We consider the system loopy.

  Alphabet:

    f : [N -> N * N] --> N 
    g : [] --> N -> N 
    h : [N] --> N 

  Rules:

    f(g, x) => h(x) 
    f(i, x) => i x 
    h(x) => f(/\y.y, x) 

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 dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] f#(g, X) =#> h#(X)   
    1] f#(F, X) =#> F(X)   
    2] h#(X) =#> f#(/\x.x, X)   

  Rules R_0:

    f(g, X) => h(X) 
    f(F, X) => F X 
    h(X) => f(/\x.x, 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).

This combination (P_0, R_0) has no formative rules!  We will name the empty set of rules:R_1.

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:

  f#(g, X) >? h#(X) 
  f#(F, X) >? F(X) 
  h#(X) >? f#(/\x.x, X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( h#(X) ) = #argfun-h##(f#(/\x.x, X)) 

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

The following interpretation satisfies the requirements:

  #argfun-h## = \y0.1 + y0 
  f# = \G0y1.3 + G0(y1) 
  g = \y0.3 + 3y0 
  h# = \y0.0 

Using this interpretation, the requirements translate to:

  [[f#(g, _x0)]] = 6 + 3x0 > 4 + x0 = [[#argfun-h##(f#(/\x.x, _x0))]] 
  [[f#(_F0, _x1)]] = 3 + F0(x1) > F0(x1) = [[_F0(_x1)]] 
  [[#argfun-h##(f#(/\x.x, _x0))]] = 4 + x0 > 3 + x0 = [[f#(/\x.x, _x0)]] 

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.