We consider the system Applicative_05__TreeHeight.

  Alphabet:

    0 : [] --> d 
    cons : [d * c] --> c 
    false : [] --> a 
    height : [] --> d -> d 
    if : [a * d] --> d 
    le : [d * d] --> a 
    map : [d -> d * c] --> c 
    maxlist : [d * c] --> d 
    nil : [] --> c 
    node : [b * c] --> d 
    s : [d] --> d 
    true : [] --> a 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    maxlist(x, cons(y, z)) => if(le(x, y), maxlist(y, z)) 
    maxlist(x, nil) => x 
    height node(x, y) => s(maxlist(0, map(height, 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]):

  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  le(0, X) >? true 
  le(s(X), 0) >? false 
  le(s(X), s(Y)) >? le(X, Y) 
  maxlist(X, cons(Y, Z)) >? if(le(X, Y), maxlist(Y, Z)) 
  maxlist(X, nil) >? X 
  height node(X, Y) >? s(maxlist(0, map(height, Y))) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  cons = Lam[y0;y1].3 + y1 + 3*y0 
  false = 0 
  height = Lam[y0].0 
  if = Lam[y0;y1].y0 + y1 
  le = Lam[y0;y1].1 + y0 + y1 
  map = Lam[G0;y1].3 + 3*y1 + 2*y1*G0(y1) + 2*G0(0) + 3*G0(y1) 
  maxlist = Lam[y0;y1].y1 + 2*y0 
  nil = 0 
  node = Lam[y0;y1].3 + y0 + 3*y1 
  s = Lam[y0].y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[map(_F0, nil)]] = 3 + 5*F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 12 + 3*x2 + 9*x1 + 2*x2*F0(3 + x2 + 3*x1) + 2*F0(0) + 6*x1*F0(3 + x2 + 3*x1) + 9*F0(3 + x2 + 3*x1) > 6 + 3*x1 + 3*x2 + 2*x2*F0(x2) + 2*F0(0) + 3*F0(x1) + 3*F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[le(0, _x0)]] = 1 + x0 > 0 = [[true]] 
  [[le(s(_x0), 0)]] = 1 + x0 > 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[le(_x0, _x1)]] 
  [[maxlist(_x0, cons(_x1, _x2))]] = 3 + x2 + 2*x0 + 3*x1 > 1 + x0 + x2 + 3*x1 = [[if(le(_x0, _x1), maxlist(_x1, _x2))]] 
  [[maxlist(_x0, nil)]] = 2*x0 >= x0 = [[_x0]] 
  [[height node(_x0, _x1)]] = 3 + x0 + 3*x1 >= 3 + 3*x1 = [[s(maxlist(0, map(height, _x1)))]] 

We can thus remove the following rules:

  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  le(0, X) => true 
  le(s(X), 0) => false 
  maxlist(X, cons(Y, Z)) => if(le(X, Y), maxlist(Y, Z)) 

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

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

  le(s(X), s(Y)) >? le(X, Y) 
  maxlist(X, nil) >? X 
  height node(X, Y) >? s(maxlist(0, map(height, Y))) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  height = Lam[y0].2 + 2*y0 
  le = Lam[y0;y1].y0 + y1 
  map = Lam[G0;y1].y1 + G0(0) 
  maxlist = Lam[y0;y1].y0 + y1 
  nil = 3 
  node = Lam[y0;y1].3 + y0 + 3*y1 
  s = Lam[y0].y0 

Using this interpretation, the requirements translate to:

  [[le(s(_x0), s(_x1))]] = x0 + x1 >= x0 + x1 = [[le(_x0, _x1)]] 
  [[maxlist(_x0, nil)]] = 3 + x0 > x0 = [[_x0]] 
  [[height node(_x0, _x1)]] = 11 + 3*x0 + 9*x1 > 2 + x1 = [[s(maxlist(0, map(height, _x1)))]] 

We can thus remove the following rules:

  maxlist(X, nil) => X 
  height node(X, Y) => s(maxlist(0, map(height, 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]):

  le(s(X), s(Y)) >? le(X, Y) 

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

The following interpretation satisfies the requirements:

  le = Lam[y0;y1].2*y0 + 2*y1 
  s = Lam[y0].3 + 3*y0 

Using this interpretation, the requirements translate to:

  [[le(s(_x0), s(_x1))]] = 12 + 6*x0 + 6*x1 > 2*x0 + 2*x1 = [[le(_x0, _x1)]] 

We can thus remove the following rules:

  le(s(X), s(Y)) => le(X, Y) 

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.