We consider the system fuhkop11frocos.

  Alphabet:

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

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    reverse(nil) => nil 
    shuffle(nil) => nil 
    shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 
    mirror(nil) => nil 
    mirror(cons(x, y)) => append(cons(x, mirror(y)), cons(x, nil)) 
    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 static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] append#(cons(X, Y), Z) =#> append#(Y, Z)   
    1] shuffle#(cons(X, Y)) =#> shuffle#(reverse(Y))   
    2] shuffle#(cons(X, Y)) =#> reverse#(Y)   
    3] mirror#(cons(X, Y)) =#> append#(cons(X, mirror(Y)), cons(X, nil))   
    4] mirror#(cons(X, Y)) =#> mirror#(Y)   
    5] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    reverse(nil) => nil 
    shuffle(nil) => nil 
    shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
    mirror(nil) => nil 
    mirror(cons(X, Y)) => append(cons(X, mirror(Y)), cons(X, nil)) 
    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, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  append#(cons(X, Y), Z) >? append#(Y, Z) 
  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  shuffle#(cons(X, Y)) >? reverse#(Y) 
  mirror#(cons(X, Y)) >? append#(cons(X, mirror(Y)), cons(X, nil)) 
  mirror#(cons(X, Y)) >? mirror#(Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  reverse(nil) >= nil 
  shuffle(nil) >= nil 
  shuffle(cons(X, Y)) >= cons(X, shuffle(reverse(Y))) 
  mirror(nil) >= nil 
  mirror(cons(X, Y)) >= append(cons(X, mirror(Y)), cons(X, nil)) 
  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:

  append = \y0y1.y1 
  append# = \y0y1.0 
  cons = \y0y1.0 
  map = \G0y1.2 + 2G0(0) + 2G0(y1) 
  map# = \G0y1.0 
  mirror = \y0.0 
  mirror# = \y0.3 
  nil = 0 
  reverse = \y0.0 
  reverse# = \y0.0 
  shuffle = \y0.2 
  shuffle# = \y0.3 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[shuffle#(cons(_x0, _x1))]] = 3 >= 3 = [[shuffle#(reverse(_x1))]] 
  [[shuffle#(cons(_x0, _x1))]] = 3 > 0 = [[reverse#(_x1)]] 
  [[mirror#(cons(_x0, _x1))]] = 3 > 0 = [[append#(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 
  [[mirror#(cons(_x0, _x1))]] = 3 >= 3 = [[mirror#(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = x2 >= 0 = [[cons(_x0, append(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(nil)]] = 2 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 >= 0 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[mirror(nil)]] = 0 >= 0 = [[nil]] 
  [[mirror(cons(_x0, _x1))]] = 0 >= 0 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 
  [[map(_F0, nil)]] = 2 + 4F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 4F0(0) >= 0 = [[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_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of:

  append#(cons(X, Y), Z) =#> append#(Y, Z)   
  shuffle#(cons(X, Y)) =#> shuffle#(reverse(Y))   
  mirror#(cons(X, Y)) =#> mirror#(Y)   
  map#(F, cons(X, Y)) =#> map#(F, Y)   

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

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  append#(cons(X, Y), Z) >? append#(Y, Z) 
  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  mirror#(cons(X, Y)) >? mirror#(Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  reverse(nil) >= nil 
  shuffle(nil) >= nil 
  shuffle(cons(X, Y)) >= cons(X, shuffle(reverse(Y))) 
  mirror(nil) >= nil 
  mirror(cons(X, Y)) >= append(cons(X, mirror(Y)), cons(X, nil)) 
  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:

