/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 : [o * o] --> o 
  0 : [] --> o 
  s : [o] --> o 

  !plus(X, 0) => X 
  !plus(X, s(Y)) => s(!plus(X, Y)) 
  !plus(0, s(X)) => s(X) 
  s(!plus(0, X)) => s(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]):

  !plus(X, 0) >? X 
  !plus(X, s(Y)) >? s(!plus(X, Y)) 
  !plus(0, s(X)) >? s(X) 
  s(!plus(0, X)) >? s(X) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[!plus(_x0, 0)]] = 9 + 3x0 > x0 = [[_x0]] 
  [[!plus(_x0, s(_x1))]] = 6 + 3x0 + 3x1 > 2 + 3x0 + 3x1 = [[s(!plus(_x0, _x1))]] 
  [[!plus(0, s(_x0))]] = 15 + 3x0 > 2 + x0 = [[s(_x0)]] 
  [[s(!plus(0, _x0))]] = 11 + 3x0 > 2 + x0 = [[s(_x0)]] 

We can thus remove the following rules:

  !plus(X, 0) => X 
  !plus(X, s(Y)) => s(!plus(X, Y)) 
  !plus(0, s(X)) => s(X) 
  s(!plus(0, X)) => s(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.