/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:

    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    merge : [list * list * list] --> list 
    nil : [] --> list 

  Rules:

    merge(nil, nil, x) => x 
    merge(nil, cons(x, y), z) => merge(y, nil, cons(x, z)) 
    merge(cons(x, y), z, u) => merge(z, y, cons(x, u)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 

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]):

  merge(nil, nil, X) >? X 
  merge(nil, cons(X, Y), Z) >? merge(Y, nil, cons(X, Z)) 
  merge(cons(X, Y), Z, U) >? merge(Z, Y, cons(X, U)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y0 + y1 
  map = \G0y1.3y1 + G0(0) + 2y1G0(y1) 
  merge = \y0y1y2.3 + y0 + y1 + y2 
  nil = 2 

Using this interpretation, the requirements translate to:

  [[merge(nil, nil, _x0)]] = 7 + x0 > x0 = [[_x0]] 
  [[merge(nil, cons(_x0, _x1), _x2)]] = 6 + x0 + x1 + x2 >= 6 + x0 + x1 + x2 = [[merge(_x1, nil, cons(_x0, _x2))]] 
  [[merge(cons(_x0, _x1), _x2, _x3)]] = 4 + x0 + x1 + x2 + x3 >= 4 + x0 + x1 + x2 + x3 = [[merge(_x2, _x1, cons(_x0, _x3))]] 
  [[map(_F0, nil)]] = 6 + F0(0) + 4F0(2) > 2 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 3 + 3x1 + 3x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 1 + x1 + 3x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 

We can thus remove the following rules:

  merge(nil, nil, X) => X 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 

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

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

  merge(nil, cons(X, Y), Z) >? merge(Y, nil, cons(X, Z)) 
  merge(cons(X, Y), Z, U) >? merge(Z, Y, cons(X, U)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + y1 + 3y0 
  merge = \y0y1y2.2y2 + 3y0 + 3y1 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[merge(nil, cons(_x0, _x1), _x2)]] = 9 + 2x2 + 3x1 + 9x0 > 6 + 2x2 + 3x1 + 6x0 = [[merge(_x1, nil, cons(_x0, _x2))]] 
  [[merge(cons(_x0, _x1), _x2, _x3)]] = 9 + 2x3 + 3x1 + 3x2 + 9x0 > 6 + 2x3 + 3x1 + 3x2 + 6x0 = [[merge(_x2, _x1, cons(_x0, _x3))]] 

We can thus remove the following rules:

  merge(nil, cons(X, Y), Z) => merge(Y, nil, cons(X, Z)) 
  merge(cons(X, Y), Z, U) => merge(Z, Y, cons(X, U)) 

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.