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


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> nat 
    rec : [nat * a * nat -> a -> a] --> a 
    s : [nat] --> nat 

  Rules:

    rec(0, x, f) => x 
    rec(s(x), y, f) => f x rec(x, y, f) 

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]):

  rec(0, X, F) >? X 
  rec(s(X), Y, F) >? F X rec(X, Y, F) 

about to try horpo

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {0, @_{o -> o -> o}, @_{o -> o}, rec, s}, and the following precedence: 0 > @_{o -> o -> o} = rec > @_{o -> o} > s

With these choices, we have:

  1] rec(0, X, F) > X  because [2], by definition 
  2] rec*(0, X, F) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] rec(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because [5], by (Star) 
  5] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because rec > @_{o -> o}, [6] and [11], by (Copy) 
  6] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X)  because rec = @_{o -> o -> o}, rec in Mul, [7] and [10], by (Stat) 
  7] s(X) > X  because [8], by definition 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] F >= F  by (Meta) 
  11] rec*(s(X), Y, F) >= rec(X, Y, F)  because rec in Mul, [7], [12] and [10], by (Stat) 
  12] Y >= Y  by (Meta) 

We can thus remove the following rules:

  rec(0, X, F) => 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]):

  rec(s(X), Y, F) >? F X rec(X, Y, F) 

about to try horpo

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {@_{o -> o -> o}, @_{o -> o}, rec, s}, and the following precedence: rec > @_{o -> o -> o} > @_{o -> o} > s

With these choices, we have:

  1] rec(s(X), Y, F) > @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because [2], by definition 
  2] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because rec > @_{o -> o}, [3] and [10], by (Copy) 
  3] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X)  because rec > @_{o -> o -> o}, [4] and [6], by (Copy) 
  4] rec*(s(X), Y, F) >= F  because [5], by (Select) 
  5] F >= F  by (Meta) 
  6] rec*(s(X), Y, F) >= X  because [7], by (Select) 
  7] s(X) >= X  because [8], by (Star) 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] rec*(s(X), Y, F) >= rec(X, Y, F)  because rec in Mul, [11], [13] and [14], by (Stat) 
  11] s(X) > X  because [12], by definition 
  12] s*(X) >= X  because [9], by (Select) 
  13] Y >= Y  by (Meta) 
  14] F >= F  by (Meta) 

We can thus remove the following rules:

  rec(s(X), Y, F) => F X rec(X, Y, F) 

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.