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

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

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

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + y1 
  0 = 1 
  fib = \y0.2 + 2y0 
  s = \y0.1 + 3y0 

Using this interpretation, the requirements translate to:

  [[fib(0)]] = 4 > 1 = [[0]] 
  [[fib(s(0))]] = 10 > 4 = [[s(0)]] 
  [[fib(s(s(_x0)))]] = 10 + 18x0 > 6 + 8x0 = [[!plus(fib(s(_x0)), fib(_x0))]] 

We can thus remove the following rules:

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