/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 
  cons : [o * o] --> o 
  filter : [o * o * o] --> o 
  n!6220!6220filter : [o * o * o] --> o 
  n!6220!6220nats : [o] --> o 
  n!6220!6220sieve : [o] --> o 
  nats : [o] --> o 
  s : [o] --> o 
  sieve : [o] --> o 
  zprimes : [] --> o 

  filter(cons(X, Y), 0, Z) => cons(0, n!6220!6220filter(activate(Y), Z, Z)) 
  filter(cons(X, Y), s(Z), U) => cons(X, n!6220!6220filter(activate(Y), Z, U)) 
  sieve(cons(0, X)) => cons(0, n!6220!6220sieve(activate(X))) 
  sieve(cons(s(X), Y)) => cons(s(X), n!6220!6220sieve(filter(activate(Y), X, X))) 
  nats(X) => cons(X, n!6220!6220nats(s(X))) 
  zprimes => sieve(nats(s(s(0)))) 
  filter(X, Y, Z) => n!6220!6220filter(X, Y, Z) 
  sieve(X) => n!6220!6220sieve(X) 
  nats(X) => n!6220!6220nats(X) 
  activate(n!6220!6220filter(X, Y, Z)) => filter(X, Y, Z) 
  activate(n!6220!6220sieve(X)) => sieve(X) 
  activate(n!6220!6220nats(X)) => nats(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] filter#(cons(X, Y), 0, Z) =#> activate#(Y)   
    1] filter#(cons(X, Y), s(Z), U) =#> activate#(Y)   
    2] sieve#(cons(0, X)) =#> activate#(X)   
    3] sieve#(cons(s(X), Y)) =#> filter#(activate(Y), X, X)   
    4] sieve#(cons(s(X), Y)) =#> activate#(Y)   
    5] zprimes# =#> sieve#(nats(s(s(0))))   
    6] zprimes# =#> nats#(s(s(0)))   
    7] activate#(n!6220!6220filter(X, Y, Z)) =#> filter#(X, Y, Z)   
    8] activate#(n!6220!6220sieve(X)) =#> sieve#(X)   
    9] activate#(n!6220!6220nats(X)) =#> nats#(X)   

  Rules R_0:

    filter(cons(X, Y), 0, Z) => cons(0, n!6220!6220filter(activate(Y), Z, Z)) 
    filter(cons(X, Y), s(Z), U) => cons(X, n!6220!6220filter(activate(Y), Z, U)) 
    sieve(cons(0, X)) => cons(0, n!6220!6220sieve(activate(X))) 
    sieve(cons(s(X), Y)) => cons(s(X), n!6220!6220sieve(filter(activate(Y), X, X))) 
    nats(X) => cons(X, n!6220!6220nats(s(X))) 
    zprimes => sieve(nats(s(s(0)))) 
    filter(X, Y, Z) => n!6220!6220filter(X, Y, Z) 
    sieve(X) => n!6220!6220sieve(X) 
    nats(X) => n!6220!6220nats(X) 
    activate(n!6220!6220filter(X, Y, Z)) => filter(X, Y, Z) 
    activate(n!6220!6220sieve(X)) => sieve(X) 
    activate(n!6220!6220nats(X)) => nats(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 : 7, 8, 9 
    * 1 : 7, 8, 9 
    * 2 : 7, 8, 9 
    * 3 : 0, 1 
    * 4 : 7, 8, 9 
    * 5 : 2, 3, 4 
    * 6 :  
    * 7 : 0, 1 
    * 8 : 2, 3, 4 
    * 9 :  

This graph has the following strongly connected components:

  P_1:

    filter#(cons(X, Y), 0, Z) =#> activate#(Y)   
    filter#(cons(X, Y), s(Z), U) =#> activate#(Y)   
    sieve#(cons(0, X)) =#> activate#(X)   
    sieve#(cons(s(X), Y)) =#> filter#(activate(Y), X, X)   
    sieve#(cons(s(X), Y)) =#> activate#(Y)   
    activate#(n!6220!6220filter(X, Y, Z)) =#> filter#(X, Y, Z)   
    activate#(n!6220!6220sieve(X)) =#> sieve#(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 with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_0) are:

  filter(cons(X, Y), 0, Z) => cons(0, n!6220!6220filter(activate(Y), Z, Z)) 
  filter(cons(X, Y), s(Z), U) => cons(X, n!6220!6220filter(activate(Y), Z, U)) 
  sieve(cons(0, X)) => cons(0, n!6220!6220sieve(activate(X))) 
  sieve(cons(s(X), Y)) => cons(s(X), n!6220!6220sieve(filter(activate(Y), X, X))) 
  nats(X) => cons(X, n!6220!6220nats(s(X))) 
  filter(X, Y, Z) => n!6220!6220filter(X, Y, Z) 
  sieve(X) => n!6220!6220sieve(X) 
  nats(X) => n!6220!6220nats(X) 
  activate(n!6220!6220filter(X, Y, Z)) => filter(X, Y, Z) 
  activate(n!6220!6220sieve(X)) => sieve(X) 
  activate(n!6220!6220nats(X)) => nats(X) 
  activate(X) => X 

