We consider the system shuffle.

  Alphabet:

    app : [list * list] --> list 
    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    nil : [] --> list 
    reverse : [list] --> list 
    shuffle : [list] --> list 

  Rules:

    app(nil, x) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    reverse(nil) => nil 
    reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) 
    shuffle(nil) => nil 
    shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 
    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] app#(cons(X, Y), Z) =#> app#(Y, Z)   
    1] reverse#(cons(X, Y)) =#> app#(reverse(Y), cons(X, nil))   
    2] reverse#(cons(X, Y)) =#> reverse#(Y)   
    3] shuffle#(cons(X, Y)) =#> shuffle#(reverse(Y))   
    4] shuffle#(cons(X, Y)) =#> reverse#(Y)   
    5] map#(F, cons(X, Y)) =#> F(X)   
    6] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    app(nil, X) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    reverse(nil) => nil 
    reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 
    shuffle(nil) => nil 
    shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
    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 : 0 
    * 2 : 1, 2 
    * 3 : 3, 4 
    * 4 : 1, 2 
    * 5 : 0, 1, 2, 3, 4, 5, 6 
    * 6 : 5, 6 

This graph has the following strongly connected components:

  P_1:

    app#(cons(X, Y), Z) =#> app#(Y, Z)   

  P_2:

    reverse#(cons(X, Y)) =#> reverse#(Y)   

  P_3:

    shuffle#(cons(X, Y)) =#> shuffle#(reverse(Y))   

  P_4:

    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), (P_3, R_0, m, f) and (P_4, 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), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_0, 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) 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(cons(X, Y)) >= app(reverse(Y), cons(X, nil)) 
  shuffle(nil) >= nil 
  shuffle(cons(X, Y)) >= cons(X, shuffle(reverse(Y))) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + y1 
  cons = \y0y1.2 + y0 + y1 
  map = \G0y1.y1 + y1G0(y1) 
  map# = \G0y1.3 + y1G0(y1) 
  nil = 0 
  reverse = \y0.y0 
  shuffle = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 3 + x2F0(x2) = [[map#(_F0, _x2)]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 4 + 2x0 + 2x1 >= 2 + x0 + 2x1 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + x2 + x2F0(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_4, R_0, minimal, formative) by (P_5, R_0, minimal, formative), where P_5 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), (P_3, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_5, R_0, 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_5, 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 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).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_3, R_0) are:

  app(nil, X) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(cons(X, Y)) >= app(reverse(Y), cons(X, nil)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 
  nil = 0 
  reverse = \y0.y0 
  shuffle# = \y0.y0 

Using this interpretation, the requirements translate to:

  [[shuffle#(cons(_x0, _x1))]] = 1 + x1 > x1 = [[shuffle#(reverse(_x1))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0).  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(reverse#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(reverse#(cons(X, Y))) = cons(X, Y) |> Y = nu(reverse#(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(app#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(app#(cons(X, Y), Z)) = cons(X, Y) |> Y = nu(app#(Y, Z)) 

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.