We consider the system 434.

  Alphabet:

    0 : [] --> nat 
    cons : [nat * natlist] --> natlist 
    foldl : [nat -> nat -> nat * nat * natlist] --> nat 
    nil : [] --> natlist 
    plus : [nat * nat] --> nat 
    sum : [natlist] --> nat 

  Rules:

    foldl(/\x./\y.X(x, y), Y, nil) => Y 
    foldl(/\x./\y.X(x, y), Y, cons(Z, U)) => foldl(/\z./\u.X(z, u), X(Y, Z), U) 
    sum(X) => foldl(/\x./\y.plus(x, y), 0, X) 

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

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> foldl#(/\z./\u.X(z, u), X(Y, Z), U)   
    1] foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(z, u)   
    2] foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(Y, Z)   
    3] sum#(X) =#> foldl#(/\x./\y.plus(x, y), 0, X)   

  Rules R_0:

    foldl(/\x./\y.X(x, y), Y, nil) => Y 
    foldl(/\x./\y.X(x, y), Y, cons(Z, U)) => foldl(/\z./\u.X(z, u), X(Y, Z), U) 
    sum(X) => foldl(/\x./\y.plus(x, y), 0, 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:

  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) >? foldl#(/\z./\u.X(z, u), X(Y, Z), U) 
  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(~c0, ~c1) 
  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(Y, Z) 
  sum#(X) >? foldl#(/\x./\y.plus-(x, y), 0, X) 
  plus-(X, Y) >= plus(X, Y) 

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

  pi( sum#(X) ) = #argfun-sum##(foldl#(/\x./\y.plus-(x, y), 0, X)) 

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]] = _|_ 
  [[foldl#(x_1, x_2, x_3)]] = foldl#(x_3, x_1, x_2) 
  [[sum#(x_1)]] = x_1 
  [[~c0]] = _|_ 
  [[~c1]] = _|_ 

We choose Lex = {foldl#} and Mul = {#argfun-sum##, cons, plus, plus-}, and the following precedence: cons > plus = plus- > foldl# > #argfun-sum##

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

  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) > foldl#(/\x./\y.X(x, y), X(Y, Z), U) 
  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_) 
  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) > X(Y, Z) 
  #argfun-sum##(foldl#(/\x./\y.plus-(x, y), _|_, X)) >= foldl#(/\x./\y.plus-(x, y), _|_, X) 
  plus-(X, Y) >= plus(X, Y) 

With these choices, we have:

  1] foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) > foldl#(/\x./\y.X(x, y), X(Y, Z), U)  because [2], by definition 
  2] foldl#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl#(/\x./\y.X(x, y), X(Y, Z), U)  because [3], [6], [12] and [20], by (Stat) 
  3] cons(Z, U) > U  because [4], by definition 
  4] cons*(Z, U) >= U  because [5], by (Select) 
  5] U >= U  by (Meta) 
  6] foldl#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= /\x./\y.X(x, y)  because [7], by (Select) 
  7] /\x./\z.X(x, z) >= /\x./\z.X(x, z)  because [8], by (Abs) 
  8] /\z.X(y, z) >= /\z.X(y, z)  because [9], by (Abs) 
  9] X(y, x) >= X(y, x)  because [10] and [11], by (Meta) 
  10] y >= y  by (Var) 
  11] x >= x  by (Var) 
  12] foldl#*(/\z./\u.X(z, u), Y, cons(Z, U)) >= X(Y, Z)  because [13], by (Select) 
  13] X(foldl#*(/\z./\u.X(z, u), Y, cons(Z, U)), foldl#*(/\v./\w.X(v, w), Y, cons(Z, U))) >= X(Y, Z)  because [14] and [16], by (Meta) 
  14] foldl#*(/\z./\u.X(z, u), Y, cons(Z, U)) >= Y  because [15], by (Select) 
  15] Y >= Y  by (Meta) 
  16] foldl#*(/\z./\u.X(z, u), Y, cons(Z, U)) >= Z  because [17], by (Select) 
  17] cons(Z, U) >= Z  because [18], by (Star) 
  18] cons*(Z, U) >= Z  because [19], by (Select) 
  19] Z >= Z  by (Meta) 
  20] foldl#*(/\z./\u.X(z, u), Y, cons(Z, U)) >= U  because [21], by (Select) 
  21] cons(Z, U) >= U  because [4], by (Star) 

