We consider the system AotoYamada_05__026.

  Alphabet:

    comp : [c -> c * c -> c] --> c -> c 
    cons : [a * b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 
    twice : [c -> c] --> c -> c 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    comp(f, g) x => f (g x) 
    twice(f) => comp(f, f) 

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]):

  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  comp(F, G) X >? F (G X) 
  twice(F) >? comp(F, F) 

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {@_{o -> o}, comp, cons, map, twice}, and the following precedence: map > cons > twice > comp > @_{o -> o}

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

  map(F, _|_) > _|_ 
  map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 
  @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X)) 
  twice(F) >= comp(F, F) 

With these choices, we have:

  1] map(F, _|_) > _|_  because [2], by definition 
  2] map*(F, _|_) >= _|_  by (Bot) 

  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] @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [18], by (Star) 
  18] @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [19], by (Select) 
  19] comp(F, G) @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [20] 
  20] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [21] and [23], by (Copy) 
  21] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= F  because [22], by (Select) 
  22] F >= F  by (Meta) 
  23] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [24] and [26], by (Copy) 
  24] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= G  because [25], by (Select) 
  25] G >= G  by (Meta) 
  26] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= X  because [27], by (Select) 
  27] @_{o -> o}*(comp(F, G), X) >= X  because [28], by (Select) 
  28] X >= X  by (Meta) 

  29] twice(F) >= comp(F, F)  because [30], by (Star) 
  30] twice*(F) >= comp(F, F)  because twice > comp, [31] and [31], by (Copy) 
  31] twice*(F) >= F  because [32], by (Select) 
  32] F >= F  by (Meta) 

We can thus remove the following rules:

  map(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]):

  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  comp(F, G) X >? F (G X) 
  twice(F) >? comp(F, F) 

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

Argument functions:

  [[map(x_1, x_2)]] = map(x_2, x_1) 

We choose Lex = {cons, map} and Mul = {@_{o -> o}, comp, twice}, and the following precedence: map > cons > twice > comp > @_{o -> o}

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

  map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) 
  @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X)) 
  twice(F) > comp(F, F) 

With these choices, we have:

  1] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y))  because [2], by definition 
  2] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [3] and [10], by (Copy) 
  3] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [4] and [6], by (Copy) 
  4] map*(F, cons(X, Y)) >= F  because [5], by (Select) 
  5] F >= F  by (Meta) 
  6] map*(F, cons(X, Y)) >= X  because [7], by (Select) 
  7] cons(X, Y) >= X  because [8], by (Star) 
  8] cons*(X, Y) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] map*(F, cons(X, Y)) >= map(F, Y)  because [11], [4], [14], [16] and [11], by (Stat) 
  11] cons(X, Y) > Y  because [12], by definition 
  12] cons*(X, Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 
  14] map*(F, cons(X, Y)) >= Y  because [15], by (Select) 
  15] cons(X, Y) >= Y  because [12], by (Star) 
  16] F >= F  by (Meta) 

  17] @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [18], by (Star) 
  18] @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [19], by (Select) 
  19] comp(F, G) @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [20] 
  20] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [21] and [23], by (Copy) 
  21] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= F  because [22], by (Select) 
  22] F >= F  by (Meta) 
  23] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [24] and [26], by (Copy) 
  24] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= G  because [25], by (Select) 
  25] G >= G  by (Meta) 
  26] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= X  because [27], by (Select) 
  27] @_{o -> o}*(comp(F, G), X) >= X  because [28], by (Select) 
  28] X >= X  by (Meta) 

  29] twice(F) > comp(F, F)  because [30], by definition 
  30] twice*(F) >= comp(F, F)  because twice > comp, [31] and [31], by (Copy) 
  31] twice*(F) >= F  because [32], by (Select) 
  32] F >= F  by (Meta) 

We can thus remove the following rules:

  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  twice(F) => comp(F, F) 

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

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

  comp(F, G, X) >? F (G X) 

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

Argument functions:

  [[comp(x_1, x_2, x_3)]] = comp(x_2, x_3, x_1) 

We choose Lex = {comp} and Mul = {@_{o -> o}}, and the following precedence: comp > @_{o -> o}

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

  comp(F, G, X) > @_{o -> o}(F, @_{o -> o}(G, X)) 

With these choices, we have:

  1] comp(F, G, X) > @_{o -> o}(F, @_{o -> o}(G, X))  because [2], by definition 
  2] comp*(F, G, X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [3] and [5], by (Copy) 
  3] comp*(F, G, X) >= F  because [4], by (Select) 
  4] F >= F  by (Meta) 
  5] comp*(F, G, X) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [6] and [8], by (Copy) 
  6] comp*(F, G, X) >= G  because [7], by (Select) 
  7] G >= G  by (Meta) 
  8] comp*(F, G, X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 

We can thus remove the following rules:

  comp(F, G, X) => F (G 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.