/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 
  dbl : [o] --> o 
  first : [o * o] --> o 
  half : [o] --> o 
  n!6220!6220first : [o * o] --> o 
  n!6220!6220s : [o] --> o 
  n!6220!6220terms : [o] --> o 
  nil : [] --> o 
  recip : [o] --> o 
  s : [o] --> o 
  sqr : [o] --> o 
  terms : [o] --> o 

  terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) => 0 
  sqr(s(X)) => s(add(sqr(X), dbl(X))) 
  dbl(0) => 0 
  dbl(s(X)) => s(s(dbl(X))) 
  add(0, X) => X 
  add(s(X), Y) => s(add(X, Y)) 
  first(0, X) => nil 
  first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
  half(0) => 0 
  half(s(0)) => 0 
  half(s(s(X))) => s(half(X)) 
  half(dbl(X)) => X 
  terms(X) => n!6220!6220terms(X) 
  s(X) => n!6220!6220s(X) 
  first(X, Y) => n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) => terms(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
  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] terms#(X) =#> sqr#(X)   
    1] sqr#(s(X)) =#> s#(add(sqr(X), dbl(X)))   
    2] sqr#(s(X)) =#> add#(sqr(X), dbl(X))   
    3] sqr#(s(X)) =#> sqr#(X)   
    4] sqr#(s(X)) =#> dbl#(X)   
    5] dbl#(s(X)) =#> s#(s(dbl(X)))   
    6] dbl#(s(X)) =#> s#(dbl(X))   
    7] dbl#(s(X)) =#> dbl#(X)   
    8] add#(s(X), Y) =#> s#(add(X, Y))   
    9] add#(s(X), Y) =#> add#(X, Y)   
    10] first#(s(X), cons(Y, Z)) =#> activate#(Z)   
    11] half#(s(s(X))) =#> s#(half(X))   
    12] half#(s(s(X))) =#> half#(X)   
    13] activate#(n!6220!6220terms(X)) =#> terms#(activate(X))   
    14] activate#(n!6220!6220terms(X)) =#> activate#(X)   
    15] activate#(n!6220!6220s(X)) =#> s#(activate(X))   
    16] activate#(n!6220!6220s(X)) =#> activate#(X)   
    17] activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
    18] activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
    19] activate#(n!6220!6220first(X, Y)) =#> activate#(Y)   

  Rules R_0:

    terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
    sqr(0) => 0 
    sqr(s(X)) => s(add(sqr(X), dbl(X))) 
    dbl(0) => 0 
    dbl(s(X)) => s(s(dbl(X))) 
    add(0, X) => X 
    add(s(X), Y) => s(add(X, Y)) 
    first(0, X) => nil 
    first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
    half(0) => 0 
    half(s(0)) => 0 
    half(s(s(X))) => s(half(X)) 
    half(dbl(X)) => X 
    terms(X) => n!6220!6220terms(X) 
    s(X) => n!6220!6220s(X) 
    first(X, Y) => n!6220!6220first(X, Y) 
    activate(n!6220!6220terms(X)) => terms(activate(X)) 
    activate(n!6220!6220s(X)) => s(activate(X)) 
    activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
    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 : 1, 2, 3, 4 
    * 1 :  
    * 2 : 8, 9 
    * 3 : 1, 2, 3, 4 
    * 4 : 5, 6, 7 
    * 5 :  
    * 6 :  
    * 7 : 5, 6, 7 
    * 8 :  
    * 9 : 8, 9 
    * 10 : 13, 14, 15, 16, 17, 18, 19 
    * 11 :  
    * 12 : 11, 12 
    * 13 : 0 
    * 14 : 13, 14, 15, 16, 17, 18, 19 
    * 15 :  
    * 16 : 13, 14, 15, 16, 17, 18, 19 
    * 17 : 10 
    * 18 : 13, 14, 15, 16, 17, 18, 19 
    * 19 : 13, 14, 15, 16, 17, 18, 19 

This graph has the following strongly connected components:

  P_1:

    sqr#(s(X)) =#> sqr#(X)   

  P_2:

    dbl#(s(X)) =#> dbl#(X)   

  P_3:

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

  P_4:

    first#(s(X), cons(Y, Z)) =#> activate#(Z)   
    activate#(n!6220!6220terms(X)) =#> activate#(X)   
    activate#(n!6220!6220s(X)) =#> activate#(X)   
    activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
    activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
    activate#(n!6220!6220first(X, Y)) =#> activate#(Y)   

  P_5:

    half#(s(s(X))) =#> half#(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) and (P_5, 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) 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(half#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(half#(s(s(X)))) = s(s(X)) |> X = nu(half#(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).

