We consider the system Applicative_05__Ex6Folding.

  Alphabet:

    0 : [] --> c 
    1 : [] --> c 
    add : [] --> c -> a -> c 
    cons : [a * b] --> b 
    fold : [c -> a -> c * b * c] --> c 
    mul : [] --> c -> a -> c 
    nil : [] --> b 
    prod : [b] --> c 
    sum : [b] --> c 

  Rules:

    fold(f, nil, x) => x 
    fold(f, cons(x, y), z) => fold(f, y, f z x) 
    sum(x) => fold(add, x, 0) 
    fold(mul, x, 1) => prod(x) 

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] fold#(F, cons(X, Y), Z) =#> fold#(F, Y, F Z X)   
    1] fold#(F, cons(X, Y), Z) =#> F(Z, X)   
    2] sum#(X) =#> fold#(add, X, 0)   

  Rules R_0:

    fold(F, nil, X) => X 
    fold(F, cons(X, Y), Z) => fold(F, Y, F Z X) 
    sum(X) => fold(add, X, 0) 
    fold(mul, X, 1) => prod(X) 

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).

This combination (P_0, R_0) has no formative rules!  We will name the empty set of rules:R_1.

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:

  fold#(F, cons(X, Y), Z) >? fold#(F, Y, F Z X) 
  fold#(F, cons(X, Y), Z) >? F(Z, X) 
  sum#(X) >? fold#(add, X, 0) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( sum#(X) ) = #argfun-sum##(fold#(add, X, 0)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[0]] = _|_ 
  [[fold#(x_1, x_2, x_3)]] = fold#(x_2, x_1, x_3) 
  [[sum#(x_1)]] = x_1 

We choose Lex = {fold#} and Mul = {#argfun-sum##, @_{o -> o -> o}, @_{o -> o}, add, cons}, and the following precedence: #argfun-sum## = add > cons > fold# > @_{o -> o -> o} > @_{o -> o}

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  fold#(F, cons(X, Y), Z) > fold#(F, Y, @_{o -> o}(@_{o -> o -> o}(F, Z), X)) 
  fold#(F, cons(X, Y), Z) > @_{o -> o}(@_{o -> o -> o}(F, Z), X) 
  #argfun-sum##(fold#(add, X, _|_)) >= fold#(add, X, _|_) 

With these choices, we have:

  1] fold#(F, cons(X, Y), Z) > fold#(F, Y, @_{o -> o}(@_{o -> o -> o}(F, Z), X))  because [2], by definition 
  2] fold#*(F, cons(X, Y), Z) >= fold#(F, Y, @_{o -> o}(@_{o -> o -> o}(F, Z), X))  because [3], [6], [8] and [10], by (Stat) 
  3] cons(X, Y) > Y  because [4], by definition 
  4] cons*(X, Y) >= Y  because [5], by (Select) 
  5] Y >= Y  by (Meta) 
  6] fold#*(F, cons(X, Y), Z) >= F  because [7], by (Select) 
  7] F >= F  by (Meta) 
  8] fold#*(F, cons(X, Y), Z) >= Y  because [9], by (Select) 
  9] cons(X, Y) >= Y  because [4], by (Star) 
  10] fold#*(F, cons(X, Y), Z) >= @_{o -> o}(@_{o -> o -> o}(F, Z), X)  because fold# > @_{o -> o}, [11] and [14], by (Copy) 
  11] fold#*(F, cons(X, Y), Z) >= @_{o -> o -> o}(F, Z)  because fold# > @_{o -> o -> o}, [6] and [12], by (Copy) 
  12] fold#*(F, cons(X, Y), Z) >= Z  because [13], by (Select) 
  13] Z >= Z  by (Meta) 
  14] fold#*(F, cons(X, Y), Z) >= X  because [15], by (Select) 
  15] cons(X, Y) >= X  because [16], by (Star) 
  16] cons*(X, Y) >= X  because [17], by (Select) 
  17] X >= X  by (Meta) 

  18] fold#(F, cons(X, Y), Z) > @_{o -> o}(@_{o -> o -> o}(F, Z), X)  because [10], by definition 

  19] #argfun-sum##(fold#(add, X, _|_)) >= fold#(add, X, _|_)  because [20], by (Star) 
  20] #argfun-sum##*(fold#(add, X, _|_)) >= fold#(add, X, _|_)  because #argfun-sum## > fold#, [21], [22] and [26], by (Copy) 
  21] #argfun-sum##*(fold#(add, X, _|_)) >= add  because #argfun-sum## = add and #argfun-sum## in Mul, by (Stat) 
  22] #argfun-sum##*(fold#(add, X, _|_)) >= X  because [23], by (Select) 
  23] fold#(add, X, _|_) >= X  because [24], by (Star) 
  24] fold#*(add, X, _|_) >= X  because [25], by (Select) 
  25] X >= X  by (Meta) 
  26] #argfun-sum##*(fold#(add, X, _|_)) >= _|_  by (Bot) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of:

  sum#(X) =#> fold#(add, X, 0)   

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, 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 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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.