We consider the system Applicative_05__BTreeMember.

  Alphabet:

    0 : [] --> a 
    eq : [a * a] --> c 
    false : [] --> c 
    fork : [b * a * b] --> b 
    if : [c * c * c] --> c 
    lt : [a * a] --> c 
    member : [a * b] --> c 
    null : [] --> b 
    s : [a] --> a 
    true : [] --> c 

  Rules:

    lt(s(x), s(y)) => lt(x, y) 
    lt(0, s(x)) => true 
    lt(x, 0) => false 
    eq(x, x) => true 
    eq(s(x), 0) => false 
    eq(0, s(x)) => false 
    member(x, null) => false 
    member(x, fork(y, z, u)) => if(lt(x, z), member(x, y), if(eq(x, z), true, member(x, u))) 

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]):

  lt(s(X), s(Y)) >? lt(X, Y) 
  lt(0, s(X)) >? true 
  lt(X, 0) >? false 
  eq(X, X) >? true 
  eq(s(X), 0) >? false 
  eq(0, s(X)) >? false 
  member(X, null) >? false 
  member(X, fork(Y, Z, U)) >? if(lt(X, Z), member(X, Y), if(eq(X, Z), true, member(X, U))) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  eq = \y0y1.y0 + y1 
  false = 0 
  fork = \y0y1y2.3 + 3y0 + 3y1 + 3y2 
  if = \y0y1y2.y0 + y1 + y2 
  lt = \y0y1.y0 + y1 
  member = \y0y1.y0 + 2y0y1 + 3y1 
  null = 3 
  s = \y0.3 + 3y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[lt(s(_x0), s(_x1))]] = 6 + 3x0 + 3x1 > x0 + x1 = [[lt(_x0, _x1)]] 
  [[lt(0, s(_x0))]] = 6 + 3x0 > 0 = [[true]] 
  [[lt(_x0, 0)]] = 3 + x0 > 0 = [[false]] 
  [[eq(_x0, _x0)]] = 2x0 >= 0 = [[true]] 
  [[eq(s(_x0), 0)]] = 6 + 3x0 > 0 = [[false]] 
  [[eq(0, s(_x0))]] = 6 + 3x0 > 0 = [[false]] 
  [[member(_x0, null)]] = 9 + 7x0 > 0 = [[false]] 
  [[member(_x0, fork(_x1, _x2, _x3))]] = 9 + 6x0x1 + 6x0x2 + 6x0x3 + 7x0 + 9x1 + 9x2 + 9x3 > 2x0x1 + 2x0x3 + 2x2 + 3x1 + 3x3 + 4x0 = [[if(lt(_x0, _x2), member(_x0, _x1), if(eq(_x0, _x2), true, member(_x0, _x3)))]] 

We can thus remove the following rules:

  lt(s(X), s(Y)) => lt(X, Y) 
  lt(0, s(X)) => true 
  lt(X, 0) => false 
  eq(s(X), 0) => false 
  eq(0, s(X)) => false 
  member(X, null) => false 
  member(X, fork(Y, Z, U)) => if(lt(X, Z), member(X, Y), if(eq(X, Z), true, member(X, U))) 

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

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

  eq(X, X) >? true 

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

The following interpretation satisfies the requirements:

  eq = \y0y1.3 + y0 + y1 
  true = 0 

Using this interpretation, the requirements translate to:

  [[eq(_x0, _x0)]] = 3 + 2x0 > 0 = [[true]] 

We can thus remove the following rules:

  eq(X, X) => true 

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.