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 rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  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)) 

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

Argument functions:

  [[0]] = _|_ 
  [[add]] = _|_ 
  [[mul]] = _|_ 

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

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

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

With these choices, we have:

  1] @_{o -> o}(fold(F, X), nil) > X  because [2], by definition 
  2] @_{o -> o}*(fold(F, X), nil) >= X  because [3], by (Select) 
  3] fold(F, X) @_{o -> o}*(fold(F, X), nil) >= X  because [4] 
  4] fold*(F, X, @_{o -> o}*(fold(F, X), nil)) >= X  because [5], by (Select) 
  5] X >= X  by (Meta) 

  6] @_{o -> o}(fold(F, X), cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [7], by definition 
  7] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [8], by (Select) 
  8] fold(F, X) @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [9] 
  9] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because fold > @_{o -> o}, [10] and [18], by (Copy) 
  10] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o -> o}(F, Y)  because fold > @_{o -> o -> o}, [11] and [13], by (Copy) 
  11] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= F  because [12], by (Select) 
  12] F >= F  by (Meta) 
  13] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= Y  because [14], by (Select) 
  14] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= Y  because [15], by (Select) 
  15] cons(Y, Z) >= Y  because [16], by (Star) 
  16] cons*(Y, Z) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 
  18] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o}(fold(F, X), Z)  because [19], by (Select) 
  19] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(fold(F, X), Z)  because @_{o -> o} in Mul, [20] and [23], by (Stat) 
  20] fold(F, X) >= fold(F, X)  because fold in Mul, [21] and [22], by (Fun) 
  21] F >= F  by (Meta) 
  22] X >= X  by (Meta) 
  23] cons(Y, Z) > Z  because [24], by definition 
  24] cons*(Y, Z) >= Z  because [25], by (Select) 
  25] Z >= Z  by (Meta) 

  26] plus(_|_, X) >= X  because [27], by (Star) 
  27] plus*(_|_, X) >= X  because [28], by (Select) 
  28] X >= X  by (Meta) 

  29] plus(s(X), Y) >= s(plus(X, Y))  because [30], by (Star) 
  30] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [31], by (Copy) 
  31] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [32] and [35], by (Stat) 
  32] s(X) > X  because [33], by definition 
  33] s*(X) >= X  because [34], by (Select) 
  34] X >= X  by (Meta) 
  35] Y >= Y  by (Meta) 

  36] times(_|_, X) >= _|_  by (Bot) 

  37] times(s(X), Y) > plus(times(X, Y), Y)  because [38], by definition 
  38] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [39] and [44], by (Copy) 
  39] times*(s(X), Y) >= times(X, Y)  because times in Mul, [40] and [43], by (Stat) 
  40] s(X) > X  because [41], by definition 
  41] s*(X) >= X  because [42], by (Select) 
  42] X >= X  by (Meta) 
  43] Y >= Y  by (Meta) 
  44] times*(s(X), Y) >= Y  because [43], by (Select) 

  45] sum >= fold(_|_, _|_)  because [46], by (Star) 
  46] sum* >= fold(_|_, _|_)  because sum > fold, [47] and [48], by (Copy) 
  47] sum* >= _|_  by (Bot) 
  48] sum* >= _|_  by (Bot) 

  49] prod >= fold(_|_, s(_|_))  because [50], by (Star) 
  50] prod* >= fold(_|_, s(_|_))  because prod > fold, [51] and [52], by (Copy) 
  51] prod* >= _|_  by (Bot) 
  52] prod* >= s(_|_)  because prod > s and [53], by (Copy) 
  53] prod* >= _|_  by (Bot) 

We can thus remove the following rules:

  fold(F, X) nil => X 
  fold(F, X) cons(Y, Z) => F Y (fold(F, X) Z) 
  times(s(X), Y) => plus(times(X, Y), Y) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  plus(0, X) >? X 
  plus(s(X), Y) >? s(plus(X, Y)) 
  times(0, X) >? 0 
  sum >? fold(add, 0) 
  prod >? fold(mul, s(0)) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  add = \y0y1.0 
  fold = \G0y1y2.y1 + G0(0,0) 
  mul = \y0y1.0 
  plus = \y0y1.3 + y1 + 3y0 
  prod = \y0.3 + 3y0 
  s = \y0.y0 
  sum = \y0.3 + 3y0 
  times = \y0y1.3 + y1 + 3y0 

Using this interpretation, the requirements translate to:

  [[plus(0, _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 3 + x1 + 3x0 >= 3 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
  [[times(0, _x0)]] = 3 + x0 > 0 = [[0]] 
  [[sum]] = \y0.3 + 3y0 > \y0.0 = [[fold(add, 0)]] 
  [[prod]] = \y0.3 + 3y0 > \y0.0 = [[fold(mul, s(0))]] 

We can thus remove the following rules:

  plus(0, X) => X 
  times(0, X) => 0 
  sum => fold(add, 0) 
  prod => fold(mul, s(0)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  plus(s(X), Y) >? s(plus(X, Y)) 

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

The following interpretation satisfies the requirements:

  plus = \y0y1.y1 + 3y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[plus(s(_x0), _x1)]] = 3 + x1 + 3x0 > 1 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 

We can thus remove the following rules:

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

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.