  22] foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_)  because [23], by (Star) 
  23] foldl#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(_|_, _|_)  because [24], by (Select) 
  24] X(foldl#*(/\x./\y.X(x, y), Y, cons(Z, U)), foldl#*(/\z./\u.X(z, u), Y, cons(Z, U))) >= X(_|_, _|_)  because [25] and [26], by (Meta) 
  25] foldl#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 
  26] foldl#*(/\x./\y.X(x, y), Y, cons(Z, U)) >= _|_  by (Bot) 

  27] foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) > X(Y, Z)  because [12], by definition 

  28] #argfun-sum##(foldl#(/\x./\y.plus-(x, y), _|_, X)) >= foldl#(/\x./\y.plus-(x, y), _|_, X)  because [29], by (Star) 
  29] #argfun-sum##*(foldl#(/\x./\y.plus-(x, y), _|_, X)) >= foldl#(/\x./\y.plus-(x, y), _|_, X)  because [30], by (Select) 
  30] foldl#(/\x./\y.plus-(x, y), _|_, X) >= foldl#(/\x./\y.plus-(x, y), _|_, X)  because [31], [36] and [37], by (Fun) 
  31] /\x./\z.plus-(x, z) >= /\x./\z.plus-(x, z)  because [32], by (Abs) 
  32] /\z.plus-(y, z) >= /\z.plus-(y, z)  because [33], by (Abs) 
  33] plus-(y, x) >= plus-(y, x)  because plus- in Mul, [34] and [35], by (Fun) 
  34] y >= y  by (Var) 
  35] x >= x  by (Var) 
  36] _|_ >= _|_  by (Bot) 
  37] X >= X  by (Meta) 

  38] plus-(X, Y) >= plus(X, Y)  because plus- = plus, plus- in Mul, [39] and [40], by (Fun) 
  39] X >= X  by (Meta) 
  40] Y >= Y  by (Meta) 

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:

  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> X(z, u)   
  sum#(X) =#> foldl#(/\x./\y.plus(x, y), 0, X)   

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 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:

  foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) >? X(~c0, ~c1) 
  sum#(X) >? foldl#(/\x./\y.plus-(x, y), 0, X) 
  plus-(X, Y) >= plus(X, Y) 

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

  pi( sum#(X) ) = #argfun-sum##(foldl#(/\x./\y.plus-(x, y), 0, X)) 

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

The following interpretation satisfies the requirements:

  #argfun-sum## = \y0.3 + y0 
  0 = 0 
  cons = \y0y1.3 
  foldl# = \G0y1y2.3 + y2y2G0(y2,y2) 
  plus = \y0y1.0 
  plus- = \y0y1.3 + 3y0 + 3y1 
  sum# = \y0.0 
  ~c0 = 0 
  ~c1 = 0 

Using this interpretation, the requirements translate to:

  [[foldl#(/\x./\y._x0(x, y), _x1, cons(_x2, _x3))]] = 3 + 9F0(3,3) > F0(0,0) = [[_x0(~c0, ~c1)]] 
  [[#argfun-sum##(foldl#(/\x./\y.plus-(x, y), 0, _x0))]] = 6 + 3x0x0 + 6x0x0x0 > 3 + 3x0x0 + 6x0x0x0 = [[foldl#(/\x./\y.plus-(x, y), 0, _x0)]] 
  [[plus-(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 0 = [[plus(_x0, _x1)]] 

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