/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 
  activate : [o] --> o 
  add : [o * o] --> o 
  cons : [o * o] --> o 
  from : [o] --> o 
  fst : [o * o] --> o 
  len : [o] --> o 
  n!6220!6220add : [o * o] --> o 
  n!6220!6220from : [o] --> o 
  n!6220!6220fst : [o * o] --> o 
  n!6220!6220len : [o] --> o 
  n!6220!6220s : [o] --> o 
  nil : [] --> o 
  s : [o] --> o 

  fst(0, X) => nil 
  fst(s(X), cons(Y, Z)) => cons(Y, n!6220!6220fst(activate(X), activate(Z))) 
  from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
  add(0, X) => X 
  add(s(X), Y) => s(n!6220!6220add(activate(X), Y)) 
  len(nil) => 0 
  len(cons(X, Y)) => s(n!6220!6220len(activate(Y))) 
  fst(X, Y) => n!6220!6220fst(X, Y) 
  from(X) => n!6220!6220from(X) 
  s(X) => n!6220!6220s(X) 
  add(X, Y) => n!6220!6220add(X, Y) 
  len(X) => n!6220!6220len(X) 
  activate(n!6220!6220fst(X, Y)) => fst(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) => from(activate(X)) 
  activate(n!6220!6220s(X)) => s(X) 
  activate(n!6220!6220add(X, Y)) => add(activate(X), activate(Y)) 
  activate(n!6220!6220len(X)) => len(activate(X)) 
  activate(X) => 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] fst#(s(X), cons(Y, Z)) =#> activate#(X)   
    1] fst#(s(X), cons(Y, Z)) =#> activate#(Z)   
    2] add#(s(X), Y) =#> s#(n!6220!6220add(activate(X), Y))   
    3] add#(s(X), Y) =#> activate#(X)   
    4] len#(cons(X, Y)) =#> s#(n!6220!6220len(activate(Y)))   
    5] len#(cons(X, Y)) =#> activate#(Y)   
    6] activate#(n!6220!6220fst(X, Y)) =#> fst#(activate(X), activate(Y))   
    7] activate#(n!6220!6220fst(X, Y)) =#> activate#(X)   
    8] activate#(n!6220!6220fst(X, Y)) =#> activate#(Y)   
    9] activate#(n!6220!6220from(X)) =#> from#(activate(X))   
    10] activate#(n!6220!6220from(X)) =#> activate#(X)   
    11] activate#(n!6220!6220s(X)) =#> s#(X)   
    12] activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   
    13] activate#(n!6220!6220add(X, Y)) =#> activate#(X)   
    14] activate#(n!6220!6220add(X, Y)) =#> activate#(Y)   
    15] activate#(n!6220!6220len(X)) =#> len#(activate(X))   
    16] activate#(n!6220!6220len(X)) =#> activate#(X)   

  Rules R_0:

    fst(0, X) => nil 
    fst(s(X), cons(Y, Z)) => cons(Y, n!6220!6220fst(activate(X), activate(Z))) 
    from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 
    add(0, X) => X 
    add(s(X), Y) => s(n!6220!6220add(activate(X), Y)) 
    len(nil) => 0 
    len(cons(X, Y)) => s(n!6220!6220len(activate(Y))) 
    fst(X, Y) => n!6220!6220fst(X, Y) 
    from(X) => n!6220!6220from(X) 
    s(X) => n!6220!6220s(X) 
    add(X, Y) => n!6220!6220add(X, Y) 
    len(X) => n!6220!6220len(X) 
    activate(n!6220!6220fst(X, Y)) => fst(activate(X), activate(Y)) 
    activate(n!6220!6220from(X)) => from(activate(X)) 
    activate(n!6220!6220s(X)) => s(X) 
    activate(n!6220!6220add(X, Y)) => add(activate(X), activate(Y)) 
    activate(n!6220!6220len(X)) => len(activate(X)) 
    activate(X) => 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 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 1 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 2 :  
    * 3 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 4 :  
    * 5 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 6 : 0, 1 
    * 7 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 8 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 9 :  
    * 10 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 11 :  
    * 12 : 2, 3 
    * 13 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 14 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 
    * 15 : 4, 5 
    * 16 : 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 

