We consider the system Applicative_05__Ex5Folding.

  Alphabet:

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

  Rules:

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

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, X) cons(Y, Z) =#> F(Y, fold(F, X) Z)   
    1] fold(F, X) cons(Y, Z) =#> fold(F, X) Z   
    2] fold(F, X) cons(Y, Z) =#> fold#(F, X)   
    3] plus#(s(X), Y) =#> plus#(X, Y)   
    4] times#(s(X), Y) =#> plus#(times(X, Y), Y)   
    5] times#(s(X), Y) =#> times#(X, Y)   
    6] sum X =#> fold(add, 0) X   
    7] sum# =#> fold#(add, 0)   
    8] prod X =#> fold(mul, s(0)) X   
    9] prod# =#> fold#(mul, s(0))   

  Rules R_0:

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

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 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 
    * 1 : 0, 1, 2 
    * 2 :  
    * 3 : 3 
    * 4 : 3 
    * 5 : 4, 5 
    * 6 : 0, 1, 2 
    * 7 :  
    * 8 : 0, 1, 2 
    * 9 :  

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    plus#(s(X), Y) =#> plus#(X, Y)   

  P_3:

    times#(s(X), Y) =#> times#(X, 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), (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(times#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(times#(s(X), Y)) = s(X) |> X = nu(times#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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 apply the subterm criterion with the following projection function:

  nu(plus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(plus#(s(X), Y)) = s(X) |> X = nu(plus#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f).  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).

This combination (P_1, 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_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

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:

  fold(F, X, cons(Y, Z)) >? F(Y, fold(F, X, Z)) 
  fold(F, X, cons(Y, Z)) >? fold(F, X, Z) 
  sum(X) >? fold(add, 0, X) 
  prod(X) >? fold(mul, s(0), X) 

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

  pi( prod(X) ) = #argfun-prod#(fold(mul, s(0), X)) 
  pi( sum(X) ) = #argfun-sum#(fold(add, 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]] = _|_ 
  [[add]] = _|_ 
  [[mul]] = _|_ 
  [[prod(x_1)]] = x_1 
  [[sum(x_1)]] = x_1 

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

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

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

With these choices, we have:

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

  16] fold(F, X, cons(Y, Z)) >= fold(F, X, Z)  because [10], by (Star) 

  17] #argfun-sum#(fold(_|_, _|_, X)) >= fold(_|_, _|_, X)  because [18], by (Star) 
  18] #argfun-sum#*(fold(_|_, _|_, X)) >= fold(_|_, _|_, X)  because #argfun-sum# > fold, [19], [20] and [21], by (Copy) 
  19] #argfun-sum#*(fold(_|_, _|_, X)) >= _|_  by (Bot) 
  20] #argfun-sum#*(fold(_|_, _|_, X)) >= _|_  by (Bot) 
  21] #argfun-sum#*(fold(_|_, _|_, X)) >= X  because [22], by (Select) 
  22] fold(_|_, _|_, X) >= X  because [23], by (Star) 
  23] fold*(_|_, _|_, X) >= X  because [24], by (Select) 
  24] X >= X  by (Meta) 

  25] #argfun-prod#(fold(_|_, s(_|_), X)) >= fold(_|_, s(_|_), X)  because [26], by (Star) 
  26] #argfun-prod#*(fold(_|_, s(_|_), X)) >= fold(_|_, s(_|_), X)  because #argfun-prod# > fold, [27], [28] and [32], by (Copy) 
  27] #argfun-prod#*(fold(_|_, s(_|_), X)) >= _|_  by (Bot) 
  28] #argfun-prod#*(fold(_|_, s(_|_), X)) >= s(_|_)  because [29], by (Select) 
  29] fold(_|_, s(_|_), X) >= s(_|_)  because [30], by (Star) 
  30] fold*(_|_, s(_|_), X) >= s(_|_)  because fold > s and [31], by (Copy) 
  31] fold*(_|_, s(_|_), X) >= _|_  by (Bot) 
  32] #argfun-prod#*(fold(_|_, s(_|_), X)) >= X  because [33], by (Select) 
  33] fold(_|_, s(_|_), X) >= X  because [34], by (Star) 
  34] fold*(_|_, s(_|_), X) >= X  because [35], by (Select) 
  35] X >= X  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, 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 : 0 
    * 1 : 0 
    * 2 : 0 

This graph has the following strongly connected components:

  P_5:

    fold(F, X) cons(Y, Z) =#> fold(F, X) Z   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_4, R_1, m, f) by (P_5, R_1, m, f).

Thus, the original system is terminating if (P_5, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_5, R_1, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(fold) = 3 

Thus, we can orient the dependency pairs as follows:

  nu(fold(F, X) cons(Y, Z)) = cons(Y, Z) |> Z = nu(fold(F, X) Z) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_1, minimal, f) by ({}, R_1, minimal, f).  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.