/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 
  a : [o] --> o 
  cs : [o * o] --> o 
  d : [o] --> o 
  f : [o * o] --> o 
  nf : [o * o] --> o 
  nil : [] --> o 
  ns : [o] --> o 
  nt : [o] --> o 
  p : [o * o] --> o 
  q : [o] --> o 
  r : [o] --> o 
  s : [o] --> o 
  t : [o] --> o 

  t(X) => cs(r(q(X)), nt(ns(X))) 
  q(0) => 0 
  q(s(X)) => s(p(q(X), d(X))) 
  d(0) => 0 
  d(s(X)) => s(s(d(X))) 
  p(0, X) => X 
  p(X, 0) => X 
  p(s(X), s(Y)) => s(s(p(X, Y))) 
  f(0, X) => nil 
  f(s(X), cs(Y, Z)) => cs(Y, nf(X, a(Z))) 
  t(X) => nt(X) 
  s(X) => ns(X) 
  f(X, Y) => nf(X, Y) 
  a(nt(X)) => t(a(X)) 
  a(ns(X)) => s(a(X)) 
  a(nf(X, Y)) => f(a(X), a(Y)) 
  a(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] t#(X) =#> q#(X)   
    1] q#(s(X)) =#> s#(p(q(X), d(X)))   
    2] q#(s(X)) =#> p#(q(X), d(X))   
    3] q#(s(X)) =#> q#(X)   
    4] q#(s(X)) =#> d#(X)   
    5] d#(s(X)) =#> s#(s(d(X)))   
    6] d#(s(X)) =#> s#(d(X))   
    7] d#(s(X)) =#> d#(X)   
    8] p#(s(X), s(Y)) =#> s#(s(p(X, Y)))   
    9] p#(s(X), s(Y)) =#> s#(p(X, Y))   
    10] p#(s(X), s(Y)) =#> p#(X, Y)   
    11] f#(s(X), cs(Y, Z)) =#> a#(Z)   
    12] a#(nt(X)) =#> t#(a(X))   
    13] a#(nt(X)) =#> a#(X)   
    14] a#(ns(X)) =#> s#(a(X))   
    15] a#(ns(X)) =#> a#(X)   
    16] a#(nf(X, Y)) =#> f#(a(X), a(Y))   
    17] a#(nf(X, Y)) =#> a#(X)   
    18] a#(nf(X, Y)) =#> a#(Y)   

  Rules R_0:

    t(X) => cs(r(q(X)), nt(ns(X))) 
    q(0) => 0 
    q(s(X)) => s(p(q(X), d(X))) 
    d(0) => 0 
    d(s(X)) => s(s(d(X))) 
    p(0, X) => X 
    p(X, 0) => X 
    p(s(X), s(Y)) => s(s(p(X, Y))) 
    f(0, X) => nil 
    f(s(X), cs(Y, Z)) => cs(Y, nf(X, a(Z))) 
    t(X) => nt(X) 
    s(X) => ns(X) 
    f(X, Y) => nf(X, Y) 
    a(nt(X)) => t(a(X)) 
    a(ns(X)) => s(a(X)) 
    a(nf(X, Y)) => f(a(X), a(Y)) 
    a(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, 10 
    * 3 : 1, 2, 3, 4 
    * 4 : 5, 6, 7 
    * 5 :  
    * 6 :  
    * 7 : 5, 6, 7 
    * 8 :  
    * 9 :  
    * 10 : 8, 9, 10 
    * 11 : 12, 13, 14, 15, 16, 17, 18 
    * 12 : 0 
    * 13 : 12, 13, 14, 15, 16, 17, 18 
    * 14 :  
    * 15 : 12, 13, 14, 15, 16, 17, 18 
    * 16 : 11 
    * 17 : 12, 13, 14, 15, 16, 17, 18 
    * 18 : 12, 13, 14, 15, 16, 17, 18 

This graph has the following strongly connected components:

  P_1:

    q#(s(X)) =#> q#(X)   

  P_2:

    d#(s(X)) =#> d#(X)   

  P_3:

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

  P_4:

    f#(s(X), cs(Y, Z)) =#> a#(Z)   
    a#(nt(X)) =#> a#(X)   
    a#(ns(X)) =#> a#(X)   
    a#(nf(X, Y)) =#> f#(a(X), a(Y))   
    a#(nf(X, Y)) =#> a#(X)   
    a#(nf(X, Y)) =#> a#(Y)   