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

  filter#(cons(X, Y), 0, Z) >? activate#(Y) 
  filter#(cons(X, Y), s(Z), U) >? activate#(Y) 
  sieve#(cons(0, X)) >? activate#(X) 
  sieve#(cons(s(X), Y)) >? filter#(activate(Y), X, X) 
  sieve#(cons(s(X), Y)) >? activate#(Y) 
  activate#(n!6220!6220filter(X, Y, Z)) >? filter#(X, Y, Z) 
  activate#(n!6220!6220sieve(X)) >? sieve#(X) 
  filter(cons(X, Y), 0, Z) >= cons(0, n!6220!6220filter(activate(Y), Z, Z)) 
  filter(cons(X, Y), s(Z), U) >= cons(X, n!6220!6220filter(activate(Y), Z, U)) 
  sieve(cons(0, X)) >= cons(0, n!6220!6220sieve(activate(X))) 
  sieve(cons(s(X), Y)) >= cons(s(X), n!6220!6220sieve(filter(activate(Y), X, X))) 
  nats(X) >= cons(X, n!6220!6220nats(s(X))) 
  filter(X, Y, Z) >= n!6220!6220filter(X, Y, Z) 
  sieve(X) >= n!6220!6220sieve(X) 
  nats(X) >= n!6220!6220nats(X) 
  activate(n!6220!6220filter(X, Y, Z)) >= filter(X, Y, Z) 
  activate(n!6220!6220sieve(X)) >= sieve(X) 
  activate(n!6220!6220nats(X)) >= nats(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 
  cons = \y0y1.y1 
  filter = \y0y1y2.y0 
  filter# = \y0y1y2.y0 
  n!6220!6220filter = \y0y1y2.y0 
  n!6220!6220nats = \y0.0 
  n!6220!6220sieve = \y0.1 + 2y0 
  nats = \y0.0 
  s = \y0.0 
  sieve = \y0.1 + 2y0 
  sieve# = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[filter#(cons(_x0, _x1), 0, _x2)]] = x1 >= x1 = [[activate#(_x1)]] 
  [[filter#(cons(_x0, _x1), s(_x2), _x3)]] = x1 >= x1 = [[activate#(_x1)]] 
  [[sieve#(cons(0, _x0))]] = 1 + x0 > x0 = [[activate#(_x0)]] 
  [[sieve#(cons(s(_x0), _x1))]] = 1 + x1 > x1 = [[filter#(activate(_x1), _x0, _x0)]] 
  [[sieve#(cons(s(_x0), _x1))]] = 1 + x1 > x1 = [[activate#(_x1)]] 
  [[activate#(n!6220!6220filter(_x0, _x1, _x2))]] = x0 >= x0 = [[filter#(_x0, _x1, _x2)]] 
  [[activate#(n!6220!6220sieve(_x0))]] = 1 + 2x0 >= 1 + x0 = [[sieve#(_x0)]] 
  [[filter(cons(_x0, _x1), 0, _x2)]] = x1 >= x1 = [[cons(0, n!6220!6220filter(activate(_x1), _x2, _x2))]] 
  [[filter(cons(_x0, _x1), s(_x2), _x3)]] = x1 >= x1 = [[cons(_x0, n!6220!6220filter(activate(_x1), _x2, _x3))]] 
  [[sieve(cons(0, _x0))]] = 1 + 2x0 >= 1 + 2x0 = [[cons(0, n!6220!6220sieve(activate(_x0)))]] 
  [[sieve(cons(s(_x0), _x1))]] = 1 + 2x1 >= 1 + 2x1 = [[cons(s(_x0), n!6220!6220sieve(filter(activate(_x1), _x0, _x0)))]] 
  [[nats(_x0)]] = 0 >= 0 = [[cons(_x0, n!6220!6220nats(s(_x0)))]] 
  [[filter(_x0, _x1, _x2)]] = x0 >= x0 = [[n!6220!6220filter(_x0, _x1, _x2)]] 
  [[sieve(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[n!6220!6220sieve(_x0)]] 
  [[nats(_x0)]] = 0 >= 0 = [[n!6220!6220nats(_x0)]] 
  [[activate(n!6220!6220filter(_x0, _x1, _x2))]] = x0 >= x0 = [[filter(_x0, _x1, _x2)]] 
  [[activate(n!6220!6220sieve(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[sieve(_x0)]] 
  [[activate(n!6220!6220nats(_x0))]] = 0 >= 0 = [[nats(_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:

  filter#(cons(X, Y), 0, Z) =#> activate#(Y)   
  filter#(cons(X, Y), s(Z), U) =#> activate#(Y)   
  activate#(n!6220!6220filter(X, Y, Z)) =#> filter#(X, Y, Z)   
  activate#(n!6220!6220sieve(X)) =#> sieve#(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 : 2, 3 
    * 1 : 2, 3 
    * 2 : 0, 1 
    * 3 :  

This graph has the following strongly connected components:

  P_3:

    filter#(cons(X, Y), 0, Z) =#> activate#(Y)   
    filter#(cons(X, Y), s(Z), U) =#> activate#(Y)   
    activate#(n!6220!6220filter(X, Y, Z)) =#> filter#(X, Y, Z)   

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 apply the subterm criterion with the following projection function:

  nu(activate#) = 1 
  nu(filter#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(cons(X, Y), 0, Z)) = cons(X, Y) |> Y = nu(activate#(Y)) 
  nu(filter#(cons(X, Y), s(Z), U)) = cons(X, Y) |> Y = nu(activate#(Y)) 
  nu(activate#(n!6220!6220filter(X, Y, Z))) = n!6220!6220filter(X, Y, Z) |> X = nu(filter#(X, Y, Z)) 

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.

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.