/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 
  add : [] --> o 
  app : [o * o] --> o 
  curry : [] --> o 
  plus : [] --> o 
  s : [] --> o 

  app(app(plus, 0), X) => X 
  app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) 
  app(app(app(curry, X), Y), Z) => app(app(X, Y), Z) 
  add => app(curry, plus) 

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

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

  app(app(plus, 0), X) >? X 
  app(app(plus, app(s, X)), Y) >? app(s, app(app(plus, X), Y)) 
  app(app(app(curry, X), Y), Z) >? app(app(X, Y), Z) 
  add >? app(curry, plus) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  add = 3 
  app = \y0y1.y0 + y1 
  curry = 0 
  plus = 0 
  s = 0 

Using this interpretation, the requirements translate to:

  [[app(app(plus, 0), _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[app(app(plus, app(s, _x0)), _x1)]] = x0 + x1 >= x0 + x1 = [[app(s, app(app(plus, _x0), _x1))]] 
  [[app(app(app(curry, _x0), _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[app(app(_x0, _x1), _x2)]] 
  [[add]] = 3 > 0 = [[app(curry, plus)]] 

We can thus remove the following rules:

  app(app(plus, 0), X) => X 
  add => app(curry, plus) 

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

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

  app(app(plus, app(s, X)), Y) >? app(s, app(app(plus, X), Y)) 
  app(app(app(curry, X), Y), Z) >? app(app(X, Y), Z) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.2 + y1 + 2y0 
  curry = 3 
  plus = 0 
  s = 0 

Using this interpretation, the requirements translate to:

  [[app(app(plus, app(s, _x0)), _x1)]] = 10 + x1 + 2x0 > 8 + x1 + 2x0 = [[app(s, app(app(plus, _x0), _x1))]] 
  [[app(app(app(curry, _x0), _x1), _x2)]] = 38 + x2 + 2x1 + 4x0 > 6 + x2 + 2x1 + 4x0 = [[app(app(_x0, _x1), _x2)]] 

We can thus remove the following rules:

  app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) 
  app(app(app(curry, X), Y), Z) => app(app(X, Y), Z) 

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.