We consider the system Applicative_first_order_05__#3.27.

  Alphabet:

    cons : [c * d] --> d 
    f : [a] --> a 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    g : [a] --> a 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    f(f(x)) => g(f(x)) 
    g(g(x)) => f(x) 
    map(h, nil) => nil 
    map(h, cons(x, y)) => cons(h x, map(h, y)) 
    filter(h, nil) => nil 
    filter(h, cons(x, y)) => filter2(h x, h, x, y) 
    filter2(true, h, x, y) => cons(x, filter(h, y)) 
    filter2(false, h, x, y) => filter(h, 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]):

  f(f(X)) >? g(f(X)) 
  g(g(X)) >? f(X) 
  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:

  [[f(x_1)]] = 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_3, x_1) 
  [[g(x_1)]] = x_1 
  [[nil]] = _|_ 

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

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

  X >= X 
  X >= X 
  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] X >= X  by (Meta) 

  2] X >= X  by (Meta) 

  3] map(F, _|_) > _|_  because [4], by definition 
  4] map*(F, _|_) >= _|_  by (Bot) 

  5] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [6], by (Star) 
  6] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [7] and [14], by (Copy) 
  7] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [8] and [10], by (Copy) 
  8] map*(F, cons(X, Y)) >= F  because [9], by (Select) 
  9] F >= F  by (Meta) 
  10] map*(F, cons(X, Y)) >= X  because [11], by (Select) 
  11] cons(X, Y) >= X  because [12], by (Star) 
  12] cons*(X, Y) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 
  14] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [15] and [16], by (Stat) 
  15] F >= F  by (Meta) 
  16] cons(X, Y) > Y  because [17], by definition 
  17] cons*(X, Y) >= Y  because [18], by (Select) 
  18] Y >= Y  by (Meta) 

  19] filter(F, _|_) >= _|_  by (Bot) 

  20] filter(F, cons(X, Y)) > filter2(@_{o -> o}(F, X), F, X, Y)  because [21], by definition 
  21] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y)  because filter = filter2, [22], [25], [26], [28] and [32], by (Stat) 
  22] cons(X, Y) > Y  because [23], by definition 
  23] cons*(X, Y) >= Y  because [24], by (Select) 
  24] Y >= Y  by (Meta) 
  25] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [26] and [28], by (Copy) 
  26] filter*(F, cons(X, Y)) >= F  because [27], by (Select) 
  27] F >= F  by (Meta) 
  28] filter*(F, cons(X, Y)) >= X  because [29], by (Select) 
  29] cons(X, Y) >= X  because [30], by (Star) 
  30] cons*(X, Y) >= X  because [31], by (Select) 
  31] X >= X  by (Meta) 
  32] filter*(F, cons(X, Y)) >= Y  because [33], by (Select) 
  33] cons(X, Y) >= Y  because [23], by (Star) 

  34] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [35], by (Star) 
  35] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [36] and [38], by (Copy) 
  36] filter2*(true, F, X, Y) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 
  38] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 = filter, [39], [40], [41] and [42], by (Stat) 
  39] F >= F  by (Meta) 
  40] Y >= Y  by (Meta) 
  41] filter2*(true, F, X, Y) >= F  because [39], by (Select) 
  42] filter2*(true, F, X, Y) >= Y  because [40], by (Select) 

  43] filter2(false, F, X, Y) >= filter(F, Y)  because [44], by (Star) 
  44] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 = filter, [45], [46], [47] and [48], by (Stat) 
  45] F >= F  by (Meta) 
  46] Y >= Y  by (Meta) 
  47] filter2*(false, F, X, Y) >= F  because [45], by (Select) 
  48] filter2*(false, F, X, Y) >= Y  because [46], by (Select) 

We can thus remove the following rules:

  map(F, nil) => nil 
  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]):

  f(f(X)) >? g(f(X)) 
  g(g(X)) >? f(X) 
  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 a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[f(x_1)]] = x_1 
  [[g(x_1)]] = x_1 
  [[nil]] = _|_ 

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

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

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

With these choices, we have:

  1] X >= X  by (Meta) 

  2] X >= X  by (Meta) 

  3] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because [4], by (Star) 
  4] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [5] and [12], by (Copy) 
  5] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [6] and [8], by (Copy) 
  6] map*(F, cons(X, Y)) >= F  because [7], by (Select) 
  7] F >= F  by (Meta) 
  8] map*(F, cons(X, Y)) >= X  because [9], by (Select) 
  9] cons(X, Y) >= X  because [10], by (Star) 
  10] cons*(X, Y) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 
  12] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [13] and [14], by (Stat) 
  13] F >= F  by (Meta) 
  14] cons(X, Y) > Y  because [15], by definition 
  15] cons*(X, Y) >= Y  because [16], by (Select) 
  16] Y >= Y  by (Meta) 