The formative rules of (P_4, R_0) are R_1 ::=

  terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) => 0 
  sqr(s(X)) => s(add(sqr(X), dbl(X))) 
  dbl(0) => 0 
  dbl(s(X)) => s(s(dbl(X))) 
  add(0, X) => X 
  add(s(X), Y) => s(add(X, Y)) 
  first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
  half(0) => 0 
  half(s(0)) => 0 
  half(s(s(X))) => s(half(X)) 
  half(dbl(X)) => X 
  terms(X) => n!6220!6220terms(X) 
  s(X) => n!6220!6220s(X) 
  first(X, Y) => n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) => terms(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
  activate(X) => X 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_4, R_0, minimal, formative) by (P_4, 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) and (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_1, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_4, R_1) are:

  terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) => 0 
  sqr(s(X)) => s(add(sqr(X), dbl(X))) 
  dbl(0) => 0 
  dbl(s(X)) => s(s(dbl(X))) 
  add(0, X) => X 
  add(s(X), Y) => s(add(X, Y)) 
  first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) => n!6220!6220terms(X) 
  s(X) => n!6220!6220s(X) 
  first(X, Y) => n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) => terms(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
  activate(X) => X 

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

  first#(s(X), cons(Y, Z)) >? activate#(Z) 
  activate#(n!6220!6220terms(X)) >? activate#(X) 
  activate#(n!6220!6220s(X)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? activate#(Y) 
  terms(X) >= cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) >= 0 
  sqr(s(X)) >= s(add(sqr(X), dbl(X))) 
  dbl(0) >= 0 
  dbl(s(X)) >= s(s(dbl(X))) 
  add(0, X) >= X 
  add(s(X), Y) >= s(add(X, Y)) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) >= n!6220!6220terms(X) 
  s(X) >= n!6220!6220s(X) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) >= terms(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(X) >= X 

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

We consider usable_rules with respect to the following argument filtering:

  recip(x_1) = recip() 

This leaves the following ordering requirements:

  first#(s(X), cons(Y, Z)) >= activate#(Z) 
  activate#(n!6220!6220terms(X)) > activate#(X) 
  activate#(n!6220!6220s(X)) >= activate#(X) 
  activate#(n!6220!6220first(X, Y)) >= first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >= activate#(X) 
  activate#(n!6220!6220first(X, Y)) >= activate#(Y) 
  terms(X) >= cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) >= n!6220!6220terms(X) 
  s(X) >= n!6220!6220s(X) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) >= terms(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(X) >= X 

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.y0 
  activate# = \y0.2y0 
  add = \y0y1.0 
  cons = \y0y1.y1 
  dbl = \y0.0 
  first = \y0y1.y0 + y1 
  first# = \y0y1.2y1 
  n!6220!6220first = \y0y1.y0 + y1 
  n!6220!6220s = \y0.y0 
  n!6220!6220terms = \y0.1 + y0 
  recip = \y0.0 
  s = \y0.y0 
  sqr = \y0.0 
  terms = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[first#(s(_x0), cons(_x1, _x2))]] = 2x2 >= 2x2 = [[activate#(_x2)]] 
  [[activate#(n!6220!6220terms(_x0))]] = 2 + 2x0 > 2x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220s(_x0))]] = 2x0 >= 2x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[first#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[activate#(_x1)]] 
  [[terms(_x0)]] = 1 + x0 >= 1 + x0 = [[cons(recip(sqr(_x0)), n!6220!6220terms(n!6220!6220s(_x0)))]] 
  [[first(s(_x0), cons(_x1, _x2))]] = x0 + x2 >= x0 + x2 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] 
  [[terms(_x0)]] = 1 + x0 >= 1 + x0 = [[n!6220!6220terms(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] 
  [[first(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[n!6220!6220first(_x0, _x1)]] 
  [[activate(n!6220!6220terms(_x0))]] = 1 + x0 >= 1 + x0 = [[terms(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(activate(_x0))]] 
  [[activate(n!6220!6220first(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[first(activate(_x0), activate(_x1))]] 
  [[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_4, R_1, minimal, formative) by (P_6, R_1, minimal, formative), where P_6 consists of:

