We consider the system 06plusmult.

  Alphabet:

    mult : [N * N] --> N 
    plus : [N * N] --> N 
    s : [N] --> N 
    z : [] --> N 

  Rules:

    plus(z, x) => x 
    plus(s(x), y) => plus(x, s(y)) 
    plus(plus(x, y), u) => plus(x, plus(y, u)) 
    mult(z, x) => z 
    mult(s(x), y) => plus(mult(x, y), y) 
    mult(plus(x, y), u) => plus(mult(x, u), mult(y, u)) 

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] plus#(s(X), Y) =#> plus#(X, s(Y))   
    1] plus#(plus(X, Y), Z) =#> plus#(X, plus(Y, Z))   
    2] plus#(plus(X, Y), Z) =#> plus#(Y, Z)   
    3] mult#(s(X), Y) =#> plus#(mult(X, Y), Y)   
    4] mult#(s(X), Y) =#> mult#(X, Y)   
    5] mult#(plus(X, Y), Z) =#> plus#(mult(X, Z), mult(Y, Z))   
    6] mult#(plus(X, Y), Z) =#> mult#(X, Z)   
    7] mult#(plus(X, Y), Z) =#> mult#(Y, Z)   

  Rules R_0:

    plus(z, X) => X 
    plus(s(X), Y) => plus(X, s(Y)) 
    plus(plus(X, Y), Z) => plus(X, plus(Y, Z)) 
    mult(z, X) => z 
    mult(s(X), Y) => plus(mult(X, Y), Y) 
    mult(plus(X, Y), Z) => plus(mult(X, Z), mult(Y, Z)) 

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 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, 2 
    * 1 : 0, 1, 2 
    * 2 : 0, 1, 2 
    * 3 : 0, 1, 2 
    * 4 : 3, 4, 5, 6, 7 
    * 5 : 0, 1, 2 
    * 6 : 3, 4, 5, 6, 7 
    * 7 : 3, 4, 5, 6, 7 

This graph has the following strongly connected components:

  P_1:

    plus#(s(X), Y) =#> plus#(X, s(Y))   
    plus#(plus(X, Y), Z) =#> plus#(X, plus(Y, Z))   
    plus#(plus(X, Y), Z) =#> plus#(Y, Z)   

  P_2:

    mult#(s(X), Y) =#> mult#(X, Y)   
    mult#(plus(X, Y), Z) =#> mult#(X, Z)   
    mult#(plus(X, Y), Z) =#> mult#(Y, Z)   

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) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (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 with usable rules [Kop12, Thm. 7.44].  (P_2, R_0) has no usable rules.

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

  mult#(s(X), Y) >? mult#(X, Y) 
  mult#(plus(X, Y), Z) >? mult#(X, Z) 
  mult#(plus(X, Y), Z) >? mult#(Y, Z) 

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

The following interpretation satisfies the requirements:

  mult# = \y0y1.y0 
  plus = \y0y1.3 + y0 + y1 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[mult#(s(_x0), _x1)]] = 3 + x0 > x0 = [[mult#(_x0, _x1)]] 
  [[mult#(plus(_x0, _x1), _x2)]] = 3 + x0 + x1 > x0 = [[mult#(_x0, _x2)]] 
  [[mult#(plus(_x0, _x1), _x2)]] = 3 + x0 + x1 > x1 = [[mult#(_x1, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, 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 (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 with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_0) are:

  plus(z, X) => X 
  plus(s(X), Y) => plus(X, s(Y)) 
  plus(plus(X, Y), Z) => plus(X, plus(Y, Z)) 

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

  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(plus(X, Y), Z) >? plus#(X, plus(Y, Z)) 
  plus#(plus(X, Y), Z) >? plus#(Y, Z) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 

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

The following interpretation satisfies the requirements:

  plus = \y0y1.1 + y0 + y1 
  plus# = \y0y1.2y0 
  s = \y0.y0 
  z = 0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 2x0 >= 2x0 = [[plus#(_x0, s(_x1))]] 
  [[plus#(plus(_x0, _x1), _x2)]] = 2 + 2x0 + 2x1 > 2x0 = [[plus#(_x0, plus(_x1, _x2))]] 
  [[plus#(plus(_x0, _x1), _x2)]] = 2 + 2x0 + 2x1 > 2x1 = [[plus#(_x1, _x2)]] 
  [[plus(z, _x0)]] = 1 + x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[plus(_x0, s(_x1))]] 
  [[plus(plus(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[plus(_x0, plus(_x1, _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_3, R_0, static, formative), where P_3 consists of:

  plus#(s(X), Y) =#> plus#(X, s(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 with usable rules [Kop12, Thm. 7.44].  (P_3, R_0) has no usable rules.

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

  plus#(s(X), Y) >? plus#(X, s(Y)) 

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

The following interpretation satisfies the requirements:

  plus# = \y0y1.3y0 
  s = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 9 + 6x0 > 3x0 = [[plus#(_x0, s(_x1))]] 

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.

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.