/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
We consider the system theBenchmark.

  Alphabet:

    a : [] --> B 
    b : [] --> B 
    c : [] --> B 
    f : [A -> B] --> B 

  Rules:

    f(/\x.a) => b 
    b => f(/\x.c) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

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

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

  f(/\x.a) >? b 
  b >? f(/\x.c) 

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

The following interpretation satisfies the requirements:

  a = 3 
  b = 2 
  c = 0 
  f = \G0.3G0(0) 

Using this interpretation, the requirements translate to:

  [[f(/\x.a)]] = 9 > 2 = [[b]] 
  [[b]] = 2 > 0 = [[f(/\x.c)]] 

We can thus remove the following rules:

  f(/\x.a) => b 
  b => f(/\x.c) 

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.