This graph has the following strongly connected components:

  P_1:

    fst#(s(X), cons(Y, Z)) =#> activate#(X)   
    fst#(s(X), cons(Y, Z)) =#> activate#(Z)   
    add#(s(X), Y) =#> activate#(X)   
    len#(cons(X, Y)) =#> activate#(Y)   
    activate#(n!6220!6220fst(X, Y)) =#> fst#(activate(X), activate(Y))   
    activate#(n!6220!6220fst(X, Y)) =#> activate#(X)   
    activate#(n!6220!6220fst(X, Y)) =#> activate#(Y)   
    activate#(n!6220!6220from(X)) =#> activate#(X)   
    activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   
    activate#(n!6220!6220add(X, Y)) =#> activate#(X)   
    activate#(n!6220!6220add(X, Y)) =#> activate#(Y)   
    activate#(n!6220!6220len(X)) =#> len#(activate(X))   
    activate#(n!6220!6220len(X)) =#> activate#(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).

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:

  fst#(s(X), cons(Y, Z)) >? activate#(X) 
  fst#(s(X), cons(Y, Z)) >? activate#(Z) 
  add#(s(X), Y) >? activate#(X) 
  len#(cons(X, Y)) >? activate#(Y) 
  activate#(n!6220!6220fst(X, Y)) >? fst#(activate(X), activate(Y)) 
  activate#(n!6220!6220fst(X, Y)) >? activate#(X) 
  activate#(n!6220!6220fst(X, Y)) >? activate#(Y) 
  activate#(n!6220!6220from(X)) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? add#(activate(X), activate(Y)) 
  activate#(n!6220!6220add(X, Y)) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? activate#(Y) 
  activate#(n!6220!6220len(X)) >? len#(activate(X)) 
  activate#(n!6220!6220len(X)) >? activate#(X) 
  fst(0, X) >= nil 
  fst(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220fst(activate(X), activate(Z))) 
  from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) 
  add(0, X) >= X 
  add(s(X), Y) >= s(n!6220!6220add(activate(X), Y)) 
  len(nil) >= 0 
  len(cons(X, Y)) >= s(n!6220!6220len(activate(Y))) 
  fst(X, Y) >= n!6220!6220fst(X, Y) 
  from(X) >= n!6220!6220from(X) 
  s(X) >= n!6220!6220s(X) 
  add(X, Y) >= n!6220!6220add(X, Y) 
  len(X) >= n!6220!6220len(X) 
  activate(n!6220!6220fst(X, Y)) >= fst(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) >= from(activate(X)) 
  activate(n!6220!6220s(X)) >= s(X) 
  activate(n!6220!6220add(X, Y)) >= add(activate(X), activate(Y)) 
  activate(n!6220!6220len(X)) >= len(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  0 = 1 
  activate = \y0.y0 
  activate# = \y0.y0 
  add = \y0y1.y1 + 2y0 
  add# = \y0y1.2y0 
  cons = \y0y1.y1 
  from = \y0.y0 
  fst = \y0y1.1 + y0 + 2y1 
  fst# = \y0y1.y0 + y1 
  len = \y0.y0 
  len# = \y0.y0 
  n!6220!6220add = \y0y1.y1 + 2y0 
  n!6220!6220from = \y0.y0 
  n!6220!6220fst = \y0y1.1 + y0 + 2y1 
  n!6220!6220len = \y0.y0 
  n!6220!6220s = \y0.y0 
  nil = 2 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[fst#(s(_x0), cons(_x1, _x2))]] = x0 + x2 >= x0 = [[activate#(_x0)]] 
  [[fst#(s(_x0), cons(_x1, _x2))]] = x0 + x2 >= x2 = [[activate#(_x2)]] 
  [[add#(s(_x0), _x1)]] = 2x0 >= x0 = [[activate#(_x0)]] 
  [[len#(cons(_x0, _x1))]] = x1 >= x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220fst(_x0, _x1))]] = 1 + x0 + 2x1 > x0 + x1 = [[fst#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220fst(_x0, _x1))]] = 1 + x0 + 2x1 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220fst(_x0, _x1))]] = 1 + x0 + 2x1 > x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220from(_x0))]] = x0 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x1 + 2x0 >= 2x0 = [[add#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x1 + 2x0 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x1 + 2x0 >= x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220len(_x0))]] = x0 >= x0 = [[len#(activate(_x0))]] 
  [[activate#(n!6220!6220len(_x0))]] = x0 >= x0 = [[activate#(_x0)]] 
  [[fst(0, _x0)]] = 2 + 2x0 >= 2 = [[nil]] 
  [[fst(s(_x0), cons(_x1, _x2))]] = 1 + x0 + 2x2 >= 1 + x0 + 2x2 = [[cons(_x1, n!6220!6220fst(activate(_x0), activate(_x2)))]] 
  [[from(_x0)]] = x0 >= x0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] 
  [[add(0, _x0)]] = 2 + x0 >= x0 = [[_x0]] 
  [[add(s(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[s(n!6220!6220add(activate(_x0), _x1))]] 
  [[len(nil)]] = 2 >= 1 = [[0]] 
  [[len(cons(_x0, _x1))]] = x1 >= x1 = [[s(n!6220!6220len(activate(_x1)))]] 
  [[fst(_x0, _x1)]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[n!6220!6220fst(_x0, _x1)]] 
  [[from(_x0)]] = x0 >= x0 = [[n!6220!6220from(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[add(_x0, _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[n!6220!6220add(_x0, _x1)]] 
  [[len(_x0)]] = x0 >= x0 = [[n!6220!6220len(_x0)]] 
  [[activate(n!6220!6220fst(_x0, _x1))]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[fst(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220from(_x0))]] = x0 >= x0 = [[from(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[activate(n!6220!6220add(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[add(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220len(_x0))]] = x0 >= x0 = [[len(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_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_2, R_0, minimal, formative), where P_2 consists of:

