/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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.

  active : [o] --> o 
  c : [] --> o 
  f : [o] --> o 
  g : [o] --> o 
  mark : [o] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  top : [o] --> o 

  active(c) => mark(f(g(c))) 
  active(f(g(X))) => mark(g(X)) 
  proper(c) => ok(c) 
  proper(f(X)) => f(proper(X)) 
  proper(g(X)) => g(proper(X)) 
  f(ok(X)) => ok(f(X)) 
  g(ok(X)) => ok(g(X)) 
  top(mark(X)) => top(proper(X)) 
  top(ok(X)) => top(active(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#(c) =#> f#(g(c))   
    1] active#(c) =#> g#(c)   
    2] active#(f(g(X))) =#> g#(X)   
    3] proper#(f(X)) =#> f#(proper(X))   
    4] proper#(f(X)) =#> proper#(X)   
    5] proper#(g(X)) =#> g#(proper(X))   
    6] proper#(g(X)) =#> proper#(X)   
    7] f#(ok(X)) =#> f#(X)   
    8] g#(ok(X)) =#> g#(X)   
    9] top#(mark(X)) =#> top#(proper(X))   
    10] top#(mark(X)) =#> proper#(X)   
    11] top#(ok(X)) =#> top#(active(X))   
    12] top#(ok(X)) =#> active#(X)   

  Rules R_0:

    active(c) => mark(f(g(c))) 
    active(f(g(X))) => mark(g(X)) 
    proper(c) => ok(c) 
    proper(f(X)) => f(proper(X)) 
    proper(g(X)) => g(proper(X)) 
    f(ok(X)) => ok(f(X)) 
    g(ok(X)) => ok(g(X)) 
    top(mark(X)) => top(proper(X)) 
    top(ok(X)) => top(active(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 
    * 1 :  
    * 2 : 8 
    * 3 : 7 
    * 4 : 3, 4, 5, 6 
    * 5 : 8 
    * 6 : 3, 4, 5, 6 
    * 7 : 7 
    * 8 : 8 
    * 9 : 11, 12 
    * 10 : 3, 4, 5, 6 
    * 11 : 9, 10 
    * 12 : 0, 1, 2 

This graph has the following strongly connected components:

  P_1:

    proper#(f(X)) =#> proper#(X)   
    proper#(g(X)) =#> proper#(X)   

  P_2:

    f#(ok(X)) =#> f#(X)   

  P_3:

    g#(ok(X)) =#> g#(X)   

  P_4:

    top#(mark(X)) =#> top#(proper(X))   
    top#(ok(X)) =#> top#(active(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) 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 ::=

  active(c) => mark(f(g(c))) 
  active(f(g(X))) => mark(g(X)) 
  proper(c) => ok(c) 
  proper(f(X)) => f(proper(X)) 
  proper(g(X)) => g(proper(X)) 
  f(ok(X)) => ok(f(X)) 
  g(ok(X)) => ok(g(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:

  top#(mark(X)) >? top#(proper(X)) 
  top#(ok(X)) >? top#(active(X)) 
  active(c) >= mark(f(g(c))) 
  active(f(g(X))) >= mark(g(X)) 
  proper(c) >= ok(c) 
  proper(f(X)) >= f(proper(X)) 
  proper(g(X)) >= g(proper(X)) 
  f(ok(X)) >= ok(f(X)) 
  g(ok(X)) >= ok(g(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  c = 3 
  f = \y0.1 
  g = \y0.0 
  mark = \y0.1 + 2y0 
  ok = \y0.y0 
  proper = \y0.y0 
  top# = \y0.y0 

Using this interpretation, the requirements translate to:

  [[top#(mark(_x0))]] = 1 + 2x0 > x0 = [[top#(proper(_x0))]] 
  [[top#(ok(_x0))]] = x0 >= x0 = [[top#(active(_x0))]] 
  [[active(c)]] = 3 >= 3 = [[mark(f(g(c)))]] 
  [[active(f(g(_x0)))]] = 1 >= 1 = [[mark(g(_x0))]] 
  [[proper(c)]] = 3 >= 3 = [[ok(c)]] 
  [[proper(f(_x0))]] = 1 >= 1 = [[f(proper(_x0))]] 
  [[proper(g(_x0))]] = 0 >= 0 = [[g(proper(_x0))]] 
  [[f(ok(_x0))]] = 1 >= 1 = [[ok(f(_x0))]] 
  [[g(ok(_x0))]] = 0 >= 0 = [[ok(g(_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:

  top#(ok(X)) =#> top#(active(X))   

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 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(g#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(g#(ok(X))) = ok(X) |> X = nu(g#(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(f#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(f#(ok(X))) = ok(X) |> X = nu(f#(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(proper#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(proper#(f(X))) = f(X) |> X = nu(proper#(X)) 
  nu(proper#(g(X))) = g(X) |> X = nu(proper#(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.