  17] filter(F, _|_) > _|_  because [18], by definition 
  18] filter*(F, _|_) >= _|_  by (Bot) 

  19] filter2(true, F, X, Y) >= cons(X, filter(F, Y))  because [20], by (Star) 
  20] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [21] and [23], by (Copy) 
  21] filter2*(true, F, X, Y) >= X  because [22], by (Select) 
  22] X >= X  by (Meta) 
  23] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 > filter, [24] and [26], by (Copy) 
  24] filter2*(true, F, X, Y) >= F  because [25], by (Select) 
  25] F >= F  by (Meta) 
  26] filter2*(true, F, X, Y) >= Y  because [27], by (Select) 
  27] Y >= Y  by (Meta) 

  28] filter2(false, F, X, Y) >= filter(F, Y)  because [29], by (Star) 
  29] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 > filter, [30] and [32], by (Copy) 
  30] filter2*(false, F, X, Y) >= F  because [31], by (Select) 
  31] F >= F  by (Meta) 
  32] filter2*(false, F, X, Y) >= Y  because [33], by (Select) 
  33] Y >= Y  by (Meta) 

We can thus remove the following rules:

  filter(F, nil) => nil 

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

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

  f(f(X)) >? g(f(X)) 
  g(g(X)) >? f(X) 
  map(F, cons(X, Y)) >? cons(F X, map(F, 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].

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

With these choices, we have:

  1] f(f(X)) >= g(f(X))  because f = g, f in Mul and [2], by (Fun) 
  2] f(X) >= f(X)  because f in Mul and [3], by (Fun) 
  3] X >= X  by (Meta) 

  4] g(g(X)) >= f(X)  because [5], by (Star) 
  5] g*(g(X)) >= f(X)  because g = f, g in Mul and [6], by (Stat) 
  6] g(X) > X  because [7], by definition 
  7] g*(X) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 

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

  23] filter2(true, F, X, Y) > cons(X, filter(F, Y))  because [24], by definition 
  24] filter2*(true, F, X, Y) >= cons(X, filter(F, Y))  because filter2 > cons, [25] and [27], by (Copy) 
  25] filter2*(true, F, X, Y) >= X  because [26], by (Select) 
  26] X >= X  by (Meta) 
  27] filter2*(true, F, X, Y) >= filter(F, Y)  because filter2 > filter, [28] and [30], by (Copy) 
  28] filter2*(true, F, X, Y) >= F  because [29], by (Select) 
  29] F >= F  by (Meta) 
  30] filter2*(true, F, X, Y) >= Y  because [31], by (Select) 
  31] Y >= Y  by (Meta) 

  32] filter2(false, F, X, Y) > filter(F, Y)  because [33], by definition 
  33] filter2*(false, F, X, Y) >= filter(F, Y)  because filter2 > filter, [34] and [36], by (Copy) 
  34] filter2*(false, F, X, Y) >= F  because [35], by (Select) 
  35] F >= F  by (Meta) 
  36] filter2*(false, F, X, Y) >= Y  because [37], by (Select) 
  37] Y >= Y  by (Meta) 

We can thus remove the following rules:

  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  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]):

  f(f(X)) >? g(f(X)) 
  g(g(X)) >? f(X) 

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

We choose Lex = {} and Mul = {f, g}, and the following precedence: f = g

With these choices, we have:

  1] f(f(X)) >= g(f(X))  because f = g, f in Mul and [2], by (Fun) 
  2] f(X) >= f(X)  because f in Mul and [3], by (Fun) 
  3] X >= X  by (Meta) 

  4] g(g(X)) > f(X)  because [5], by definition 
  5] g*(g(X)) >= f(X)  because [6], by (Select) 
  6] g(X) >= f(X)  because g = f, g in Mul and [7], by (Fun) 
  7] X >= X  by (Meta) 

We can thus remove the following rules:

  g(g(X)) => f(X) 

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

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

  f(f(X)) >? g(f(X)) 

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

Argument functions:

  [[g(x_1)]] = x_1 

We choose Lex = {} and Mul = {f}, and the following precedence: f

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

  f(f(X)) > f(X) 

With these choices, we have:

  1] f(f(X)) > f(X)  because [2], by definition 
  2] f*(f(X)) >= f(X)  because [3], by (Select) 
  3] f(X) >= f(X)  because f in Mul and [4], by (Fun) 
  4] X >= X  by (Meta) 

We can thus remove the following rules:

  f(f(X)) => g(f(X)) 

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.