  fst#(s(X), cons(Y, Z)) =#> activate#(X)   
  fst#(s(X), cons(Y, Z)) =#> activate#(Z)   
  add#(s(X), Y) =#> activate#(X)   
  len#(cons(X, Y)) =#> activate#(Y)   
  activate#(n!6220!6220from(X)) =#> activate#(X)   
  activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   
  activate#(n!6220!6220add(X, Y)) =#> activate#(X)   
  activate#(n!6220!6220add(X, Y)) =#> activate#(Y)   
  activate#(n!6220!6220len(X)) =#> len#(activate(X))   
  activate#(n!6220!6220len(X)) =#> activate#(X)   

Thus, the original system is terminating if (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

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

    * 0 : 4, 5, 6, 7, 8, 9 
    * 1 : 4, 5, 6, 7, 8, 9 
    * 2 : 4, 5, 6, 7, 8, 9 
    * 3 : 4, 5, 6, 7, 8, 9 
    * 4 : 4, 5, 6, 7, 8, 9 
    * 5 : 2 
    * 6 : 4, 5, 6, 7, 8, 9 
    * 7 : 4, 5, 6, 7, 8, 9 
    * 8 : 3 
    * 9 : 4, 5, 6, 7, 8, 9 

This graph has the following strongly connected components:

  P_3:

    add#(s(X), Y) =#> activate#(X)   
    len#(cons(X, Y)) =#> activate#(Y)   
    activate#(n!6220!6220from(X)) =#> activate#(X)   
    activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   
    activate#(n!6220!6220add(X, Y)) =#> activate#(X)   
    activate#(n!6220!6220add(X, Y)) =#> activate#(Y)   
    activate#(n!6220!6220len(X)) =#> len#(activate(X))   
    activate#(n!6220!6220len(X)) =#> activate#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_2, R_0, m, f) by (P_3, R_0, m, f).

