We consider the system apply.

  Alphabet:

    cons : [a -> a * listf] --> listf 
    dapply : [a * a -> a * a -> a] --> a 
    lapply : [a * listf] --> a 
    nil : [] --> listf 

  Rules:

    dapply(x, f, g) => f (g x) 
    lapply(x, nil) => x 
    lapply(x, cons(f, y)) => f lapply(x, 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 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] dapply#(X, F, G) =#> F(G X)   
    1] dapply#(X, F, G) =#> G(X)   
    2] lapply#(X, cons(F, Y)) =#> F(lapply(X, Y))   
    3] lapply#(X, cons(F, Y)) =#> lapply#(X, Y)   

  Rules R_0:

    dapply(X, F, G) => F (G X) 
    lapply(X, nil) => X 
    lapply(X, cons(F, Y)) => F lapply(X, Y) 

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:

  dapply#(X, F, G) >? F(G X) 
  dapply#(X, F, G) >? G(X) 
  lapply#(X, cons(F, Y)) >? F(lapply(X, Y)) 
  lapply#(X, cons(F, Y)) >? lapply#(X, Y) 

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

  pi( dapply#(X, F, G) ) = #argfun-dapply##(F (G X), G X) 

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

The following interpretation satisfies the requirements:

  #argfun-dapply## = \y0y1.3 + max(y0, y1) 
  cons = \G0y1.3 + 2y1 + G0(0) 
  dapply# = \y0G1G2.0 
  lapply = \y0y1.0 
  lapply# = \y0y1.3 + y1 

Using this interpretation, the requirements translate to:

  [[#argfun-dapply##(_F0 (_F1 _x2), _F1 _x2)]] = 3 + max(x2, F0(max(x2, F1(x2))), F1(x2)) > F0(max(x2, F1(x2))) = [[_F0(_F1 _x2)]] 
  [[#argfun-dapply##(_F0 (_F1 _x2), _F1 _x2)]] = 3 + max(x2, F0(max(x2, F1(x2))), F1(x2)) > F1(x2) = [[_F1(_x2)]] 
  [[lapply#(_x0, cons(_F1, _x2))]] = 6 + 2x2 + F1(0) > F1(0) = [[_F1(lapply(_x0, _x2))]] 
  [[lapply#(_x0, cons(_F1, _x2))]] = 6 + 2x2 + F1(0) > 3 + x2 = [[lapply#(_x0, _x2)]] 

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.