  first#(s(X), cons(Y, Z)) =#> activate#(Z)   
  activate#(n!6220!6220s(X)) =#> activate#(X)   
  activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
  activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
  activate#(n!6220!6220first(X, Y)) =#> activate#(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_6, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_6, R_1, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_6, R_1) are:

  terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) => 0 
  sqr(s(X)) => s(add(sqr(X), dbl(X))) 
  dbl(0) => 0 
  dbl(s(X)) => s(s(dbl(X))) 
  add(0, X) => X 
  add(s(X), Y) => s(add(X, Y)) 
  first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) => n!6220!6220terms(X) 
  s(X) => n!6220!6220s(X) 
  first(X, Y) => n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) => terms(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
  activate(X) => X 

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

  first#(s(X), cons(Y, Z)) >? activate#(Z) 
  activate#(n!6220!6220s(X)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? activate#(Y) 
  terms(X) >= cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) >= 0 
  sqr(s(X)) >= s(add(sqr(X), dbl(X))) 
  dbl(0) >= 0 
  dbl(s(X)) >= s(s(dbl(X))) 
  add(0, X) >= X 
  add(s(X), Y) >= s(add(X, Y)) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) >= n!6220!6220terms(X) 
  s(X) >= n!6220!6220s(X) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) >= terms(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(X) >= X 

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

We consider usable_rules with respect to the following argument filtering:

  recip(x_1) = recip() 

This leaves the following ordering requirements:

  first#(s(X), cons(Y, Z)) >= activate#(Z) 
  activate#(n!6220!6220s(X)) > activate#(X) 
  activate#(n!6220!6220first(X, Y)) >= first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >= activate#(X) 
  activate#(n!6220!6220first(X, Y)) >= activate#(Y) 
  terms(X) >= cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) >= n!6220!6220terms(X) 
  s(X) >= n!6220!6220s(X) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) >= terms(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(X) >= X 

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.y0 
  activate# = \y0.y0 
  add = \y0y1.0 
  cons = \y0y1.2y1 
  dbl = \y0.0 
  first = \y0y1.y0 + y1 
  first# = \y0y1.y1 
  n!6220!6220first = \y0y1.y0 + y1 
  n!6220!6220s = \y0.1 + 2y0 
  n!6220!6220terms = \y0.0 
  recip = \y0.0 
  s = \y0.1 + 2y0 
  sqr = \y0.0 
  terms = \y0.0 

Using this interpretation, the requirements translate to:

  [[first#(s(_x0), cons(_x1, _x2))]] = 2x2 >= x2 = [[activate#(_x2)]] 
  [[activate#(n!6220!6220s(_x0))]] = 1 + 2x0 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x0 + x1 >= x1 = [[first#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x0 + x1 >= x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = x0 + x1 >= x1 = [[activate#(_x1)]] 
  [[terms(_x0)]] = 0 >= 0 = [[cons(recip(sqr(_x0)), n!6220!6220terms(n!6220!6220s(_x0)))]] 
  [[first(s(_x0), cons(_x1, _x2))]] = 1 + 2x0 + 2x2 >= 2x0 + 2x2 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] 
  [[terms(_x0)]] = 0 >= 0 = [[n!6220!6220terms(_x0)]] 
  [[s(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[n!6220!6220s(_x0)]] 
  [[first(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[n!6220!6220first(_x0, _x1)]] 
  [[activate(n!6220!6220terms(_x0))]] = 0 >= 0 = [[terms(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[s(activate(_x0))]] 
  [[activate(n!6220!6220first(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[first(activate(_x0), activate(_x1))]] 
  [[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_1, minimal, formative) by (P_7, R_1, minimal, formative), where P_7 consists of:

  first#(s(X), cons(Y, Z)) =#> activate#(Z)   
  activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y))   
  activate#(n!6220!6220first(X, Y)) =#> activate#(X)   
  activate#(n!6220!6220first(X, Y)) =#> activate#(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_7, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_7, R_1, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_7, R_1) are:

  terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) => 0 
  sqr(s(X)) => s(add(sqr(X), dbl(X))) 
  dbl(0) => 0 
  dbl(s(X)) => s(s(dbl(X))) 
  add(0, X) => X 
  add(s(X), Y) => s(add(X, Y)) 
  first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) => n!6220!6220terms(X) 
  s(X) => n!6220!6220s(X) 
  first(X, Y) => n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) => terms(activate(X)) 
  activate(n!6220!6220s(X)) => s(activate(X)) 
  activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 
  activate(X) => X 

