We consider the system AotoYamada_05__021.

  Alphabet:

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

  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 
    inc(x) => map(plus(s(0)), x) 
    double(x) => map(times(s(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 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] inc#(X) =#> map#(plus(s(0)), X)   
    7] inc#(X) =#> plus#(s(0))   
    8] double#(X) =#> map#(times(s(s(0))), X)   
    9] double#(X) =#> times#(s(s(0)))   
    10] map#(F, cons(X, Y)) =#> F(X)   
    11] map#(F, cons(X, Y)) =#> map#(F, Y)   

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

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

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:

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

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 ::=

  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)) 

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:

  inc#(X) >? map#(plus(s(0)), X) 
  double#(X) >? map#(times(s(s(0))), X) 
  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)) 

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

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

Using this interpretation, the requirements translate to:

  [[#argfun-inc##(map#(plus(s(0)), _x0))]] = 6 > 3 = [[map#(plus(s(0)), _x0)]] 
  [[#argfun-double##(map#(times(s(s(0))), _x0))]] = 6 > 3 = [[map#(times(s(s(0))), _x0)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(3 + x1 + 2x2) + 4x2F0(3 + x1 + 2x2) + 7F0(3 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(3 + x1 + 2x2) + 4x2F0(3 + x1 + 2x2) + 7F0(3 + x1 + 2x2) >= 3 + F0(x2) + 2x2F0(x2) = [[map#(_F0, _x2)]] 
  [[#argfun-inc#(map(plus(s(0)), _x0))]] = 3 + x0 >= x0 = [[map(plus(s(0)), _x0)]] 
  [[#argfun-double#(map(times(s(s(0))), _x0))]] = 3 + x0 >= x0 = [[map(times(s(s(0))), _x0)]] 
  [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + 2x2 + 3x1F0(3 + x1 + 2x2) + 6x2F0(3 + x1 + 2x2) + 9F0(3 + x1 + 2x2) >= 3 + 2x2 + 6x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of:

  map#(F, cons(X, Y)) =#> map#(F, Y)   

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_1, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_1, minimal, f) by ({}, R_1, minimal, f).  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.