0.00/0.20	YES
0.00/0.21	We consider the system theBenchmark.
0.00/0.21	
0.00/0.21	  Alphabet:
0.00/0.21	
0.00/0.21	    cons : [b * c] --> c 
0.00/0.21	    false : [] --> a 
0.00/0.21	    filter : [b -> a * c] --> c 
0.00/0.21	    filtersub : [a * b -> a * c] --> c 
0.00/0.21	    nil : [] --> c 
0.00/0.21	    true : [] --> a 
0.00/0.21	
0.00/0.21	  Rules:
0.00/0.21	
0.00/0.21	    filter(f, nil) => nil 
0.00/0.21	    filter(f, cons(x, y)) => filtersub(f x, f, cons(x, y)) 
0.00/0.21	    filtersub(true, f, cons(x, y)) => cons(x, filter(f, y)) 
0.00/0.21	    filtersub(false, f, cons(x, y)) => filter(f, y) 
0.00/0.21	
0.00/0.21	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.21	
0.00/0.21	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]).
0.00/0.21	
0.00/0.21	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
0.00/0.21	
0.00/0.21	  Dependency Pairs P_0:
0.00/0.21	
0.00/0.21	    0] filter#(F, cons(X, Y)) =#> filtersub#(F X, F, cons(X, Y))   
0.00/0.21	    1] filtersub#(true, F, cons(X, Y)) =#> filter#(F, Y)   
0.00/0.21	    2] filtersub#(false, F, cons(X, Y)) =#> filter#(F, Y)   
0.00/0.21	
0.00/0.21	  Rules R_0:
0.00/0.21	
0.00/0.21	    filter(F, nil) => nil 
0.00/0.21	    filter(F, cons(X, Y)) => filtersub(F X, F, cons(X, Y)) 
0.00/0.21	    filtersub(true, F, cons(X, Y)) => cons(X, filter(F, Y)) 
0.00/0.21	    filtersub(false, F, cons(X, Y)) => filter(F, Y) 
0.00/0.21	
0.00/0.21	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
0.00/0.21	
0.00/0.21	We consider the dependency pair problem (P_0, R_0, static, formative).
0.00/0.21	
0.00/0.21	We apply the subterm criterion with the following projection function:
0.00/0.21	
0.00/0.21	  nu(filter#) = 2 
0.00/0.21	  nu(filtersub#) = 3 
0.00/0.21	
0.00/0.21	Thus, we can orient the dependency pairs as follows:
0.00/0.21	
0.00/0.21	  nu(filter#(F, cons(X, Y))) = cons(X, Y) = cons(X, Y) = nu(filtersub#(F X, F, cons(X, Y))) 
0.00/0.21	  nu(filtersub#(true, F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 
0.00/0.21	  nu(filtersub#(false, F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 
0.00/0.21	
0.00/0.21	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_0, R_0, static, f) by (P_1, R_0, static, f), where P_1 contains:
0.00/0.21	
0.00/0.21	  filter#(F, cons(X, Y)) =#> filtersub#(F X, F, cons(X, Y))   
0.00/0.21	
0.00/0.21	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
0.00/0.21	
0.00/0.21	We consider the dependency pair problem (P_1, R_0, static, formative).
0.00/0.21	
0.00/0.21	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.21	
0.00/0.21	    * 0 :  
0.00/0.21	
0.00/0.21	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
0.00/0.21	
0.00/0.21	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.00/0.21	
0.00/0.21	
0.00/0.21	+++ Citations +++
0.00/0.21	
0.00/0.21	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.21	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
0.00/0.21	[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.
0.00/0.21	EOF