We consider the system fuhkop12rta1.

  Alphabet:

    app : [list * list] --> list 
    cons : [nat * list] --> list 
    hshuffle : [nat -> nat * list] --> list 
    nil : [] --> list 
    reverse : [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)) 
    hshuffle(f, nil) => nil 
    hshuffle(f, cons(x, y)) => cons(f x, hshuffle(f, reverse(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]):

  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)) 
  hshuffle(F, nil) >? nil 
  hshuffle(F, cons(X, Y)) >? cons(F X, hshuffle(F, reverse(Y))) 

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

The following interpretation satisfies the requirements:

  app = Lam[y0;y1].y0 + y1 
  cons = Lam[y0;y1].3 + y0 + y1 
  hshuffle = Lam[G0;y1].2 + 3*y1 + G0(0) + 2*y1*G0(y1) + 3*G0(y1) 
  nil = 0 
  reverse = Lam[y0].2 + y0 

Using this interpretation, the requirements translate to:

  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 3 + x0 + x1 + x2 >= 3 + x0 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 2 > 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 5 + x0 + x1 >= 5 + x0 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[hshuffle(_F0, nil)]] = 2 + 4*F0(0) > 0 = [[nil]] 
  [[hshuffle(_F0, cons(_x1, _x2))]] = 11 + 3*x1 + 3*x2 + F0(0) + 2*x1*F0(3 + x1 + x2) + 2*x2*F0(3 + x1 + x2) + 9*F0(3 + x1 + x2) >= 11 + x1 + 3*x2 + F0(0) + F0(x1) + 2*x2*F0(2 + x2) + 7*F0(2 + x2) = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 

We can thus remove the following rules:

  reverse(nil) => nil 
  hshuffle(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]):

  app(nil, X) >? X 
  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
  reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 
  hshuffle(F, cons(X, Y)) >? cons(F X, hshuffle(F, reverse(Y))) 

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

The following interpretation satisfies the requirements:

  app = Lam[y0;y1].y0 + y1 
  cons = Lam[y0;y1].3 + y1 + 2*y0 
  hshuffle = Lam[G0;y1].2*y1 + G0(0) + 3*y1*G0(y1) 
  nil = 0 
  reverse = Lam[y0].y0 

Using this interpretation, the requirements translate to:

  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 3 + x1 + x2 + 2*x0 >= 3 + x1 + x2 + 2*x0 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(cons(_x0, _x1))]] = 3 + x1 + 2*x0 >= 3 + x1 + 2*x0 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[hshuffle(_F0, cons(_x1, _x2))]] = 6 + 2*x2 + 4*x1 + F0(0) + 3*x2*F0(3 + x2 + 2*x1) + 6*x1*F0(3 + x2 + 2*x1) + 9*F0(3 + x2 + 2*x1) > 3 + 2*x1 + 2*x2 + F0(0) + 2*F0(x1) + 3*x2*F0(x2) = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 

We can thus remove the following rules:

  hshuffle(F, cons(X, Y)) => cons(F X, hshuffle(F, reverse(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]):

  app(nil, X) >? X 
  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
  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 = Lam[y0;y1].1 + y0 + 2*y1 
  cons = Lam[y0;y1].3 + y0 + y1 
  nil = 1 
  reverse = Lam[y0].2 + 3*y0 

Using this interpretation, the requirements translate to:

  [[app(nil, _x0)]] = 2 + 2*x0 > x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 4 + x0 + x1 + 2*x2 >= 4 + x0 + x1 + 2*x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(cons(_x0, _x1))]] = 11 + 3*x0 + 3*x1 >= 11 + 2*x0 + 3*x1 = [[app(reverse(_x1), cons(_x0, nil))]] 

We can thus remove the following rules:

  app(nil, X) => 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]):

  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
  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 = Lam[y0;y1].1 + y0 + y1 
  cons = Lam[y0;y1].1 + y1 + 2*y0 
  nil = 0 
  reverse = Lam[y0].3*y0 

Using this interpretation, the requirements translate to:

  [[app(cons(_x0, _x1), _x2)]] = 2 + x1 + x2 + 2*x0 >= 2 + x1 + x2 + 2*x0 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(cons(_x0, _x1))]] = 3 + 3*x1 + 6*x0 > 2 + 2*x0 + 3*x1 = [[app(reverse(_x1), cons(_x0, nil))]] 

We can thus remove the following rules:

  reverse(cons(X, Y)) => app(reverse(Y), cons(X, 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]):

  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 

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

The following interpretation satisfies the requirements:

  app = Lam[y0;y1].2*y1 + 3*y0 
  cons = Lam[y0;y1].2 + y1 + 2*y0 

Using this interpretation, the requirements translate to:

  [[app(cons(_x0, _x1), _x2)]] = 6 + 2*x2 + 3*x1 + 6*x0 > 2 + 2*x0 + 2*x2 + 3*x1 = [[cons(_x0, app(_x1, _x2))]] 

We can thus remove the following rules:

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

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.