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

  first#(s(X), cons(Y, Z)) >? activate#(Z) 
  activate#(n!6220!6220first(X, Y)) >? first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >? activate#(X) 
  activate#(n!6220!6220first(X, Y)) >? activate#(Y) 
  terms(X) >= cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  sqr(0) >= 0 
  sqr(s(X)) >= s(add(sqr(X), dbl(X))) 
  dbl(0) >= 0 
  dbl(s(X)) >= s(s(dbl(X))) 
  add(0, X) >= X 
  add(s(X), Y) >= s(add(X, Y)) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) >= n!6220!6220terms(X) 
  s(X) >= n!6220!6220s(X) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) >= terms(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(X) >= X 

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

We consider usable_rules with respect to the following argument filtering:

  recip(x_1) = recip() 

This leaves the following ordering requirements:

  first#(s(X), cons(Y, Z)) >= activate#(Z) 
  activate#(n!6220!6220first(X, Y)) > first#(activate(X), activate(Y)) 
  activate#(n!6220!6220first(X, Y)) >= activate#(X) 
  activate#(n!6220!6220first(X, Y)) >= activate#(Y) 
  terms(X) >= cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) 
  first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) 
  terms(X) >= n!6220!6220terms(X) 
  s(X) >= n!6220!6220s(X) 
  first(X, Y) >= n!6220!6220first(X, Y) 
  activate(n!6220!6220terms(X)) >= terms(activate(X)) 
  activate(n!6220!6220s(X)) >= s(activate(X)) 
  activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) 
  activate(X) >= X 

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.y0 
  activate# = \y0.y0 
  add = \y0y1.1 
  cons = \y0y1.2y1 
  dbl = \y0.0 
  first = \y0y1.2 + y0 + y1 
  first# = \y0y1.y1 
  n!6220!6220first = \y0y1.2 + y0 + y1 
  n!6220!6220s = \y0.2 + 2y0 
  n!6220!6220terms = \y0.0 
  recip = \y0.0 
  s = \y0.2 + 2y0 
  sqr = \y0.0 
  terms = \y0.0 

Using this interpretation, the requirements translate to:

  [[first#(s(_x0), cons(_x1, _x2))]] = 2x2 >= x2 = [[activate#(_x2)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2 + x0 + x1 > x1 = [[first#(activate(_x0), activate(_x1))]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2 + x0 + x1 > x0 = [[activate#(_x0)]] 
  [[activate#(n!6220!6220first(_x0, _x1))]] = 2 + x0 + x1 > x1 = [[activate#(_x1)]] 
  [[terms(_x0)]] = 0 >= 0 = [[cons(recip(sqr(_x0)), n!6220!6220terms(n!6220!6220s(_x0)))]] 
  [[first(s(_x0), cons(_x1, _x2))]] = 4 + 2x0 + 2x2 >= 4 + 2x0 + 2x2 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] 
  [[terms(_x0)]] = 0 >= 0 = [[n!6220!6220terms(_x0)]] 
  [[s(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[n!6220!6220s(_x0)]] 
  [[first(_x0, _x1)]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[n!6220!6220first(_x0, _x1)]] 
  [[activate(n!6220!6220terms(_x0))]] = 0 >= 0 = [[terms(activate(_x0))]] 
  [[activate(n!6220!6220s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[s(activate(_x0))]] 
  [[activate(n!6220!6220first(_x0, _x1))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[first(activate(_x0), activate(_x1))]] 
  [[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_7, R_1, minimal, formative) by (P_8, R_1, minimal, formative), where P_8 consists of:

  first#(s(X), cons(Y, Z)) =#> activate#(Z)   

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_8, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_8, 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 :  

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

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(add#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(add#(s(X), Y)) = s(X) |> X = nu(add#(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(dbl#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(dbl#(s(X))) = s(X) |> X = nu(dbl#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(sqr#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sqr#(s(X))) = s(X) |> X = nu(sqr#(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.