We consider the system AotoYamada_05__010.

  Alphabet:

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

  Rules:

    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 
    curry(f, x) y => f x y 
    map(f) nil => nil 
    map(f) cons(x, y) => cons(f x, map(f) y) 
    inc => map(curry(plus, s(0))) 
    double => map(curry(times, s(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 
    double : [] --> a -> a 
    inc : [] --> a -> a 
    map : [b -> b] --> a -> a 
    nil : [] --> a 
    plus : [b] --> b -> b 
    s : [b] --> b 
    times : [b] --> b -> b 

  Rules:

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

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

Argument functions:

  [[0]] = _|_ 
  [[nil]] = _|_ 

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

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(s(X)), Y) >= s(@_{o -> o}(plus(X), Y)) 
  @_{o -> o}(times(_|_), X) >= _|_ 
  @_{o -> o}(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), 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(s(_|_))) 
  double > map(times(s(s(_|_)))) 

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

  16] @_{o -> o}(times(_|_), X) >= _|_  by (Bot) 

  17] @_{o -> o}(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y)  because [18], by (Star) 
  18] @_{o -> o}*(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y)  because [19], by (Select) 
  19] times(s(X)) @_{o -> o}*(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y)  because [20] 
  20] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y)  because times > @_{o -> o}, [21] and [27], by (Copy) 
  21] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= plus(@_{o -> o}(times(X), Y))  because times > plus and [22], by (Copy) 
  22] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= @_{o -> o}(times(X), Y)  because times > @_{o -> o}, [23] and [27], by (Copy) 
  23] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= times(X)  because times in Mul and [24], by (Stat) 
  24] s(X) > X  because [25], by definition 
  25] s*(X) >= X  because [26], by (Select) 
  26] X >= X  by (Meta) 
  27] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= Y  because [28], by (Select) 
  28] @_{o -> o}*(times(s(X)), Y) >= Y  because [29], by (Select) 
  29] Y >= Y  by (Meta) 

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

  31] @_{o -> o}(map(F), cons(X, Y)) > cons(@_{o -> o}(F, X), @_{o -> o}(map(F), Y))  because [32], by definition 
  32] @_{o -> o}*(map(F), cons(X, Y)) >= cons(@_{o -> o}(F, X), @_{o -> o}(map(F), Y))  because @_{o -> o} > cons, [33] and [43], by (Copy) 
  33] @_{o -> o}*(map(F), cons(X, Y)) >= @_{o -> o}(F, X)  because [34], by (Select) 
  34] map(F) @_{o -> o}*(map(F), cons(X, Y)) >= @_{o -> o}(F, X)  because [35] 
  35] map*(F, @_{o -> o}*(map(F), cons(X, Y))) >= @_{o -> o}(F, X)  because [36], by (Select) 
  36] @_{o -> o}*(map(F), cons(X, Y)) >= @_{o -> o}(F, X)  because @_{o -> o} in Mul, [37] and [40], by (Stat) 
  37] map(F) >= F  because [38], by (Star) 
  38] map*(F) >= F  because [39], by (Select) 
  39] F >= F  by (Meta) 
  40] cons(X, Y) > X  because [41], by definition 
  41] cons*(X, Y) >= X  because [42], by (Select) 
  42] X >= X  by (Meta) 
  43] @_{o -> o}*(map(F), cons(X, Y)) >= @_{o -> o}(map(F), Y)  because @_{o -> o} in Mul, [44] and [46], by (Stat) 
  44] map(F) >= map(F)  because map in Mul and [45], by (Fun) 
  45] F >= F  by (Meta) 
  46] cons(X, Y) > Y  because [47], by definition 
  47] cons*(X, Y) >= Y  because [48], by (Select) 
  48] Y >= Y  by (Meta) 

  49] inc > map(plus(s(_|_)))  because [50], by definition 
  50] inc* >= map(plus(s(_|_)))  because inc > map and [51], by (Copy) 
  51] inc* >= plus(s(_|_))  because inc > plus and [52], by (Copy) 
  52] inc* >= s(_|_)  because inc > s and [53], by (Copy) 
  53] inc* >= _|_  by (Bot) 

  54] double > map(times(s(s(_|_))))  because [55], by definition 
  55] double* >= map(times(s(s(_|_))))  because double > map and [56], by (Copy) 
  56] double* >= times(s(s(_|_)))  because double > times and [57], by (Copy) 
  57] double* >= s(s(_|_))  because double > s and [58], by (Copy) 
  58] double* >= s(_|_)  because double > s and [59], by (Copy) 
  59] double* >= _|_  by (Bot) 

