/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.

  !plus!plus : [o * o] --> o 
  a : [] --> o 
  b : [] --> o 
  rev : [o] --> o 

  rev(a) => a 
  rev(b) => b 
  rev(!plus!plus(X, Y)) => !plus!plus(rev(Y), rev(X)) 
  rev(!plus!plus(X, X)) => rev(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]):

  rev(a) >? a 
  rev(b) >? b 
  rev(!plus!plus(X, Y)) >? !plus!plus(rev(Y), rev(X)) 
  rev(!plus!plus(X, X)) >? rev(X) 

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

The following interpretation satisfies the requirements:

  !plus!plus = \y0y1.y0 + y1 
  a = 2 
  b = 0 
  rev = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[rev(a)]] = 4 > 2 = [[a]] 
  [[rev(b)]] = 0 >= 0 = [[b]] 
  [[rev(!plus!plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!plus!plus(rev(_x1), rev(_x0))]] 
  [[rev(!plus!plus(_x0, _x0))]] = 4x0 >= 2x0 = [[rev(_x0)]] 

We can thus remove the following rules:

  rev(a) => a 

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

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

  rev(b) >? b 
  rev(!plus!plus(X, Y)) >? !plus!plus(rev(Y), rev(X)) 
  rev(!plus!plus(X, X)) >? rev(X) 

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

The following interpretation satisfies the requirements:

  !plus!plus = \y0y1.y0 + y1 
  b = 2 
  rev = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[rev(b)]] = 4 > 2 = [[b]] 
  [[rev(!plus!plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!plus!plus(rev(_x1), rev(_x0))]] 
  [[rev(!plus!plus(_x0, _x0))]] = 4x0 >= 2x0 = [[rev(_x0)]] 

We can thus remove the following rules:

  rev(b) => b 

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

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

  rev(!plus!plus(X, Y)) >? !plus!plus(rev(Y), rev(X)) 
  rev(!plus!plus(X, X)) >? rev(X) 

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

The following interpretation satisfies the requirements:

  !plus!plus = \y0y1.2 + y0 + y1 
  rev = \y0.2y0 

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

  rev(!plus!plus(X, Y)) => !plus!plus(rev(Y), rev(X)) 
  rev(!plus!plus(X, X)) => rev(X) 

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.