We consider the system AotoYamada_05__009.

  Alphabet:

    and : [c * c] --> c 
    cons : [a * b] --> b 
    false : [] --> c 
    forall : [a -> c * b] --> c 
    forsome : [a -> c * b] --> c 
    nil : [] --> b 
    or : [c * c] --> c 
    true : [] --> c 

  Rules:

    and(true, true) => true 
    and(true, false) => false 
    and(false, true) => false 
    and(false, false) => false 
    or(true, true) => true 
    or(true, false) => true 
    or(false, true) => true 
    or(false, false) => false 
    forall(f, nil) => true 
    forall(f, cons(x, y)) => and(f x, forall(f, y)) 
    forsome(f, nil) => false 
    forsome(f, cons(x, y)) => or(f x, forsome(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] forall#(F, cons(X, Y)) =#> and#(F X, forall(F, Y))   
    1] forall#(F, cons(X, Y)) =#> forall#(F, Y)   
    2] forsome#(F, cons(X, Y)) =#> or#(F X, forsome(F, Y))   
    3] forsome#(F, cons(X, Y)) =#> forsome#(F, Y)   

  Rules R_0:

    and(true, true) => true 
    and(true, false) => false 
    and(false, true) => false 
    and(false, false) => false 
    or(true, true) => true 
    or(true, false) => true 
    or(false, true) => true 
    or(false, false) => false 
    forall(F, nil) => true 
    forall(F, cons(X, Y)) => and(F X, forall(F, Y)) 
    forsome(F, nil) => false 
    forsome(F, cons(X, Y)) => or(F X, forsome(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:

  forall#(F, cons(X, Y)) >? and#(F X, forall(F, Y)) 
  forall#(F, cons(X, Y)) >? forall#(F, Y) 
  forsome#(F, cons(X, Y)) >? or#(F X, forsome(F, Y)) 
  forsome#(F, cons(X, Y)) >? forsome#(F, Y) 
  and(true, true) >= true 
  and(true, false) >= false 
  and(false, true) >= false 
  and(false, false) >= false 
  or(true, true) >= true 
  or(true, false) >= true 
  or(false, true) >= true 
  or(false, false) >= false 
  forall(F, nil) >= true 
  forall(F, cons(X, Y)) >= and(F X, forall(F, Y)) 
  forsome(F, nil) >= false 
  forsome(F, cons(X, Y)) >= or(F X, forsome(F, Y)) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.0 
  and# = \y0y1.0 
  cons = \y0y1.3 + 2y1 
  false = 0 
  forall = \G0y1.0 
  forall# = \G0y1.3 + y1 + y1G0(y1) 
  forsome = \G0y1.2G0(y1) + 3y1G0(y1) 
  forsome# = \G0y1.3 + y1 + y1G0(y1) 
  nil = 3 
  or = \y0y1.0 
  or# = \y0y1.0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[forall#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 0 = [[and#(_F0 _x1, forall(_F0, _x2))]] 
  [[forall#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 3 + x2 + x2F0(x2) = [[forall#(_F0, _x2)]] 
  [[forsome#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 0 = [[or#(_F0 _x1, forsome(_F0, _x2))]] 
  [[forsome#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 3 + x2 + x2F0(x2) = [[forsome#(_F0, _x2)]] 
  [[and(true, true)]] = 0 >= 0 = [[true]] 
  [[and(true, false)]] = 0 >= 0 = [[false]] 
  [[and(false, true)]] = 0 >= 0 = [[false]] 
  [[and(false, false)]] = 0 >= 0 = [[false]] 
  [[or(true, true)]] = 0 >= 0 = [[true]] 
  [[or(true, false)]] = 0 >= 0 = [[true]] 
  [[or(false, true)]] = 0 >= 0 = [[true]] 
  [[or(false, false)]] = 0 >= 0 = [[false]] 
  [[forall(_F0, nil)]] = 0 >= 0 = [[true]] 
  [[forall(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[and(_F0 _x1, forall(_F0, _x2))]] 
  [[forsome(_F0, nil)]] = 11F0(3) >= 0 = [[false]] 
  [[forsome(_F0, cons(_x1, _x2))]] = 6x2F0(3 + 2x2) + 11F0(3 + 2x2) >= 0 = [[or(_F0 _x1, forsome(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, 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.