We can thus remove the following rules:

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

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

Argument functions:

  [[0]] = _|_ 
  [[nil]] = _|_ 

We choose Lex = {} and Mul = {map, plus, s, times}, and the following precedence: map > times > plus > s

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

  plus(_|_, X) >= X 
  plus(s(X), Y) > s(plus(X, Y)) 
  times(_|_, X) >= _|_ 
  times(s(X), Y) >= plus(times(X, Y), Y) 
  map(F, _|_) >= _|_ 

With these choices, we have:

  1] plus(_|_, X) >= X  because [2], by (Star) 
  2] plus*(_|_, 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] times(_|_, X) >= _|_  by (Bot) 

  12] times(s(X), Y) >= plus(times(X, Y), Y)  because [13], by (Star) 
  13] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [14] and [19], by (Copy) 
  14] times*(s(X), Y) >= times(X, Y)  because times in Mul, [15] and [18], by (Stat) 
  15] s(X) > X  because [16], by definition 
  16] s*(X) >= X  because [17], by (Select) 
  17] X >= X  by (Meta) 
  18] Y >= Y  by (Meta) 
  19] times*(s(X), Y) >= Y  because [18], by (Select) 

  20] map(F, _|_) >= _|_  by (Bot) 

We can thus remove the following rules:

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

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

Argument functions:

  [[0]] = _|_ 
  [[nil]] = _|_ 

We choose Lex = {} and Mul = {map, plus, s, times}, and the following precedence: s > map > times > plus

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

  plus(_|_, X) >= X 
  times(_|_, X) > _|_ 
  times(s(X), Y) >= plus(times(X, Y), Y) 
  map(F, _|_) >= _|_ 

With these choices, we have:

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

  4] times(_|_, X) > _|_  because [5], by definition 
  5] times*(_|_, X) >= _|_  by (Bot) 

  6] times(s(X), Y) >= plus(times(X, Y), Y)  because [7], by (Star) 
  7] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [8] and [13], by (Copy) 
  8] times*(s(X), Y) >= times(X, Y)  because times in Mul, [9] and [12], by (Stat) 
  9] s(X) > X  because [10], by definition 
  10] s*(X) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 
  12] Y >= Y  by (Meta) 
  13] times*(s(X), Y) >= Y  because [12], by (Select) 

  14] map(F, _|_) >= _|_  by (Bot) 

We can thus remove the following rules:

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

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {0, map, plus, s, times}, and the following precedence: 0 > times > map > plus > 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 
  times(s(X), Y) >= plus(times(X, Y), Y) 
  map(F, _|_) >= _|_ 

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) 

  4] times(s(X), Y) >= plus(times(X, Y), Y)  because [5], by (Star) 
  5] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [6] and [11], by (Copy) 
  6] times*(s(X), Y) >= times(X, Y)  because times 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] times*(s(X), Y) >= Y  because [10], by (Select) 

  12] map(F, _|_) >= _|_  by (Bot) 

We can thus remove the following rules:

  plus(0, X) => 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]):

  times(s(X), Y) >? plus(times(X, Y), Y) 
  map(F, nil) >? nil 

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

Argument functions:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {map, plus, s, times}, and the following precedence: s > map > times > plus

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

  times(s(X), Y) >= plus(times(X, Y), Y) 
  map(F, _|_) > _|_ 

With these choices, we have:

  1] times(s(X), Y) >= plus(times(X, Y), Y)  because [2], by (Star) 
  2] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [3] and [8], by (Copy) 
  3] times*(s(X), Y) >= times(X, Y)  because times in Mul, [4] and [7], by (Stat) 
  4] s(X) > X  because [5], by definition 
  5] s*(X) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 
  7] Y >= Y  by (Meta) 
  8] times*(s(X), Y) >= Y  because [7], by (Select) 

  9] map(F, _|_) > _|_  because [10], by definition 
  10] map*(F, _|_) >= _|_  by (Bot) 

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]):

  times(s(X), Y) >? plus(times(X, Y), Y) 

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

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

With these choices, we have:

  1] times(s(X), Y) > plus(times(X, Y), Y)  because [2], by definition 
  2] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [3] and [8], by (Copy) 
  3] times*(s(X), Y) >= times(X, Y)  because times in Mul, [4] and [7], by (Stat) 
  4] s(X) > X  because [5], by definition 
  5] s*(X) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 
  7] Y >= Y  by (Meta) 
  8] times*(s(X), Y) >= Y  because [7], by (Select) 

We can thus remove the following rules:

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