We consider the system AotoYamada_05__011.

  Alphabet:

    0 : [] --> b 
    cons : [b * a] --> a 
    curry : [b -> b -> b * b] --> b -> b 
    inc : [] --> a -> a 
    map : [b -> b] --> a -> a 
    nil : [] --> a 
    plus : [] --> b -> b -> b 
    s : [b] --> b 

  Rules:

    plus 0 x => x 
    plus s(x) y => s(plus x y) 
    map(f) nil => nil 
    map(f) cons(x, y) => cons(f x, map(f) y) 
    curry(f, x) y => f x y 
    inc => map(curry(plus, s(0))) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

Symbol curry is an encoding for application that is only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing it.  This gives:

  Alphabet:

    0 : [] --> b 
    cons : [b * a] --> a 
    inc : [] --> a -> a 
    map : [b -> b] --> a -> a 
    nil : [] --> a 
    plus : [b] --> b -> b 
    s : [b] --> b 

  Rules:

    plus(0) X => X 
    plus(s(X)) Y => s(plus(X) Y) 
    map(F) nil => nil 
    map(F) cons(X, Y) => cons(F X, map(F) Y) 
    inc => map(plus(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(0) X >? X 
  plus(s(X)) Y >? s(plus(X) Y) 
  map(F) nil >? nil 
  map(F) cons(X, Y) >? cons(F X, map(F) Y) 
  inc >? map(plus(s(0))) 

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

Argument functions:

  [[0]] = _|_ 
  [[nil]] = _|_ 
  [[s(x_1)]] = x_1 

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

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

  @_{o -> o}(plus(_|_), X) >= X 
  @_{o -> o}(plus(X), Y) >= @_{o -> o}(plus(X), Y) 
  @_{o -> o}(map(F), _|_) >= _|_ 
  @_{o -> o}(map(F), cons(X, Y)) >= cons(@_{o -> o}(F, X), @_{o -> o}(map(F), Y)) 
  inc > map(plus(_|_)) 

With these choices, we have:

  1] @_{o -> o}(plus(_|_), X) >= X  because [2], by (Star) 
  2] @_{o -> o}*(plus(_|_), X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] @_{o -> o}(plus(X), Y) >= @_{o -> o}(plus(X), Y)  because @_{o -> o} in Mul, [5] and [7], by (Fun) 
  5] plus(X) >= plus(X)  because plus in Mul and [6], by (Fun) 
  6] X >= X  by (Meta) 
  7] Y >= Y  by (Meta) 

  8] @_{o -> o}(map(F), _|_) >= _|_  by (Bot) 

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

  24] inc > map(plus(_|_))  because [25], by definition 
  25] inc* >= map(plus(_|_))  because inc > map and [26], by (Copy) 
  26] inc* >= plus(_|_)  because inc > plus and [27], by (Copy) 
  27] inc* >= _|_  by (Bot) 

We can thus remove the following rules:

  inc => map(plus(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(0, X) >? X 
  plus(s(X), Y) >? s(plus(X, Y)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 

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

Argument functions:

  [[nil]] = _|_ 

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

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

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

With these choices, we have:

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

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

  11] map(F, _|_) > _|_  because [12], by definition 
  12] map*(F, _|_) >= _|_  by (Bot) 

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

We can thus remove the following rules:

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

  plus(0, X) >? X 

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

We choose Lex = {} and Mul = {0, plus}, and the following precedence: 0 > plus

With these choices, we have:

  1] plus(0, X) > X  because [2], by definition 
  2] plus*(0, X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

We can thus remove the following rules:

  plus(0, X) => 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.