We consider the system merge.

  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 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] merge#(nil, cons(X, Y), Z) =#> merge#(Y, nil, cons(X, Z))   
    1] merge#(cons(X, Y), Z, U) =#> merge#(Z, Y, cons(X, U))   
    2] map#(F, cons(X, Y)) =#> F(X)   
    3] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    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)) 

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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 1 
    * 1 : 0, 1 
    * 2 : 0, 1, 2, 3 
    * 3 : 2, 3 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

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

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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:

  map#(F, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  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 = \G0y1.2y1 + y1G0(y1) 
  map# = \G0y1.3 + y1 + y1G0(y1) 
  merge = \y0y1y2.y0 + y1 + y2 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 3 + x2 + x2F0(x2) = [[map#(_F0, _x2)]] 
  [[merge(nil, nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[merge(nil, cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[merge(_x1, nil, cons(_x0, _x2))]] 
  [[merge(cons(_x0, _x1), _x2, _x3)]] = 1 + x0 + x1 + x2 + x3 >= 1 + x0 + x1 + x2 + x3 = [[merge(_x2, _x1, cons(_x0, _x3))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + 2x2 + x2F0(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_2, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_1, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  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 + 3y0 + 3y1 
  merge# = \y0y1y2.3y0 + 3y1 
  nil = 0 

Using this interpretation, the requirements translate to:

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

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, 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.