0.00/0.19	YES
0.00/0.20	We consider the system theBenchmark.
0.00/0.20	
0.00/0.20	  Alphabet:
0.00/0.20	
0.00/0.20	    !plus : [nat * nat] --> nat 
0.00/0.20	    cons : [nat * list] --> list 
0.00/0.20	    map : [nat -> nat * list] --> list 
0.00/0.20	    nil : [] --> list 
0.00/0.20	    ps : [list] --> list 
0.00/0.20	
0.00/0.20	  Rules:
0.00/0.20	
0.00/0.20	    map(f, nil) => nil 
0.00/0.20	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
0.00/0.20	    ps(nil) => nil 
0.00/0.20	    ps(cons(x, y)) => cons(x, ps(map(/\z.!plus(x, z), y))) 
0.00/0.20	
0.00/0.20	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.20	
0.00/0.20	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]).
0.00/0.20	
0.00/0.20	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
0.00/0.20	
0.00/0.20	  Dependency Pairs P_0:
0.00/0.20	
0.00/0.20	    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
0.00/0.20	    1] ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y))   
0.00/0.20	    2] ps#(cons(X, Y)) =#> map#(/\x.!plus(X, x), Y)   
0.00/0.20	
0.00/0.20	  Rules R_0:
0.00/0.20	
0.00/0.20	    map(F, nil) => nil 
0.00/0.20	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.00/0.20	    ps(nil) => nil 
0.00/0.20	    ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) 
0.00/0.20	
0.00/0.20	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
0.00/0.20	
0.00/0.20	We consider the dependency pair problem (P_0, R_0, static, formative).
0.00/0.20	
0.00/0.20	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.20	
0.00/0.20	    * 0 : 0 
0.00/0.20	    * 1 : 1, 2 
0.00/0.20	    * 2 : 0 
0.00/0.20	
0.00/0.20	This graph has the following strongly connected components:
0.00/0.20	
0.00/0.20	  P_1:
0.00/0.20	
0.00/0.20	    map#(F, cons(X, Y)) =#> map#(F, Y)   
0.00/0.20	
0.00/0.20	  P_2:
0.00/0.20	
0.00/0.20	    ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y))   
0.00/0.20	
0.00/0.20	By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f).
0.00/0.20	
0.00/0.20	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.
0.00/0.20	
0.00/0.20	We consider the dependency pair problem (P_2, R_0, static, formative).
0.00/0.20	
0.00/0.20	The formative rules of (P_2, R_0) are R_1 ::=
0.00/0.20	
0.00/0.20	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.00/0.20	  ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) 
0.00/0.20	
0.00/0.20	By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, static, formative) by (P_2, R_1, static, formative).
0.00/0.20	
0.00/0.20	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_1, static, formative) is finite.
0.00/0.20	
0.00/0.20	We consider the dependency pair problem (P_2, R_1, static, formative).
0.00/0.20	
0.00/0.20	We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:
0.00/0.20	
0.00/0.20	  ps#(cons(X, Y)) >? ps#(map(/\x.!plus(X, x), Y)) 
0.00/0.20	  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
0.00/0.20	  ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) 
0.00/0.20	
0.00/0.20	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.20	
0.00/0.20	The following interpretation satisfies the requirements:
0.00/0.20	
0.00/0.20	  !plus = \y0y1.0 
0.00/0.20	  cons = \y0y1.1 + y1 
0.00/0.20	  map = \G0y1.y1 
0.00/0.20	  ps = \y0.y0 
0.00/0.20	  ps# = \y0.2y0 
0.00/0.20	
0.00/0.20	Using this interpretation, the requirements translate to:
0.00/0.20	
0.00/0.20	  [[ps#(cons(_x0, _x1))]] = 2 + 2x1 > 2x1 = [[ps#(map(/\x.!plus(_x0, x), _x1))]] 
0.00/0.20	  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
0.00/0.20	  [[ps(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[cons(_x0, ps(map(/\x.!plus(_x0, x), _x1)))]] 
0.00/0.20	
0.00/0.20	By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_1) by ({}, R_1).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
0.00/0.20	
0.00/0.20	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
0.00/0.20	
0.00/0.20	We consider the dependency pair problem (P_1, R_0, static, formative).
0.00/0.20	
0.00/0.20	We apply the subterm criterion with the following projection function:
0.00/0.20	
0.00/0.20	  nu(map#) = 2 
0.00/0.20	
0.00/0.20	Thus, we can orient the dependency pairs as follows:
0.00/0.20	
0.00/0.20	  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 
0.00/0.20	
0.00/0.20	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
0.00/0.20	
0.00/0.20	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.00/0.20	
0.00/0.20	
0.00/0.20	+++ Citations +++
0.00/0.20	
0.00/0.20	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.20	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
0.00/0.20	[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.
0.00/0.20	EOF