We consider the system AotoYamada_05__024.

  Alphabet:

    cons : [a * b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 

  Rules:

    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 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] map#(F, cons(X, Y)) =#> F(X)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

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

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).

The formative rules of (P_0, R_0) are R_1 ::=

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

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_1, 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:

  map#(F, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  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 + 2y1 
  map = \G0y1.y1 + 2y1G0(y1) 
  map# = \G0y1.3 + 2y1 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 5 + 2x1 + 4x2 + F0(1 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 5 + 2x1 + 4x2 + F0(1 + x1 + 2x2) > 3 + 2x2 + F0(x2) = [[map#(_F0, _x2)]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x1 + 2x2 + 2x1F0(1 + x1 + 2x2) + 2F0(1 + x1 + 2x2) + 4x2F0(1 + x1 + 2x2) >= 1 + 2x2 + 4x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_1) by ({}, R_1).  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.