Thus, the original system is terminating if (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, 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:

  add#(s(X), Y) >? activate#(X) 
  len#(cons(X, Y)) >? activate#(Y) 
  activate#(n!6220!6220from(X)) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? add#(activate(X), activate(Y)) 
  activate#(n!6220!6220add(X, Y)) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? activate#(Y) 
  activate#(n!6220!6220len(X)) >? len#(activate(X)) 
  activate#(n!6220!6220len(X)) >? activate#(X) 
  fst(0, X) >= nil 
  fst(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220fst(activate(X), activate(Z))) 
  from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) 
  add(0, X) >= X 
  add(s(X), Y) >= s(n!6220!6220add(activate(X), Y)) 
  len(nil) >= 0 
  len(cons(X, Y)) >= s(n!6220!6220len(activate(Y))) 
  fst(X, Y) >= n!6220!6220fst(X, Y) 
  from(X) >= n!6220!6220from(X) 
  s(X) >= n!6220!6220s(X) 
  add(X, Y) >= n!6220!6220add(X, Y) 
  len(X) >= n!6220!6220len(X) 
  activate(n!6220!6220fst(X, Y)) >= fst(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) >= from(activate(X)) 
  activate(n!6220!6220s(X)) >= s(X) 
  activate(n!6220!6220add(X, Y)) >= add(activate(X), activate(Y)) 
  activate(n!6220!6220len(X)) >= len(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.y0 
  activate# = \y0.y0 
  add = \y0y1.y0 + y1 
  add# = \y0y1.y0 
  cons = \y0y1.y1 
  from = \y0.2y0 
  fst = \y0y1.0 
  len = \y0.1 + y0 
  len# = \y0.y0 
  n!6220!6220add = \y0y1.y0 + y1 
  n!6220!6220from = \y0.2y0 
  n!6220!6220fst = \y0y1.0 
  n!6220!6220len = \y0.1 + y0 
  n!6220!6220s = \y0.y0 
  nil = 0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[add#(s(_x0), _x1)]] = x0 >= x0 = [[activate#(_x0)]] 
  [[len#(cons(_x0, _x1))]] = x1 >= x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220from(_x0))]] = 2x0 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x0 + x1 >= x0 = [[add#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x0 + x1 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x0 + x1 >= x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220len(_x0))]] = 1 + x0 > x0 = [[len#(activate(_x0))]] 
  [[activate#(n!6220!6220len(_x0))]] = 1 + x0 > x0 = [[activate#(_x0)]] 
  [[fst(0, _x0)]] = 0 >= 0 = [[nil]] 
  [[fst(s(_x0), cons(_x1, _x2))]] = 0 >= 0 = [[cons(_x1, n!6220!6220fst(activate(_x0), activate(_x2)))]] 
  [[from(_x0)]] = 2x0 >= 2x0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] 
  [[add(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[add(s(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[s(n!6220!6220add(activate(_x0), _x1))]] 
  [[len(nil)]] = 1 >= 0 = [[0]] 
  [[len(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[s(n!6220!6220len(activate(_x1)))]] 
  [[fst(_x0, _x1)]] = 0 >= 0 = [[n!6220!6220fst(_x0, _x1)]] 
  [[from(_x0)]] = 2x0 >= 2x0 = [[n!6220!6220from(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[add(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[n!6220!6220add(_x0, _x1)]] 
  [[len(_x0)]] = 1 + x0 >= 1 + x0 = [[n!6220!6220len(_x0)]] 
  [[activate(n!6220!6220fst(_x0, _x1))]] = 0 >= 0 = [[fst(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220from(_x0))]] = 2x0 >= 2x0 = [[from(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[activate(n!6220!6220add(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[add(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220len(_x0))]] = 1 + x0 >= 1 + x0 = [[len(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_4, R_0, minimal, formative), where P_4 consists of:

  add#(s(X), Y) =#> activate#(X)   
  len#(cons(X, Y)) =#> activate#(Y)   
  activate#(n!6220!6220from(X)) =#> activate#(X)   
  activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   
  activate#(n!6220!6220add(X, Y)) =#> activate#(X)   
  activate#(n!6220!6220add(X, Y)) =#> activate#(Y)   