  append = \y0y1.y0 + y1 
  append# = \y0y1.0 
  cons = \y0y1.1 
  map = \G0y1.1 
  map# = \G0y1.0 
  mirror = \y0.1 + y0 
  mirror# = \y0.0 
  nil = 0 
  reverse = \y0.0 
  shuffle = \y0.2 
  shuffle# = \y0.y0 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[shuffle#(cons(_x0, _x1))]] = 1 > 0 = [[shuffle#(reverse(_x1))]] 
  [[mirror#(cons(_x0, _x1))]] = 0 >= 0 = [[mirror#(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x2 >= 1 = [[cons(_x0, append(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(nil)]] = 2 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 >= 1 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[mirror(nil)]] = 1 >= 0 = [[nil]] 
  [[mirror(cons(_x0, _x1))]] = 2 >= 2 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 
  [[map(_F0, nil)]] = 1 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 >= 1 = [[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_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

  append#(cons(X, Y), Z) =#> append#(Y, Z)   
  mirror#(cons(X, Y)) =#> mirror#(Y)   
  map#(F, cons(X, Y)) =#> map#(F, Y)   

Thus, the original system is terminating if (P_2, R_0, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  append#(cons(X, Y), Z) >? append#(Y, Z) 
  mirror#(cons(X, Y)) >? mirror#(Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  reverse(nil) >= nil 
  shuffle(nil) >= nil 
  shuffle(cons(X, Y)) >= cons(X, shuffle(reverse(Y))) 
  mirror(nil) >= nil 
  mirror(cons(X, Y)) >= append(cons(X, mirror(Y)), cons(X, nil)) 
  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:

  append = \y0y1.1 + y0 + y1 
  append# = \y0y1.2y0 
  cons = \y0y1.1 + y1 
  map = \G0y1.y1 
  map# = \G0y1.0 
  mirror = \y0.1 + 3y0 
  mirror# = \y0.0 
  nil = 0 
  reverse = \y0.0 
  shuffle = \y0.y0 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 2 + 2x1 > 2x1 = [[append#(_x1, _x2)]] 
  [[mirror#(cons(_x0, _x1))]] = 0 >= 0 = [[mirror#(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = 1 + x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 1 + x1 >= 1 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[mirror(nil)]] = 1 >= 0 = [[nil]] 
  [[mirror(cons(_x0, _x1))]] = 4 + 3x1 >= 4 + 3x1 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[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_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of:

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

Thus, the original system is terminating if (P_3, R_0, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  mirror#(cons(X, Y)) >? mirror#(Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  reverse(nil) >= nil 
  shuffle(nil) >= nil 
  shuffle(cons(X, Y)) >= cons(X, shuffle(reverse(Y))) 
  mirror(nil) >= nil 
  mirror(cons(X, Y)) >= append(cons(X, mirror(Y)), cons(X, nil)) 
  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:

  append = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 
  map = \G0y1.y1 
  map# = \G0y1.0 
  mirror = \y0.1 + 3y0 
  mirror# = \y0.y0 
  nil = 0 
  reverse = \y0.0 
  shuffle = \y0.y0 

Using this interpretation, the requirements translate to:

  [[mirror#(cons(_x0, _x1))]] = 1 + x1 > x1 = [[mirror#(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 1 + x1 >= 1 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[mirror(nil)]] = 1 >= 0 = [[nil]] 
  [[mirror(cons(_x0, _x1))]] = 4 + 3x1 >= 3 + 3x1 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[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_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_0, static, formative) is finite.

We consider the dependency pair problem (P_4, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  map#(F, cons(X, Y)) >? map#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  reverse(nil) >= nil 
  shuffle(nil) >= nil 
  shuffle(cons(X, Y)) >= cons(X, shuffle(reverse(Y))) 
  mirror(nil) >= nil 
  mirror(cons(X, Y)) >= append(cons(X, mirror(Y)), cons(X, nil)) 
  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:

  append = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 
  map = \G0y1.y1 
  map# = \G0y1.y1 
  mirror = \y0.2y0 
  nil = 0 
  reverse = \y0.0 
  shuffle = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 + 2x1 >= 1 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[mirror(nil)]] = 0 >= 0 = [[nil]] 
  [[mirror(cons(_x0, _x1))]] = 2 + 2x1 >= 2 + 2x1 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0).  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.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.