formative), where P_16 consists of:

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

Thus, the original system is terminating if (P_16, R_3, minimal, formative) is finite.

We consider the dependency pair problem (P_16, R_3, 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 : 0, 1 
    * 1 :  

This graph has the following strongly connected components:

  P_17:

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

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_16, R_3, m, f) by (P_17, R_3, m, f).

Thus, the original system is terminating if (P_17, R_3, minimal, formative) is finite.

We consider the dependency pair problem (P_17, R_3, 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)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_17, R_3, minimal, f) by ({}, R_3, 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.