We consider the system 428.

  Alphabet:

    0 : [] --> nat 
    a : [nat * nat] --> nat 
    s : [nat] --> nat 
    sum : [nat * nat -> nat] --> nat 

  Rules:

    sum(0, /\x.X(x)) => X(0) 
    sum(s(X), /\x.Y(x)) => a(sum(X, /\y.Y(y)), Y(s(X))) 

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

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

  Dependency Pairs P_0:

    0] sum#(0, /\x.X(x)) =#> X(0)   
    1] sum#(s(X), /\x.Y(x)) =#> sum#(X, /\y.Y(y))   
    2] sum#(s(X), /\x.Y(x)) =#> Y(y)   
    3] sum#(s(X), /\x.Y(x)) =#> Y(s(X))   

  Rules R_0:

    sum(0, /\x.X(x)) => X(0) 
    sum(s(X), /\x.Y(x)) => a(sum(X, /\y.Y(y)), Y(s(X))) 

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).

The formative rules of (P_0, R_0) are R_1 ::=

  sum(0, /\x.X(x)) => X(0) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_1, 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:

  sum#(0, /\x.X(x)) >? X(0) 
  sum#(s(X), /\x.Y(x)) >? sum#(X, /\y.Y(y)) 
  sum#(s(X), /\x.Y(x)) >? Y(~c0) 
  sum#(s(X), /\x.Y(x)) >? Y(s(X)) 
  sum(0, /\x.X(x)) >= X(0) 

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

The following interpretation satisfies the requirements:

  0 = 2 
  s = \y0.1 + 2y0 
  sum = \y0G1.y0G1(y0) 
  sum# = \y0G1.3 + y0 + y0G1(y0) 
  ~c0 = 0 

Using this interpretation, the requirements translate to:

  [[sum#(0, /\x._x0(x))]] = 5 + 2F0(2) > F0(2) = [[_x0(0)]] 
  [[sum#(s(_x0), /\x._x1(x))]] = 4 + 2x0 + F1(1 + 2x0) + 2x0F1(1 + 2x0) > 3 + x0 + x0F1(x0) = [[sum#(_x0, /\x._x1(x))]] 
  [[sum#(s(_x0), /\x._x1(x))]] = 4 + 2x0 + F1(1 + 2x0) + 2x0F1(1 + 2x0) > F1(0) = [[_x1(~c0)]] 
  [[sum#(s(_x0), /\x._x1(x))]] = 4 + 2x0 + F1(1 + 2x0) + 2x0F1(1 + 2x0) > F1(1 + 2x0) = [[_x1(s(_x0))]] 
  [[sum(0, /\x._x0(x))]] = 2F0(2) >= F0(2) = [[_x0(0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_1) by ({}, R_1).  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.