/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.

  active : [o] --> o 
  c : [o] --> o 
  d : [o] --> o 
  f : [o] --> o 
  g : [o] --> o 
  h : [o] --> o 
  mark : [o] --> o 

  active(f(f(X))) => mark(c(f(g(f(X))))) 
  active(c(X)) => mark(d(X)) 
  active(h(X)) => mark(c(d(X))) 
  mark(f(X)) => active(f(mark(X))) 
  mark(c(X)) => active(c(X)) 
  mark(g(X)) => active(g(X)) 
  mark(d(X)) => active(d(X)) 
  mark(h(X)) => active(h(mark(X))) 
  f(mark(X)) => f(X) 
  f(active(X)) => f(X) 
  c(mark(X)) => c(X) 
  c(active(X)) => c(X) 
  g(mark(X)) => g(X) 
  g(active(X)) => g(X) 
  d(mark(X)) => d(X) 
  d(active(X)) => d(X) 
  h(mark(X)) => h(X) 
  h(active(X)) => h(X) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  active(f(f(X))) >? mark(c(f(g(f(X))))) 
  active(c(X)) >? mark(d(X)) 
  active(h(X)) >? mark(c(d(X))) 
  mark(f(X)) >? active(f(mark(X))) 
  mark(c(X)) >? active(c(X)) 
  mark(g(X)) >? active(g(X)) 
  mark(d(X)) >? active(d(X)) 
  mark(h(X)) >? active(h(mark(X))) 
  f(mark(X)) >? f(X) 
  f(active(X)) >? f(X) 
  c(mark(X)) >? c(X) 
  c(active(X)) >? c(X) 
  g(mark(X)) >? g(X) 
  g(active(X)) >? g(X) 
  d(mark(X)) >? d(X) 
  d(active(X)) >? d(X) 
  h(mark(X)) >? h(X) 
  h(active(X)) >? h(X) 

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

The following interpretation satisfies the requirements:

  active = \y0.y0 
  c = \y0.y0 
  d = \y0.y0 
  f = \y0.y0 
  g = \y0.y0 
  h = \y0.1 + y0 
  mark = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active(f(f(_x0)))]] = x0 >= x0 = [[mark(c(f(g(f(_x0)))))]] 
  [[active(c(_x0))]] = x0 >= x0 = [[mark(d(_x0))]] 
  [[active(h(_x0))]] = 1 + x0 > x0 = [[mark(c(d(_x0)))]] 
  [[mark(f(_x0))]] = x0 >= x0 = [[active(f(mark(_x0)))]] 
  [[mark(c(_x0))]] = x0 >= x0 = [[active(c(_x0))]] 
  [[mark(g(_x0))]] = x0 >= x0 = [[active(g(_x0))]] 
  [[mark(d(_x0))]] = x0 >= x0 = [[active(d(_x0))]] 
  [[mark(h(_x0))]] = 1 + x0 >= 1 + x0 = [[active(h(mark(_x0)))]] 
  [[f(mark(_x0))]] = x0 >= x0 = [[f(_x0)]] 
  [[f(active(_x0))]] = x0 >= x0 = [[f(_x0)]] 
  [[c(mark(_x0))]] = x0 >= x0 = [[c(_x0)]] 
  [[c(active(_x0))]] = x0 >= x0 = [[c(_x0)]] 
  [[g(mark(_x0))]] = x0 >= x0 = [[g(_x0)]] 
  [[g(active(_x0))]] = x0 >= x0 = [[g(_x0)]] 
  [[d(mark(_x0))]] = x0 >= x0 = [[d(_x0)]] 
  [[d(active(_x0))]] = x0 >= x0 = [[d(_x0)]] 
  [[h(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[h(_x0)]] 
  [[h(active(_x0))]] = 1 + x0 >= 1 + x0 = [[h(_x0)]] 

We can thus remove the following rules:

