We consider the system Applicative_first_order_05__#3.16.

  Alphabet:

    0 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    plus : [a * a] --> a 
    s : [a] --> a 
    times : [a * a] --> a 
    true : [] --> b 

  Rules:

    times(x, 0) => 0 
    times(x, s(y)) => plus(times(x, y), x) 
    plus(x, 0) => x 
    plus(0, x) => x 
    plus(x, s(y)) => s(plus(x, y)) 
    plus(s(x), y) => s(plus(x, y)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

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]):

  times(X, 0) >? 0 
  times(X, s(Y)) >? plus(times(X, Y), X) 
  plus(X, 0) >? X 
  plus(0, X) >? X 
  plus(X, s(Y)) >? s(plus(X, Y)) 
  plus(s(X), Y) >? s(plus(X, Y)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  filter(F, nil) >? nil 
  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >? filter(F, Y) 

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

Argument functions:

  [[cons(x_1, x_2)]] = cons(x_2, x_1) 
  [[filter(x_1, x_2)]] = filter(x_2, x_1) 
  [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_1, x_3) 
  [[nil]] = _|_ 

We choose Lex = {cons, false, filter, filter2, true} and Mul = {0, @_{o -> o}, map, plus, s, times}, and the following precedence: 0 > false > filter = filter2 > map > @_{o -> o} > times > plus > s > cons = true

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

  times(X, 0) >= 0 
  times(X, s(Y)) > plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(0, X) > X 
  plus(X, s(Y)) >= s(plus(X, Y)) 
  plus(s(X), Y) >= s(plus(X, Y)) 
  map(F, _|_) > _|_ 
  map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) 
  filter(F, _|_) > _|_ 
  filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 
  filter2(true, F, X, Y) > cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) > filter(F, Y) 

With these choices, we have:

  1] times(X, 0) >= 0  because [2], by (Star) 
  2] times*(X, 0) >= 0  because [3], by (Select) 
  3] 0 >= 0  by (Fun) 

  4] times(X, s(Y)) > plus(times(X, Y), X)  because [5], by definition 
  5] times*(X, s(Y)) >= plus(times(X, Y), X)  because times > plus, [6] and [11], by (Copy) 
  6] times*(X, s(Y)) >= times(X, Y)  because times in Mul, [7] and [8], by (Stat) 
  7] X >= X  by (Meta) 
  8] s(Y) > Y  because [9], by definition 
  9] s*(Y) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 
  11] times*(X, s(Y)) >= X  because [7], by (Select) 

  12] plus(X, 0) >= X  because [13], by (Star) 
  13] plus*(X, 0) >= X  because [14], by (Select) 
  14] X >= X  by (Meta) 

  15] plus(0, X) > X  because [16], by definition 
  16] plus*(0, X) >= X  because [17], by (Select) 
  17] X >= X  by (Meta) 

  18] plus(X, s(Y)) >= s(plus(X, Y))  because [19], by (Star) 
  19] plus*(X, s(Y)) >= s(plus(X, Y))  because plus > s and [20], by (Copy) 
  20] plus*(X, s(Y)) >= plus(X, Y)  because plus in Mul, [21] and [22], by (Stat) 
  21] X >= X  by (Meta) 
  22] s(Y) > Y  because [23], by definition 
  23] s*(Y) >= Y  because [24], by (Select) 
  24] Y >= Y  by (Meta) 

  25] plus(s(X), Y) >= s(plus(X, Y))  because [26], by (Star) 
  26] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [27], by (Copy) 
  27] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [28] and [31], by (Stat) 
  28] s(X) > X  because [29], by definition 
  29] s*(X) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 
  31] Y >= Y  by (Meta) 

  32] map(F, _|_) > _|_  because [33], by definition 
  33] map*(F, _|_) >= _|_  by (Bot) 

  34] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y))  because [35], by definition 
  35] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [36] and [43], by (Copy) 
  36] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [37] and [39], by (Copy) 
  37] map*(F, cons(X, Y)) >= F  because [38], by (Select) 
  38] F >= F  by (Meta) 
  39] map*(F, cons(X, Y)) >= X  because [40], by (Select) 
  40] cons(X, Y) >= X  because [41], by (Star) 
  41] cons*(X, Y) >= X  because [42], by (Select) 
  42] X >= X  by (Meta) 
  43] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [44] and [45], by (Stat) 
  44] F >= F  by (Meta) 
  45] cons(X, Y) > Y  because [46], by definition 
  46] cons*(X, Y) >= Y  because [47], by (Select) 
  47] Y >= Y  by (Meta) 

