/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 : [o * o] --> o 
  !times : [o * o] --> o 
  0 : [] --> o 
  s : [o] --> o 
  sqr : [o] --> o 
  sum : [o] --> o 

  sum(0) => 0 
  sum(s(X)) => !plus(sqr(s(X)), sum(X)) 
  sqr(X) => !times(X, X) 
  sum(s(X)) => !plus(!times(s(X), s(X)), sum(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]):

  sum(0) >? 0 
  sum(s(X)) >? !plus(sqr(s(X)), sum(X)) 
  sqr(X) >? !times(X, X) 
  sum(s(X)) >? !plus(!times(s(X), s(X)), sum(X)) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + y1 
  !times = \y0y1.y0 + y1 
  0 = 0 
  s = \y0.2 + 3y0 
  sqr = \y0.2 + 2y0 
  sum = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[sum(0)]] = 3 > 0 = [[0]] 
  [[sum(s(_x0))]] = 9 + 9x0 >= 9 + 9x0 = [[!plus(sqr(s(_x0)), sum(_x0))]] 
  [[sqr(_x0)]] = 2 + 2x0 > 2x0 = [[!times(_x0, _x0)]] 
  [[sum(s(_x0))]] = 9 + 9x0 > 7 + 9x0 = [[!plus(!times(s(_x0), s(_x0)), sum(_x0))]] 

We can thus remove the following rules:

  sum(0) => 0 
  sqr(X) => !times(X, X) 
  sum(s(X)) => !plus(!times(s(X), s(X)), sum(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]):

  sum(s(X)) >? !plus(sqr(s(X)), sum(X)) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + y1 
  s = \y0.2 + 3y0 
  sqr = \y0.y0 
  sum = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[sum(s(_x0))]] = 6 + 9x0 > 2 + 6x0 = [[!plus(sqr(s(_x0)), sum(_x0))]] 

We can thus remove the following rules:

  sum(s(X)) => !plus(sqr(s(X)), sum(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.