We consider the system AotoYamada_05__027.

  Alphabet:

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

  Rules:

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

  plus(0) X >? X 
  plus(s(X)) Y >? s(plus(X) Y) 
  inc(X) >? map(plus(s(0)), X) 
  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:

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

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

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

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] inc(X) >= map(plus(s(_|_)), X)  because [17], by (Star) 
  17] inc*(X) >= map(plus(s(_|_)), X)  because inc > map, [18] and [21], by (Copy) 
  18] inc*(X) >= plus(s(_|_))  because inc > plus and [19], by (Copy) 
  19] inc*(X) >= s(_|_)  because inc > s and [20], by (Copy) 
  20] inc*(X) >= _|_  by (Bot) 
  21] inc*(X) >= X  because [22], by (Select) 
  22] X >= X  by (Meta) 

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

  24] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y))  because [25], by definition 
  25] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y))  because map > cons, [26] and [33], by (Copy) 
  26] map*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because map > @_{o -> o}, [27] and [29], by (Copy) 
  27] map*(F, cons(X, Y)) >= F  because [28], by (Select) 
  28] F >= F  by (Meta) 
  29] map*(F, cons(X, Y)) >= X  because [30], by (Select) 
  30] cons(X, Y) >= X  because [31], by (Star) 
  31] cons*(X, Y) >= X  because [32], by (Select) 
  32] X >= X  by (Meta) 
  33] map*(F, cons(X, Y)) >= map(F, Y)  because map in Mul, [34] and [35], by (Stat) 
  34] F >= F  by (Meta) 
  35] cons(X, Y) > Y  because [36], by definition 
  36] cons*(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)) 

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

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

Argument functions:

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

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

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) 
  inc(X) >= map(plus(_|_), X) 
  map(F, nil) > nil 

With these choices, we have:

  1] @_{o -> o}(plus(_|_), X) > X  because [2], by definition 
  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] inc(X) >= map(plus(_|_), X)  because [9], by (Star) 
  9] inc*(X) >= map(plus(_|_), X)  because inc > map, [10] and [12], by (Copy) 
  10] inc*(X) >= plus(_|_)  because inc > plus and [11], by (Copy) 
  11] inc*(X) >= _|_  by (Bot) 
  12] inc*(X) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 

  14] map(F, nil) > nil  because [15], by definition 
  15] map*(F, nil) >= nil  because [16], by (Select) 
  16] nil >= nil  by (Fun) 

We can thus remove the following rules:

  plus(0) X => X 
  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]):

  plus(s(X)) Y >? s(plus(X) Y) 
  inc(X) >? map(plus(s(0)), X) 

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

Argument functions:

  [[0]] = _|_ 

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

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

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

With these choices, we have:

  1] @_{o -> o}(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y))  because [2], by (Star) 
  2] @_{o -> o}*(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y))  because [3], by (Select) 
  3] plus(s(X)) @_{o -> o}*(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y))  because [4] 
  4] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= s(@_{o -> o}(plus(X), Y))  because plus > s and [5], by (Copy) 
  5] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= @_{o -> o}(plus(X), Y)  because plus > @_{o -> o}, [6] and [10], by (Copy) 
  6] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= plus(X)  because plus in Mul and [7], by (Stat) 
  7] s(X) > X  because [8], by definition 
  8] s*(X) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= Y  because [11], by (Select) 
  11] @_{o -> o}*(plus(s(X)), Y) >= Y  because [12], by (Select) 
  12] Y >= Y  by (Meta) 

  13] inc(X) > map(plus(s(_|_)), X)  because [14], by definition 
  14] inc*(X) >= map(plus(s(_|_)), X)  because inc > map, [15] and [18], by (Copy) 
  15] inc*(X) >= plus(s(_|_))  because inc > plus and [16], by (Copy) 
  16] inc*(X) >= s(_|_)  because inc > s and [17], by (Copy) 
  17] inc*(X) >= _|_  by (Bot) 
  18] inc*(X) >= X  because [19], by (Select) 
  19] X >= X  by (Meta) 

We can thus remove the following rules:

  inc(X) => map(plus(s(0)), 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]):

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

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

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

With these choices, we have:

  1] plus(s(X), Y) > s(plus(X, Y))  because [2], by definition 
  2] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [3], by (Copy) 
  3] plus*(s(X), Y) >= plus(X, Y)  because plus 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) 

We can thus remove the following rules:

  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.