Thus, the original system is terminating if (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, 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 : 2, 3, 4, 5 
    * 1 : 2, 3, 4, 5 
    * 2 : 2, 3, 4, 5 
    * 3 : 0 
    * 4 : 2, 3, 4, 5 
    * 5 : 2, 3, 4, 5 

This graph has the following strongly connected components:

  P_5:

    add#(s(X), Y) =#> activate#(X)   
    activate#(n!6220!6220from(X)) =#> activate#(X)   
    activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   
    activate#(n!6220!6220add(X, Y)) =#> activate#(X)   
    activate#(n!6220!6220add(X, Y)) =#> activate#(Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_4, R_0, m, f) by (P_5, R_0, m, f).

Thus, the original system is terminating if (P_5, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_5, 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:

  add#(s(X), Y) >? activate#(X) 
  activate#(n!6220!6220from(X)) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? add#(activate(X), activate(Y)) 
  activate#(n!6220!6220add(X, Y)) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? activate#(Y) 
  fst(0, X) >= nil 
  fst(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220fst(activate(X), activate(Z))) 
  from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) 
  add(0, X) >= X 
  add(s(X), Y) >= s(n!6220!6220add(activate(X), Y)) 
  len(nil) >= 0 
  len(cons(X, Y)) >= s(n!6220!6220len(activate(Y))) 
  fst(X, Y) >= n!6220!6220fst(X, Y) 
  from(X) >= n!6220!6220from(X) 
  s(X) >= n!6220!6220s(X) 
  add(X, Y) >= n!6220!6220add(X, Y) 
  len(X) >= n!6220!6220len(X) 
  activate(n!6220!6220fst(X, Y)) >= fst(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) >= from(activate(X)) 
  activate(n!6220!6220s(X)) >= s(X) 
  activate(n!6220!6220add(X, Y)) >= add(activate(X), activate(Y)) 
  activate(n!6220!6220len(X)) >= len(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.y0 
  activate# = \y0.y0 
  add = \y0y1.y0 + 2y1 
  add# = \y0y1.y0 
  cons = \y0y1.0 
  from = \y0.1 + 2y0 
  fst = \y0y1.0 
  len = \y0.0 
  n!6220!6220add = \y0y1.y0 + 2y1 
  n!6220!6220from = \y0.1 + 2y0 
  n!6220!6220fst = \y0y1.0 
  n!6220!6220len = \y0.0 
  n!6220!6220s = \y0.y0 
  nil = 0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[add#(s(_x0), _x1)]] = x0 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220from(_x0))]] = 1 + 2x0 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x0 + 2x1 >= x0 = [[add#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x0 + 2x1 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = x0 + 2x1 >= x1 = [[activate#(_x1)]] 
  [[fst(0, _x0)]] = 0 >= 0 = [[nil]] 
  [[fst(s(_x0), cons(_x1, _x2))]] = 0 >= 0 = [[cons(_x1, n!6220!6220fst(activate(_x0), activate(_x2)))]] 
  [[from(_x0)]] = 1 + 2x0 >= 0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] 
  [[add(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[add(s(_x0), _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[s(n!6220!6220add(activate(_x0), _x1))]] 
  [[len(nil)]] = 0 >= 0 = [[0]] 
  [[len(cons(_x0, _x1))]] = 0 >= 0 = [[s(n!6220!6220len(activate(_x1)))]] 
  [[fst(_x0, _x1)]] = 0 >= 0 = [[n!6220!6220fst(_x0, _x1)]] 
  [[from(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[n!6220!6220from(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[add(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[n!6220!6220add(_x0, _x1)]] 
  [[len(_x0)]] = 0 >= 0 = [[n!6220!6220len(_x0)]] 
  [[activate(n!6220!6220fst(_x0, _x1))]] = 0 >= 0 = [[fst(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220from(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[from(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[activate(n!6220!6220add(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[add(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220len(_x0))]] = 0 >= 0 = [[len(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, minimal, formative) by (P_6, R_0, minimal, formative), where P_6 consists of:

  add#(s(X), Y) =#> activate#(X)   
  activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   
  activate#(n!6220!6220add(X, Y)) =#> activate#(X)   
  activate#(n!6220!6220add(X, Y)) =#> activate#(Y)   

