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 = {s} and Mul = {@_{o -> o -> o}, @_{o -> o}, cons, fold, nil, plus, prod, sum, times}, and the following precedence: cons > nil > prod > sum > fold > @_{o -> o -> o} > @_{o -> o} > 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 (Star) 
  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] @_{o -> o}*(fold(F, X), nil) >= X  because [6], by (Select) 
  6] fold(F, X) @_{o -> o}*(fold(F, X), nil) >= X  because [7] 
  7] fold*(F, X, @_{o -> o}*(fold(F, X), nil)) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 

  9] @_{o -> o}(fold(F, X), cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [10], by definition 
  10] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [11], by (Select) 
  11] 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 [12] 
  12] 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}, [13] and [21], by (Copy) 
  13] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o -> o}(F, Y)  because fold > @_{o -> o -> o}, [14] and [16], by (Copy) 
  14] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= F  because [15], by (Select) 
  15] F >= F  by (Meta) 
  16] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= Y  because [17], by (Select) 
  17] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= Y  because [18], by (Select) 
  18] cons(Y, Z) >= Y  because [19], by (Star) 
  19] cons*(Y, Z) >= Y  because [20], by (Select) 
  20] Y >= Y  by (Meta) 
  21] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o}(fold(F, X), Z)  because [22], by (Select) 
  22] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(fold(F, X), Z)  because @_{o -> o} in Mul, [23] and [26], by (Stat) 
  23] fold(F, X) >= fold(F, X)  because fold in Mul, [24] and [25], by (Fun) 
  24] F >= F  by (Meta) 
  25] X >= X  by (Meta) 
  26] cons(Y, Z) > Z  because [27], by definition 
  27] cons*(Y, Z) >= Z  because [28], by (Select) 
  28] Z >= Z  by (Meta) 

  29] plus(_|_, X) > X  because [30], by definition 
  30] plus*(_|_, X) >= X  because [31], by (Select) 
  31] X >= X  by (Meta) 

  32] plus(s(X), Y) >= s(plus(X, Y))  because [33], by (Star) 
  33] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [34], by (Copy) 
  34] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [35] and [38], by (Stat) 
  35] s(X) > X  because [36], by definition 
  36] s*(X) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 
  38] Y >= Y  by (Meta) 

  39] times(_|_, X) > _|_  because [40], by definition 
  40] times*(_|_, X) >= _|_  by (Bot) 

  41] times(s(X), Y) > plus(times(X, Y), Y)  because [42], by definition 
  42] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [43] and [48], by (Copy) 
  43] times*(s(X), Y) >= times(X, Y)  because times in Mul, [44] and [47], by (Stat) 
  44] s(X) > X  because [45], by definition 
  45] s*(X) >= X  because [46], by (Select) 
  46] X >= X  by (Meta) 
  47] Y >= Y  by (Meta) 
  48] times*(s(X), Y) >= Y  because [47], by (Select) 

  49] sum > fold(_|_, _|_)  because [50], by definition 
  50] sum* >= fold(_|_, _|_)  because sum > fold, [51] and [52], by (Copy) 
  51] sum* >= _|_  by (Bot) 
  52] sum* >= _|_  by (Bot) 

  53] prod >= fold(_|_, s(_|_))  because [54], by (Star) 
  54] prod* >= fold(_|_, s(_|_))  because prod > fold, [55] and [56], by (Copy) 
  55] prod* >= _|_  by (Bot) 
  56] prod* >= s(_|_)  because prod > s and [57], by (Copy) 
  57] prod* >= _|_  by (Bot) 

We can thus remove the following rules:

  fold(F, X) cons(Y, Z) => F Y (fold(F, X) Z) 
  plus(0, X) => X 
  times(0, X) => 0 
  times(s(X), Y) => plus(times(X, Y), Y) 
  sum => fold(add, 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 
  plus(s(X), Y) >? s(plus(X, Y)) 
  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 
  fold = Lam[G0;y1;y2].2 + y1 + 3*y2 + G0(0,0) + 2*G0(y1,y2) + 2*G0(y2,y2) 
  mul = Lam[y0;y1].0 
  nil = 3 
  plus = Lam[y0;y1].y1 + 3*y0 
  prod = Lam[y0].3 + 3*y0 
  s = Lam[y0].y0 

Using this interpretation, the requirements translate to:

  [[fold(_F0, _x1) nil]] = 14 + x1 + F0(0,0) + 2*F0(3,3) + 2*F0(x1,3) > x1 = [[_x1]] 
  [[plus(s(_x0), _x1)]] = x1 + 3*x0 >= x1 + 3*x0 = [[s(plus(_x0, _x1))]] 
  [[prod]] = Lam[y0].3 + 3*y0 > Lam[y0].2 + 3*y0 = [[fold(mul, s(0))]] 

We can thus remove the following rules:

  fold(F, X) nil => X 
  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 = Lam[y0;y1].2*y1 + 3*y0 
  s = Lam[y0].2 + y0 

Using this interpretation, the requirements translate to:

  [[plus(s(_x0), _x1)]] = 6 + 2*x1 + 3*x0 > 2 + 2*x1 + 3*x0 = [[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.