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) and (P_4, 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) 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 ::=

  t(X) => cs(r(q(X)), nt(ns(X))) 
  q(0) => 0 
  q(s(X)) => s(p(q(X), d(X))) 
  d(0) => 0 
  d(s(X)) => s(s(d(X))) 
  p(0, X) => X 
  p(X, 0) => X 
  p(s(X), s(Y)) => s(s(p(X, Y))) 
  f(s(X), cs(Y, Z)) => cs(Y, nf(X, a(Z))) 
  t(X) => nt(X) 
  s(X) => ns(X) 
  f(X, Y) => nf(X, Y) 
  a(nt(X)) => t(a(X)) 
  a(ns(X)) => s(a(X)) 
  a(nf(X, Y)) => f(a(X), a(Y)) 
  a(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 [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(s(X), cs(Y, Z)) >? a#(Z) 
  a#(nt(X)) >? a#(X) 
  a#(ns(X)) >? a#(X) 
  a#(nf(X, Y)) >? f#(a(X), a(Y)) 
  a#(nf(X, Y)) >? a#(X) 
  a#(nf(X, Y)) >? a#(Y) 
  t(X) >= cs(r(q(X)), nt(ns(X))) 
  q(0) >= 0 
  q(s(X)) >= s(p(q(X), d(X))) 
  d(0) >= 0 
  d(s(X)) >= s(s(d(X))) 
  p(0, X) >= X 
  p(X, 0) >= X 
  p(s(X), s(Y)) >= s(s(p(X, Y))) 
  f(s(X), cs(Y, Z)) >= cs(Y, nf(X, a(Z))) 
  t(X) >= nt(X) 
  s(X) >= ns(X) 
  f(X, Y) >= nf(X, Y) 
  a(nt(X)) >= t(a(X)) 
  a(ns(X)) >= s(a(X)) 
  a(nf(X, Y)) >= f(a(X), a(Y)) 
  a(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:

  r(x_1) = r() 

This leaves the following ordering requirements:

  f#(s(X), cs(Y, Z)) >= a#(Z) 
  a#(nt(X)) > a#(X) 
  a#(ns(X)) >= a#(X) 
  a#(nf(X, Y)) >= f#(a(X), a(Y)) 
  a#(nf(X, Y)) >= a#(X) 
  a#(nf(X, Y)) >= a#(Y) 
  t(X) >= cs(r(q(X)), nt(ns(X))) 
  f(s(X), cs(Y, Z)) >= cs(Y, nf(X, a(Z))) 
  t(X) >= nt(X) 
  s(X) >= ns(X) 
  f(X, Y) >= nf(X, Y) 
  a(nt(X)) >= t(a(X)) 
  a(ns(X)) >= s(a(X)) 
  a(nf(X, Y)) >= f(a(X), a(Y)) 
  a(X) >= X 

The following interpretation satisfies the requirements:

  0 = 0 
  a = \y0.y0 
  a# = \y0.y0 
  cs = \y0y1.y1 
  d = \y0.0 
  f = \y0y1.y0 + y1 
  f# = \y0y1.y1 
  nf = \y0y1.y0 + y1 
  ns = \y0.y0 
  nt = \y0.1 + y0 
  p = \y0y1.0 
  q = \y0.0 
  r = \y0.0 
  s = \y0.y0 
  t = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0), cs(_x1, _x2))]] = x2 >= x2 = [[a#(_x2)]] 
  [[a#(nt(_x0))]] = 1 + x0 > x0 = [[a#(_x0)]] 
  [[a#(ns(_x0))]] = x0 >= x0 = [[a#(_x0)]] 
  [[a#(nf(_x0, _x1))]] = x0 + x1 >= x1 = [[f#(a(_x0), a(_x1))]] 
  [[a#(nf(_x0, _x1))]] = x0 + x1 >= x0 = [[a#(_x0)]] 
  [[a#(nf(_x0, _x1))]] = x0 + x1 >= x1 = [[a#(_x1)]] 
  [[t(_x0)]] = 1 + x0 >= 1 + x0 = [[cs(r(q(_x0)), nt(ns(_x0)))]] 
  [[f(s(_x0), cs(_x1, _x2))]] = x0 + x2 >= x0 + x2 = [[cs(_x1, nf(_x0, a(_x2)))]] 
  [[t(_x0)]] = 1 + x0 >= 1 + x0 = [[nt(_x0)]] 
  [[s(_x0)]] = x0 >= x0 = [[ns(_x0)]] 
  [[f(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[nf(_x0, _x1)]] 
  [[a(nt(_x0))]] = 1 + x0 >= 1 + x0 = [[t(a(_x0))]] 
  [[a(ns(_x0))]] = x0 >= x0 = [[s(a(_x0))]] 
  [[a(nf(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[f(a(_x0), a(_x1))]] 
  [[a(_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_5, R_1, minimal, formative), where P_5 consists of:

  f#(s(X), cs(Y, Z)) =#> a#(Z)   
  a#(ns(X)) =#> a#(X)   
  a#(nf(X, Y)) =#> f#(a(X), a(Y))   
  a#(nf(X, Y)) =#> a#(X)   
  a#(nf(X, Y)) =#> a#(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_5, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_5, 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:

