/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.

  !minus : [o * o] --> o 
  0 : [] --> o 
  double : [o] --> o 
  half : [o] --> o 
  if : [o * o * o] --> o 
  s : [o] --> o 

  double(0) => 0 
  double(s(X)) => s(s(double(X))) 
  half(0) => 0 
  half(s(0)) => 0 
  half(s(s(X))) => s(half(X)) 
  !minus(X, 0) => X 
  !minus(s(X), s(Y)) => !minus(X, Y) 
  if(0, X, Y) => X 
  if(s(X), Y, Z) => Z 
  half(double(X)) => 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]):

  double(0) >? 0 
  double(s(X)) >? s(s(double(X))) 
  half(0) >? 0 
  half(s(0)) >? 0 
  half(s(s(X))) >? s(half(X)) 
  !minus(X, 0) >? X 
  !minus(s(X), s(Y)) >? !minus(X, Y) 
  if(0, X, Y) >? X 
  if(s(X), Y, Z) >? Z 
  half(double(X)) >? X 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.3 + y0 + y1 
  0 = 0 
  double = \y0.y0 
  half = \y0.1 + 2y0 
  if = \y0y1y2.3 + y0 + y1 + y2 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[double(0)]] = 0 >= 0 = [[0]] 
  [[double(s(_x0))]] = x0 >= x0 = [[s(s(double(_x0)))]] 
  [[half(0)]] = 1 > 0 = [[0]] 
  [[half(s(0))]] = 1 > 0 = [[0]] 
  [[half(s(s(_x0)))]] = 1 + 2x0 >= 1 + 2x0 = [[s(half(_x0))]] 
  [[!minus(_x0, 0)]] = 3 + x0 > x0 = [[_x0]] 
  [[!minus(s(_x0), s(_x1))]] = 3 + x0 + x1 >= 3 + x0 + x1 = [[!minus(_x0, _x1)]] 
  [[if(0, _x0, _x1)]] = 3 + x0 + x1 > x0 = [[_x0]] 
  [[if(s(_x0), _x1, _x2)]] = 3 + x0 + x1 + x2 > x2 = [[_x2]] 
  [[half(double(_x0))]] = 1 + 2x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  half(0) => 0 
  half(s(0)) => 0 
  !minus(X, 0) => X 
  if(0, X, Y) => X 
  if(s(X), Y, Z) => Z 
  half(double(X)) => 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]):

  double(0) >? 0 
  double(s(X)) >? s(s(double(X))) 
  half(s(s(X))) >? s(half(X)) 
  !minus(s(X), s(Y)) >? !minus(X, Y) 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 + y1 
  0 = 3 
  double = \y0.1 + y0 
  half = \y0.2y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[double(0)]] = 4 > 3 = [[0]] 
  [[double(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(s(double(_x0)))]] 
  [[half(s(s(_x0)))]] = 2x0 >= 2x0 = [[s(half(_x0))]] 
  [[!minus(s(_x0), s(_x1))]] = x0 + x1 >= x0 + x1 = [[!minus(_x0, _x1)]] 

We can thus remove the following rules:

  double(0) => 0 

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

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

  double(s(X)) >? s(s(double(X))) 
  half(s(s(X))) >? s(half(X)) 
  !minus(s(X), s(Y)) >? !minus(X, Y) 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 + y1 
  double = \y0.3y0 
  half = \y0.y0 
  s = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[double(s(_x0))]] = 6 + 3x0 > 4 + 3x0 = [[s(s(double(_x0)))]] 
  [[half(s(s(_x0)))]] = 4 + x0 > 2 + x0 = [[s(half(_x0))]] 
  [[!minus(s(_x0), s(_x1))]] = 4 + x0 + x1 > x0 + x1 = [[!minus(_x0, _x1)]] 

We can thus remove the following rules:

  double(s(X)) => s(s(double(X))) 
  half(s(s(X))) => s(half(X)) 
  !minus(s(X), s(Y)) => !minus(X, Y) 

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.