0.00/0.03	YES
0.00/0.03	We consider the system theBenchmark.
0.00/0.03	
0.00/0.03	  Alphabet:
0.00/0.03	
0.00/0.03	    fapp : [o * o] --> o 
0.00/0.03	    h : [o] --> o 
0.00/0.03	    lam : [o -> o] --> o 
0.00/0.03	
0.00/0.03	  Rules:
0.00/0.03	
0.00/0.03	    fapp(lam(f), x) => f h(x) 
0.00/0.03	
0.00/0.03	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.03	
0.00/0.03	We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.
0.00/0.03	
0.00/0.03	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):
0.00/0.03	
0.00/0.03	  Dependency Pairs P_0:
0.00/0.03	
0.00/0.03	    0] fapp#(lam(F), X) =#> F(h(X))   
0.00/0.03	
0.00/0.03	  Rules R_0:
0.00/0.03	
0.00/0.03	    fapp(lam(F), X) => F h(X) 
0.00/0.03	
0.00/0.03	Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.
0.00/0.03	
0.00/0.03	We consider the dependency pair problem (P_0, R_0, minimal, formative).
0.00/0.03	
0.00/0.03	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:
0.00/0.03	
0.00/0.03	  fapp#(lam(F), X) >? F(h(X)) 
0.00/0.03	  fapp(lam(F), X) >= F h(X) 
0.00/0.03	
0.00/0.03	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.03	
0.00/0.03	The following interpretation satisfies the requirements:
0.00/0.03	
0.00/0.03	  fapp = \y0y1.3 + 3y0 
0.00/0.03	  fapp# = \y0y1.3 + y0 
0.00/0.03	  h = \y0.0 
0.00/0.03	  lam = \G0.3 + G0(0) 
0.00/0.03	
0.00/0.03	Using this interpretation, the requirements translate to:
0.00/0.03	
0.00/0.03	  [[fapp#(lam(_F0), _x1)]] = 6 + F0(0) > F0(0) = [[_F0(h(_x1))]] 
0.00/0.03	  [[fapp(lam(_F0), _x1)]] = 12 + 3F0(0) >= F0(0) = [[_F0 h(_x1)]] 
0.00/0.03	
0.00/0.03	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.
0.00/0.03	
0.00/0.03	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.00/0.03	
0.00/0.03	
0.00/0.03	+++ Citations +++
0.00/0.03	
0.00/0.03	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.03	EOF