/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 
  a!6220!6220tail : [o] --> o 
  a!6220!6220zeros : [] --> o 
  cons : [o * o] --> o 
  mark : [o] --> o 
  tail : [o] --> o 
  zeros : [] --> o 

  a!6220!6220zeros => cons(0, zeros) 
  a!6220!6220tail(cons(X, Y)) => mark(Y) 
  mark(zeros) => a!6220!6220zeros 
  mark(tail(X)) => a!6220!6220tail(mark(X)) 
  mark(cons(X, Y)) => cons(mark(X), Y) 
  mark(0) => 0 
  a!6220!6220zeros => zeros 
  a!6220!6220tail(X) => tail(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]):

  a!6220!6220zeros >? cons(0, zeros) 
  a!6220!6220tail(cons(X, Y)) >? mark(Y) 
  mark(zeros) >? a!6220!6220zeros 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  a!6220!6220zeros >? zeros 
  a!6220!6220tail(X) >? tail(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220tail = \y0.3 + 2y0 
  a!6220!6220zeros = 0 
  cons = \y0y1.y0 + y1 
  mark = \y0.2y0 
  tail = \y0.2 + 2y0 
  zeros = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220zeros]] = 0 >= 0 = [[cons(0, zeros)]] 
  [[a!6220!6220tail(cons(_x0, _x1))]] = 3 + 2x0 + 2x1 > 2x1 = [[mark(_x1)]] 
  [[mark(zeros)]] = 0 >= 0 = [[a!6220!6220zeros]] 
  [[mark(tail(_x0))]] = 4 + 4x0 > 3 + 4x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[zeros]] 
  [[a!6220!6220tail(_x0)]] = 3 + 2x0 > 2 + 2x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  a!6220!6220tail(cons(X, Y)) => mark(Y) 
  mark(tail(X)) => a!6220!6220tail(mark(X)) 
  a!6220!6220tail(X) => tail(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]):

  a!6220!6220zeros >? cons(0, zeros) 
  mark(zeros) >? a!6220!6220zeros 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  a!6220!6220zeros >? zeros 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220zeros = 2 
  cons = \y0y1.y0 + y1 
  mark = \y0.3 + 3y0 
  zeros = 1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220zeros]] = 2 > 1 = [[cons(0, zeros)]] 
  [[mark(zeros)]] = 6 > 2 = [[a!6220!6220zeros]] 
  [[mark(cons(_x0, _x1))]] = 3 + 3x0 + 3x1 >= 3 + x1 + 3x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 3 > 0 = [[0]] 
  [[a!6220!6220zeros]] = 2 > 1 = [[zeros]] 

We can thus remove the following rules:

  a!6220!6220zeros => cons(0, zeros) 
  mark(zeros) => a!6220!6220zeros 
  mark(0) => 0 
  a!6220!6220zeros => zeros 

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

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

  mark(cons(X, Y)) >? cons(mark(X), Y) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y0 + y1 
  mark = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[mark(cons(_x0, _x1))]] = 3 + 3x0 + 3x1 > 1 + x1 + 3x0 = [[cons(mark(_x0), _x1)]] 

We can thus remove the following rules:

  mark(cons(X, Y)) => cons(mark(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.