  f#(s(X), cs(Y, Z)) >? a#(Z) 
  a#(ns(X)) >? a#(X) 
  a#(nf(X, Y)) >? f#(a(X), a(Y)) 
  a#(nf(X, Y)) >? a#(X) 
  a#(nf(X, Y)) >? a#(Y) 
  t(X) >= cs(r(q(X)), nt(ns(X))) 
  q(0) >= 0 
  q(s(X)) >= s(p(q(X), d(X))) 
  d(0) >= 0 
  d(s(X)) >= s(s(d(X))) 
  p(0, X) >= X 
  p(X, 0) >= X 
  p(s(X), s(Y)) >= s(s(p(X, Y))) 
  f(s(X), cs(Y, Z)) >= cs(Y, nf(X, a(Z))) 
  t(X) >= nt(X) 
  s(X) >= ns(X) 
  f(X, Y) >= nf(X, Y) 
  a(nt(X)) >= t(a(X)) 
  a(ns(X)) >= s(a(X)) 
  a(nf(X, Y)) >= f(a(X), a(Y)) 
  a(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:

  r(x_1) = r() 

This leaves the following ordering requirements:

  f#(s(X), cs(Y, Z)) > a#(Z) 
  a#(ns(X)) >= a#(X) 
  a#(nf(X, Y)) >= f#(a(X), a(Y)) 
  a#(nf(X, Y)) >= a#(X) 
  a#(nf(X, Y)) >= a#(Y) 
  t(X) >= cs(r(q(X)), nt(ns(X))) 
  f(s(X), cs(Y, Z)) >= cs(Y, nf(X, a(Z))) 
  t(X) >= nt(X) 
  s(X) >= ns(X) 
  f(X, Y) >= nf(X, Y) 
  a(nt(X)) >= t(a(X)) 
  a(ns(X)) >= s(a(X)) 
  a(nf(X, Y)) >= f(a(X), a(Y)) 
  a(X) >= X 

The following interpretation satisfies the requirements:

  0 = 0 
  a = \y0.y0 
  a# = \y0.y0 
  cs = \y0y1.y1 
  d = \y0.0 
  f = \y0y1.y1 + 2y0 
  f# = \y0y1.y1 + 2y0 
  nf = \y0y1.y1 + 2y0 
  ns = \y0.2 + 2y0 
  nt = \y0.0 
  p = \y0y1.0 
  q = \y0.0 
  r = \y0.0 
  s = \y0.2 + 2y0 
  t = \y0.0 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0), cs(_x1, _x2))]] = 4 + x2 + 4x0 > x2 = [[a#(_x2)]] 
  [[a#(ns(_x0))]] = 2 + 2x0 > x0 = [[a#(_x0)]] 
  [[a#(nf(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[f#(a(_x0), a(_x1))]] 
  [[a#(nf(_x0, _x1))]] = x1 + 2x0 >= x0 = [[a#(_x0)]] 
  [[a#(nf(_x0, _x1))]] = x1 + 2x0 >= x1 = [[a#(_x1)]] 
  [[t(_x0)]] = 0 >= 0 = [[cs(r(q(_x0)), nt(ns(_x0)))]] 
  [[f(s(_x0), cs(_x1, _x2))]] = 4 + x2 + 4x0 >= x2 + 2x0 = [[cs(_x1, nf(_x0, a(_x2)))]] 
  [[t(_x0)]] = 0 >= 0 = [[nt(_x0)]] 
  [[s(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[ns(_x0)]] 
  [[f(_x0, _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[nf(_x0, _x1)]] 
  [[a(nt(_x0))]] = 0 >= 0 = [[t(a(_x0))]] 
  [[a(ns(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[s(a(_x0))]] 
  [[a(nf(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[f(a(_x0), a(_x1))]] 
  [[a(_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_1, minimal, formative) by (P_6, R_1, minimal, formative), where P_6 consists of:

  a#(nf(X, Y)) =#> f#(a(X), a(Y))   
  a#(nf(X, Y)) =#> a#(X)   
  a#(nf(X, Y)) =#> a#(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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 : 0, 1, 2 
    * 2 : 0, 1, 2 

This graph has the following strongly connected components:

  P_7:

    a#(nf(X, Y)) =#> a#(X)   
    a#(nf(X, Y)) =#> a#(Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_6, R_1, m, f) by (P_7, R_1, 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) and (P_7, R_1, minimal, formative) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(a#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(a#(nf(X, Y))) = nf(X, Y) |> X = nu(a#(X)) 
  nu(a#(nf(X, Y))) = nf(X, Y) |> Y = nu(a#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_7, R_1, minimal, f) by ({}, R_1, 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(p#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(p#(s(X), s(Y))) = s(X) |> X = nu(p#(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(d#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(d#(s(X))) = s(X) |> X = nu(d#(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(q#) = 1 

Thus, we can orient the dependency pairs as follows:

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