We consider the system AotoYamada_05__014.

  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)))) 

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

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] plus(s(X)) Y =#> plus(X) Y   
    1] plus(s(X)) Y =#> plus#(X)   
    2] times(s(X)) Y =#> plus(times(X) Y) Y   
    3] times(s(X)) Y =#> plus#(times(X) Y)   
    4] times(s(X)) Y =#> times(X) Y   
    5] times(s(X)) Y =#> times#(X)   
    6] map(F) cons(X, Y) =#> F(X)   
    7] map(F) cons(X, Y) =#> map(F) Y   
    8] map(F) cons(X, Y) =#> map#(F)   
    9] inc X =#> map(plus(s(0))) X   
    10] inc# =#> map#(plus(s(0)))   
    11] inc# =#> plus#(s(0))   
    12] double X =#> map(times(s(s(0)))) X   
    13] double# =#> map#(times(s(s(0))))   
    14] double# =#> times#(s(s(0)))   

  Rules R_0:

    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)))) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0, 1 
    * 1 :  
    * 2 : 0, 1 
    * 3 :  
    * 4 : 2, 3, 4, 5 
    * 5 :  
    * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 7 : 6, 7, 8 
    * 8 :  
    * 9 : 6, 7, 8 
    * 10 :  
    * 11 :  
    * 12 : 6, 7, 8 
    * 13 :  
    * 14 :  

This graph has the following strongly connected components:

  P_1:

    plus(s(X)) Y =#> plus(X) Y   

  P_2:

    times(s(X)) Y =#> times(X) Y   

  P_3:

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

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

The formative rules of (P_3, R_0) are R_1 ::=

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

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_3, R_1, minimal, formative).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

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

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( double(X) ) = #argfun-double#(map(times(s(s(0))), X), map(times(s(s(0))), X)) 
  pi( inc(X) ) = #argfun-inc#(map(plus(s(0)), X), map(plus(s(0)), X)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  #argfun-double# = \y0y1.3 + max(y0, y1) 
  #argfun-inc# = \y0y1.3 + max(y0, y1) 
  0 = 0 
  cons = \y0y1.2 + y0 + y1 
  double = \y0.0 
  inc = \y0.0 
  map = \G0y1.y1 + G0(y1) + 3y1G0(y1) + 3G0(0) 
  plus = \y0y1.0 
  s = \y0.0 
  times = \y0y1.0 

Using this interpretation, the requirements translate to:

  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 3x1F0(2 + x1 + x2) + 3x2F0(2 + x1 + x2) + 3F0(0) + 7F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 3x1F0(2 + x1 + x2) + 3x2F0(2 + x1 + x2) + 3F0(0) + 7F0(2 + x1 + x2) > x2 + F0(x2) + 3x2F0(x2) + 3F0(0) = [[map(_F0, _x2)]] 
  [[#argfun-inc#(map(plus(s(0)), _x0), map(plus(s(0)), _x0))]] = 3 + x0 > x0 = [[map(plus(s(0)), _x0)]] 
  [[#argfun-double#(map(times(s(s(0))), _x0), map(times(s(s(0))), _x0))]] = 3 + x0 > x0 = [[map(times(s(s(0))), _x0)]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 3x1F0(2 + x1 + x2) + 3x2F0(2 + x1 + x2) + 3F0(0) + 7F0(2 + x1 + x2) >= 2 + x2 + F0(x2) + 3x2F0(x2) + 3F0(0) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[#argfun-inc#(map(plus(s(0)), _x0), map(plus(s(0)), _x0))]] = 3 + x0 >= x0 = [[map(plus(s(0)), _x0)]] 
  [[#argfun-double#(map(times(s(s(0))), _x0), map(times(s(s(0))), _x0))]] = 3 + x0 >= x0 = [[map(times(s(s(0))), _x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_1) by ({}, R_1).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(times) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(times(s(X)) Y) = s(X) |> X = nu(times(X) Y) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(plus) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(plus(s(X)) Y) = s(X) |> X = nu(plus(X) Y) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.