/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.

  a!6220!6220c : [] --> o 
  a!6220!6220g : [o] --> o 
  a!6220!6220h : [o] --> o 
  c : [] --> o 
  d : [] --> o 
  g : [o] --> o 
  h : [o] --> o 
  mark : [o] --> o 

  a!6220!6220g(X) => a!6220!6220h(X) 
  a!6220!6220c => d 
  a!6220!6220h(d) => a!6220!6220g(c) 
  mark(g(X)) => a!6220!6220g(X) 
  mark(h(X)) => a!6220!6220h(X) 
  mark(c) => a!6220!6220c 
  mark(d) => d 
  a!6220!6220g(X) => g(X) 
  a!6220!6220h(X) => h(X) 
  a!6220!6220c => c 

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

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

  a!6220!6220g(X) >? a!6220!6220h(X) 
  a!6220!6220c >? d 
  a!6220!6220h(d) >? a!6220!6220g(c) 
  mark(g(X)) >? a!6220!6220g(X) 
  mark(h(X)) >? a!6220!6220h(X) 
  mark(c) >? a!6220!6220c 
  mark(d) >? d 
  a!6220!6220g(X) >? g(X) 
  a!6220!6220h(X) >? h(X) 
  a!6220!6220c >? c 

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

The following interpretation satisfies the requirements:

  a!6220!6220c = 1 
  a!6220!6220g = \y0.1 + 2y0 
  a!6220!6220h = \y0.1 + y0 
  c = 0 
  d = 0 
  g = \y0.y0 
  h = \y0.y0 
  mark = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220g(_x0)]] = 1 + 2x0 >= 1 + x0 = [[a!6220!6220h(_x0)]] 
  [[a!6220!6220c]] = 1 > 0 = [[d]] 
  [[a!6220!6220h(d)]] = 1 >= 1 = [[a!6220!6220g(c)]] 
  [[mark(g(_x0))]] = 3 + 3x0 > 1 + 2x0 = [[a!6220!6220g(_x0)]] 
  [[mark(h(_x0))]] = 3 + 3x0 > 1 + x0 = [[a!6220!6220h(_x0)]] 
  [[mark(c)]] = 3 > 1 = [[a!6220!6220c]] 
  [[mark(d)]] = 3 > 0 = [[d]] 
  [[a!6220!6220g(_x0)]] = 1 + 2x0 > x0 = [[g(_x0)]] 
  [[a!6220!6220h(_x0)]] = 1 + x0 > x0 = [[h(_x0)]] 
  [[a!6220!6220c]] = 1 > 0 = [[c]] 

We can thus remove the following rules:

  a!6220!6220c => d 
  mark(g(X)) => a!6220!6220g(X) 
  mark(h(X)) => a!6220!6220h(X) 
  mark(c) => a!6220!6220c 
  mark(d) => d 
  a!6220!6220g(X) => g(X) 
  a!6220!6220h(X) => h(X) 
  a!6220!6220c => c 

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

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

  a!6220!6220g(X) >? a!6220!6220h(X) 
  a!6220!6220h(d) >? a!6220!6220g(c) 

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

The following interpretation satisfies the requirements:

  a!6220!6220g = \y0.2 + 3y0 
  a!6220!6220h = \y0.3y0 
  c = 0 
  d = 3 

Using this interpretation, the requirements translate to:

  [[a!6220!6220g(_x0)]] = 2 + 3x0 > 3x0 = [[a!6220!6220h(_x0)]] 
  [[a!6220!6220h(d)]] = 9 > 2 = [[a!6220!6220g(c)]] 

We can thus remove the following rules:

  a!6220!6220g(X) => a!6220!6220h(X) 
  a!6220!6220h(d) => a!6220!6220g(c) 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.