  active(h(X)) => mark(c(d(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#(f(f(X))) =#> mark#(c(f(g(f(X)))))   
    1] active#(f(f(X))) =#> c#(f(g(f(X))))   
    2] active#(f(f(X))) =#> f#(g(f(X)))   
    3] active#(f(f(X))) =#> g#(f(X))   
    4] active#(f(f(X))) =#> f#(X)   
    5] active#(c(X)) =#> mark#(d(X))   
    6] active#(c(X)) =#> d#(X)   
    7] mark#(f(X)) =#> active#(f(mark(X)))   
    8] mark#(f(X)) =#> f#(mark(X))   
    9] mark#(f(X)) =#> mark#(X)   
    10] mark#(c(X)) =#> active#(c(X))   
    11] mark#(c(X)) =#> c#(X)   
    12] mark#(g(X)) =#> active#(g(X))   
    13] mark#(g(X)) =#> g#(X)   
    14] mark#(d(X)) =#> active#(d(X))   
    15] mark#(d(X)) =#> d#(X)   
    16] mark#(h(X)) =#> active#(h(mark(X)))   
    17] mark#(h(X)) =#> h#(mark(X))   
    18] mark#(h(X)) =#> mark#(X)   
    19] f#(mark(X)) =#> f#(X)   
    20] f#(active(X)) =#> f#(X)   
    21] c#(mark(X)) =#> c#(X)   
    22] c#(active(X)) =#> c#(X)   
    23] g#(mark(X)) =#> g#(X)   
    24] g#(active(X)) =#> g#(X)   
    25] d#(mark(X)) =#> d#(X)   
    26] d#(active(X)) =#> d#(X)   
    27] h#(mark(X)) =#> h#(X)   
    28] h#(active(X)) =#> h#(X)   

  Rules R_0:

    active(f(f(X))) => mark(c(f(g(f(X))))) 
    active(c(X)) => mark(d(X)) 
    mark(f(X)) => active(f(mark(X))) 
    mark(c(X)) => active(c(X)) 
    mark(g(X)) => active(g(X)) 
    mark(d(X)) => active(d(X)) 
    mark(h(X)) => active(h(mark(X))) 
    f(mark(X)) => f(X) 
    f(active(X)) => f(X) 
    c(mark(X)) => c(X) 
    c(active(X)) => c(X) 
    g(mark(X)) => g(X) 
    g(active(X)) => g(X) 
    d(mark(X)) => d(X) 
    d(active(X)) => d(X) 
    h(mark(X)) => h(X) 
    h(active(X)) => h(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 : 10, 11 
    * 1 :  
    * 2 :  
    * 3 :  
    * 4 : 19, 20 
    * 5 : 14, 15 
    * 6 : 25, 26 
    * 7 : 0, 1, 2, 3, 4 
    * 8 : 19, 20 
    * 9 : 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 
    * 10 : 5, 6 
    * 11 : 21, 22 
    * 12 :  
    * 13 : 23, 24 
    * 14 :  
    * 15 : 25, 26 
    * 16 :  
    * 17 : 27, 28 
    * 18 : 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 
    * 19 : 19, 20 
    * 20 : 19, 20 
    * 21 : 21, 22 
    * 22 : 21, 22 
    * 23 : 23, 24 
    * 24 : 23, 24 
    * 25 : 25, 26 
    * 26 : 25, 26 
    * 27 : 27, 28 
    * 28 : 27, 28 

This graph has the following strongly connected components:

  P_1:

    mark#(f(X)) =#> mark#(X)   
    mark#(h(X)) =#> mark#(X)   

  P_2:

    f#(mark(X)) =#> f#(X)   
    f#(active(X)) =#> f#(X)   

  P_3:

    c#(mark(X)) =#> c#(X)   
    c#(active(X)) =#> c#(X)   

  P_4:

    g#(mark(X)) =#> g#(X)   
    g#(active(X)) =#> g#(X)   

  P_5:

    d#(mark(X)) =#> d#(X)   
    d#(active(X)) =#> d#(X)   

  P_6:

    h#(mark(X)) =#> h#(X)   
    h#(active(X)) =#> h#(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), (P_5, R_0, m, f) and (P_6, 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), (P_5, R_0, minimal, formative) and (P_6, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_6, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(h#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(h#(mark(X))) = mark(X) |> X = nu(h#(X)) 
  nu(h#(active(X))) = active(X) |> X = nu(h#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, 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), (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(d#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(d#(mark(X))) = mark(X) |> X = nu(d#(X)) 
  nu(d#(active(X))) = active(X) |> X = nu(d#(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).

We apply the subterm criterion with the following projection function:

  nu(g#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(g#(mark(X))) = mark(X) |> X = nu(g#(X)) 
  nu(g#(active(X))) = active(X) |> X = nu(g#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, 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) 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(c#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(c#(mark(X))) = mark(X) |> X = nu(c#(X)) 
  nu(c#(active(X))) = active(X) |> X = nu(c#(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#(mark(X))) = mark(X) |> X = nu(f#(X)) 
  nu(f#(active(X))) = active(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(mark#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(mark#(f(X))) = f(X) |> X = nu(mark#(X)) 
  nu(mark#(h(X))) = h(X) |> X = nu(mark#(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.