/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.

  !plus!plus : [o * o] --> o 
  merge : [o * o] --> o 
  nil : [] --> o 
  u : [] --> o 
  v : [] --> o 

  merge(X, nil) => X 
  merge(nil, X) => X 
  merge(!plus!plus(X, Y), !plus!plus(u, v)) => !plus!plus(X, merge(Y, !plus!plus(u, v))) 
  merge(!plus!plus(X, Y), !plus!plus(u, v)) => !plus!plus(u, merge(!plus!plus(X, Y), v)) 

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

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

  merge(X, nil) >? X 
  merge(nil, X) >? X 
  merge(!plus!plus(X, Y), !plus!plus(u, v)) >? !plus!plus(X, merge(Y, !plus!plus(u, v))) 
  merge(!plus!plus(X, Y), !plus!plus(u, v)) >? !plus!plus(u, merge(!plus!plus(X, Y), v)) 

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

The following interpretation satisfies the requirements:

  !plus!plus = \y0y1.y0 + y1 
  merge = \y0y1.y1 + 2y0 
  nil = 3 
  u = 0 
  v = 0 

Using this interpretation, the requirements translate to:

  [[merge(_x0, nil)]] = 3 + 2x0 > x0 = [[_x0]] 
  [[merge(nil, _x0)]] = 6 + x0 > x0 = [[_x0]] 
  [[merge(!plus!plus(_x0, _x1), !plus!plus(u, v))]] = 2x0 + 2x1 >= x0 + 2x1 = [[!plus!plus(_x0, merge(_x1, !plus!plus(u, v)))]] 
  [[merge(!plus!plus(_x0, _x1), !plus!plus(u, v))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!plus!plus(u, merge(!plus!plus(_x0, _x1), v))]] 

We can thus remove the following rules:

  merge(X, nil) => X 
  merge(nil, 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]):

  merge(!plus!plus(X, Y), !plus!plus(u, v)) >? !plus!plus(X, merge(Y, !plus!plus(u, v))) 
  merge(!plus!plus(X, Y), !plus!plus(u, v)) >? !plus!plus(u, merge(!plus!plus(X, Y), v)) 

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

The following interpretation satisfies the requirements:

  !plus!plus = \y0y1.y0 + y1 
  merge = \y0y1.y0 + 2y1 
  u = 2 
  v = 0 

Using this interpretation, the requirements translate to:

  [[merge(!plus!plus(_x0, _x1), !plus!plus(u, v))]] = 4 + x0 + x1 >= 4 + x0 + x1 = [[!plus!plus(_x0, merge(_x1, !plus!plus(u, v)))]] 
  [[merge(!plus!plus(_x0, _x1), !plus!plus(u, v))]] = 4 + x0 + x1 > 2 + x0 + x1 = [[!plus!plus(u, merge(!plus!plus(_x0, _x1), v))]] 

We can thus remove the following rules:

  merge(!plus!plus(X, Y), !plus!plus(u, v)) => !plus!plus(u, merge(!plus!plus(X, Y), v)) 

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

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

  merge(!plus!plus(X, Y), !plus!plus(u, v)) >? !plus!plus(X, merge(Y, !plus!plus(u, v))) 

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

The following interpretation satisfies the requirements:

  !plus!plus = \y0y1.1 + y0 + y1 
  merge = \y0y1.y1 + 3y0 
  u = 0 
  v = 0 

Using this interpretation, the requirements translate to:

  [[merge(!plus!plus(_x0, _x1), !plus!plus(u, v))]] = 4 + 3x0 + 3x1 > 2 + x0 + 3x1 = [[!plus!plus(_x0, merge(_x1, !plus!plus(u, v)))]] 

We can thus remove the following rules:

  merge(!plus!plus(X, Y), !plus!plus(u, v)) => !plus!plus(X, merge(Y, !plus!plus(u, v))) 

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.