Thus, the original system is terminating if (P_6, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_6, 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:

  add#(s(X), Y) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? add#(activate(X), activate(Y)) 
  activate#(n!6220!6220add(X, Y)) >? activate#(X) 
  activate#(n!6220!6220add(X, Y)) >? activate#(Y) 
  fst(0, X) >= nil 
  fst(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220fst(activate(X), activate(Z))) 
  from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) 
  add(0, X) >= X 
  add(s(X), Y) >= s(n!6220!6220add(activate(X), Y)) 
  len(nil) >= 0 
  len(cons(X, Y)) >= s(n!6220!6220len(activate(Y))) 
  fst(X, Y) >= n!6220!6220fst(X, Y) 
  from(X) >= n!6220!6220from(X) 
  s(X) >= n!6220!6220s(X) 
  add(X, Y) >= n!6220!6220add(X, Y) 
  len(X) >= n!6220!6220len(X) 
  activate(n!6220!6220fst(X, Y)) >= fst(activate(X), activate(Y)) 
  activate(n!6220!6220from(X)) >= from(activate(X)) 
  activate(n!6220!6220s(X)) >= s(X) 
  activate(n!6220!6220add(X, Y)) >= add(activate(X), activate(Y)) 
  activate(n!6220!6220len(X)) >= len(activate(X)) 
  activate(X) >= X 

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

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.y0 
  activate# = \y0.y0 
  add = \y0y1.1 + y0 + y1 
  add# = \y0y1.1 + y0 
  cons = \y0y1.0 
  from = \y0.0 
  fst = \y0y1.0 
  len = \y0.0 
  n!6220!6220add = \y0y1.1 + y0 + y1 
  n!6220!6220from = \y0.0 
  n!6220!6220fst = \y0y1.0 
  n!6220!6220len = \y0.0 
  n!6220!6220s = \y0.y0 
  nil = 0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[add#(s(_x0), _x1)]] = 1 + x0 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 = [[add#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = 1 + x0 + x1 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220add(_x0, _x1))]] = 1 + x0 + x1 > x1 = [[activate#(_x1)]] 
  [[fst(0, _x0)]] = 0 >= 0 = [[nil]] 
  [[fst(s(_x0), cons(_x1, _x2))]] = 0 >= 0 = [[cons(_x1, n!6220!6220fst(activate(_x0), activate(_x2)))]] 
  [[from(_x0)]] = 0 >= 0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] 
  [[add(0, _x0)]] = 1 + x0 >= x0 = [[_x0]] 
  [[add(s(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(n!6220!6220add(activate(_x0), _x1))]] 
  [[len(nil)]] = 0 >= 0 = [[0]] 
  [[len(cons(_x0, _x1))]] = 0 >= 0 = [[s(n!6220!6220len(activate(_x1)))]] 
  [[fst(_x0, _x1)]] = 0 >= 0 = [[n!6220!6220fst(_x0, _x1)]] 
  [[from(_x0)]] = 0 >= 0 = [[n!6220!6220from(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[add(_x0, _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[n!6220!6220add(_x0, _x1)]] 
  [[len(_x0)]] = 0 >= 0 = [[n!6220!6220len(_x0)]] 
  [[activate(n!6220!6220fst(_x0, _x1))]] = 0 >= 0 = [[fst(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220from(_x0))]] = 0 >= 0 = [[from(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(_x0)]] 
  [[activate(n!6220!6220add(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[add(activate(_x0), activate(_x1))]] 
  [[activate(n!6220!6220len(_x0))]] = 0 >= 0 = [[len(activate(_x0))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_0, minimal, formative) by (P_7, R_0, minimal, formative), where P_7 consists of:

  activate#(n!6220!6220add(X, Y)) =#> add#(activate(X), activate(Y))   

Thus, the original system is terminating if (P_7, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_7, 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 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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.