We consider the system AotoYamada_05__006.

  Alphabet:

    branch : [a * b * b] --> b 
    leaf : [a] --> b 
    mapbt : [a -> a * b] --> b 

  Rules:

    mapbt(f, leaf(x)) => leaf(f x) 
    mapbt(f, branch(x, y, z)) => branch(f x, mapbt(f, y), mapbt(f, z)) 

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

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] mapbt#(F, leaf(X)) =#> F(X)   
    1] mapbt#(F, branch(X, Y, Z)) =#> F(X)   
    2] mapbt#(F, branch(X, Y, Z)) =#> mapbt#(F, Y)   
    3] mapbt#(F, branch(X, Y, Z)) =#> mapbt#(F, Z)   

  Rules R_0:

    mapbt(F, leaf(X)) => leaf(F X) 
    mapbt(F, branch(X, Y, Z)) => branch(F X, mapbt(F, Y), mapbt(F, Z)) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  mapbt#(F, leaf(X)) >? F(X) 
  mapbt#(F, branch(X, Y, Z)) >? F(X) 
  mapbt#(F, branch(X, Y, Z)) >? mapbt#(F, Y) 
  mapbt#(F, branch(X, Y, Z)) >? mapbt#(F, Z) 
  mapbt(F, leaf(X)) >= leaf(F X) 
  mapbt(F, branch(X, Y, Z)) >= branch(F X, mapbt(F, Y), mapbt(F, Z)) 

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

The following interpretation satisfies the requirements:

  branch = \y0y1y2.3 + y0 + y1 + y2 
  leaf = \y0.3 + y0 
  mapbt = \G0y1.2y1 + 3y1G0(y1) 
  mapbt# = \G0y1.3 + y1 + y1G0(y1) 

Using this interpretation, the requirements translate to:

  [[mapbt#(_F0, leaf(_x1))]] = 6 + x1 + 3F0(3 + x1) + x1F0(3 + x1) > F0(x1) = [[_F0(_x1)]] 
  [[mapbt#(_F0, branch(_x1, _x2, _x3))]] = 6 + x1 + x2 + x3 + 3F0(3 + x1 + x2 + x3) + x1F0(3 + x1 + x2 + x3) + x2F0(3 + x1 + x2 + x3) + x3F0(3 + x1 + x2 + x3) > F0(x1) = [[_F0(_x1)]] 
  [[mapbt#(_F0, branch(_x1, _x2, _x3))]] = 6 + x1 + x2 + x3 + 3F0(3 + x1 + x2 + x3) + x1F0(3 + x1 + x2 + x3) + x2F0(3 + x1 + x2 + x3) + x3F0(3 + x1 + x2 + x3) > 3 + x2 + x2F0(x2) = [[mapbt#(_F0, _x2)]] 
  [[mapbt#(_F0, branch(_x1, _x2, _x3))]] = 6 + x1 + x2 + x3 + 3F0(3 + x1 + x2 + x3) + x1F0(3 + x1 + x2 + x3) + x2F0(3 + x1 + x2 + x3) + x3F0(3 + x1 + x2 + x3) > 3 + x3 + x3F0(x3) = [[mapbt#(_F0, _x3)]] 
  [[mapbt(_F0, leaf(_x1))]] = 6 + 2x1 + 3x1F0(3 + x1) + 9F0(3 + x1) >= 3 + max(x1, F0(x1)) = [[leaf(_F0 _x1)]] 
  [[mapbt(_F0, branch(_x1, _x2, _x3))]] = 6 + 2x1 + 2x2 + 2x3 + 3x1F0(3 + x1 + x2 + x3) + 3x2F0(3 + x1 + x2 + x3) + 3x3F0(3 + x1 + x2 + x3) + 9F0(3 + x1 + x2 + x3) >= 3 + 2x2 + 2x3 + 3x2F0(x2) + 3x3F0(x3) + max(x1, F0(x1)) = [[branch(_F0 _x1, mapbt(_F0, _x2), mapbt(_F0, _x3))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.