  48] filter(F, _|_) > _|_  because [49], by definition 
  49] filter*(F, _|_) >= _|_  by (Bot) 

  50] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because [51], by (Star) 
  51] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [52], [55], [56], [58] and [62], by (Stat) 
  52] cons(X, Y) > Y  because [53], by definition 
  53] cons*(X, Y) >= Y  because [54], by (Select) 
  54] Y >= Y  by (Meta) 
  55] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [56] and [58], by (Copy) 
  56] filter*(F, cons(X, Y)) >= F  because [57], by (Select) 
  57] F >= F  by (Meta) 
  58] filter*(F, cons(X, Y)) >= X  because [59], by (Select) 
  59] cons(X, Y) >= X  because [60], by (Star) 
  60] cons*(X, Y) >= X  because [61], by (Select) 
  61] X >= X  by (Meta) 
  62] filter*(F, cons(X, Y)) >= Y  because [63], by (Select) 
  63] cons(X, Y) >= Y  because [53], by (Star) 

  64] filter2(true, F, X, Y) > cons(X, filter(F, Y))  because [65], by definition 
  65] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [66] and [68], by (Copy) 
  66] filter2*(true, F, X, Y) >= X  because [67], by (Select) 
  67] X >= X  by (Meta) 
  68] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [69], [70], [71], [72], [69] and [70], by (Stat) 
  69] F >= F  by (Meta) 
  70] Y >= Y  by (Meta) 
  71] filter2*(true, F, X, Y) >= F  because [69], by (Select) 
  72] filter2*(true, F, X, Y) >= Y  because [70], by (Select) 

  73] filter2(false, F, X, Y) > filter(F, Y)  because [74], by definition 
  74] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [75], [76], [77], [78], [75] and [76], by (Stat) 
  75] F >= F  by (Meta) 
  76] Y >= Y  by (Meta) 
  77] filter2*(false, F, X, Y) >= F  because [75], by (Select) 
  78] filter2*(false, F, X, Y) >= Y  because [76], by (Select) 

We can thus remove the following rules:

  times(X, s(Y)) => plus(times(X, Y), X) 
  plus(0, X) => X 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  filter(F, nil) => nil 
  filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) => filter(F, 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]):

  times(X, 0) >? 0 
  plus(X, 0) >? X 
  plus(X, s(Y)) >? s(plus(X, Y)) 
  plus(s(X), Y) >? s(plus(X, Y)) 
  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 2 
  cons = Lam[y0;y1].3 + 3*y0 + 3*y1 
  filter = Lam[G0;y1].3 + 3*y1 + 3*y1*G0(y1) + 3*G0(0) + 3*G0(y1) 
  filter2 = Lam[y0;G1;y2;y3].y0 + y2 + y3 + 2*G1(y2) 
  plus = Lam[y0;y1].3 + 3*y0 + 3*y1 
  s = Lam[y0].y0 
  times = Lam[y0;y1].3 + y0 + 3*y1 

Using this interpretation, the requirements translate to:

  [[times(_x0, 0)]] = 9 + x0 > 2 = [[0]] 
  [[plus(_x0, 0)]] = 9 + 3*x0 > x0 = [[_x0]] 
  [[plus(_x0, s(_x1))]] = 3 + 3*x0 + 3*x1 >= 3 + 3*x0 + 3*x1 = [[s(plus(_x0, _x1))]] 
  [[plus(s(_x0), _x1)]] = 3 + 3*x0 + 3*x1 >= 3 + 3*x0 + 3*x1 = [[s(plus(_x0, _x1))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 12 + 9*x1 + 9*x2 + 3*F0(0) + 9*x1*F0(3 + 3*x1 + 3*x2) + 9*x2*F0(3 + 3*x1 + 3*x2) + 12*F0(3 + 3*x1 + 3*x2) > x2 + 2*x1 + 3*F0(x1) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 

We can thus remove the following rules:

  times(X, 0) => 0 
  plus(X, 0) => X 
  filter(F, cons(X, Y)) => filter2(F X, F, X, 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(X, s(Y)) >? s(plus(X, Y)) 
  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].3*y0 + 3*y1 
  s = Lam[y0].2 + y0 

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

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