0.91/0.96	YES
0.98/1.02	We consider the system theBenchmark.
0.98/1.02	
0.98/1.02	  Alphabet:
0.98/1.02	
0.98/1.02	    !faccolon : [] --> b -> b -> b 
0.98/1.02	    !facplus : [] --> b -> b -> b 
0.98/1.02	    a : [] --> b 
0.98/1.02	    cons : [] --> d -> e -> e 
0.98/1.02	    false : [] --> c 
0.98/1.02	    filter : [] --> (d -> c) -> e -> e 
0.98/1.02	    filter2 : [] --> c -> (d -> c) -> d -> e -> e 
0.98/1.02	    g : [] --> b -> a -> b 
0.98/1.02	    map : [] --> (d -> d) -> e -> e 
0.98/1.02	    nil : [] --> e 
0.98/1.02	    true : [] --> c 
0.98/1.02	
0.98/1.02	  Rules:
0.98/1.02	
0.98/1.02	    !faccolon (!faccolon x y) z => !faccolon x (!faccolon y z) 
0.98/1.02	    !faccolon (!facplus x y) z => !facplus (!faccolon x z) (!faccolon y z) 
0.98/1.02	    !faccolon x (!facplus y (f z)) => !faccolon (g x z) (!facplus y a) 
0.98/1.02	    map f nil => nil 
0.98/1.02	    map f (cons x y) => cons (f x) (map f y) 
0.98/1.02	    filter f nil => nil 
0.98/1.02	    filter f (cons x y) => filter2 (f x) f x y 
0.98/1.02	    filter2 true f x y => cons x (filter f y) 
0.98/1.02	    filter2 false f x y => filter f y 
0.98/1.02	
0.98/1.02	Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term.
0.98/1.02	
0.98/1.02	We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0).  This leads to the following AFSM:
0.98/1.02	
0.98/1.02	  Alphabet:
0.98/1.02	
0.98/1.02	    !faccolon : [b * b] --> b 
0.98/1.02	    !facplus : [b * b] --> b 
0.98/1.02	    a : [] --> b 
0.98/1.02	    cons : [d * e] --> e 
0.98/1.02	    false : [] --> c 
0.98/1.02	    filter : [d -> c * e] --> e 
0.98/1.02	    filter2 : [c * d -> c * d * e] --> e 
0.98/1.02	    g : [b] --> a -> b 
0.98/1.02	    map : [d -> d * e] --> e 
0.98/1.02	    nil : [] --> e 
0.98/1.02	    true : [] --> c 
0.98/1.02	    ~AP1 : [a -> b * a] --> b 
0.98/1.02	
0.98/1.02	  Rules:
0.98/1.02	
0.98/1.02	    !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) 
0.98/1.02	    !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
0.98/1.02	    !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	    map(F, nil) => nil 
0.98/1.02	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.98/1.02	    filter(F, nil) => nil 
0.98/1.02	    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
0.98/1.02	    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
0.98/1.02	    filter2(false, F, X, Y) => filter(F, Y) 
0.98/1.02	    ~AP1(F, X) => F X 
0.98/1.02	
0.98/1.02	We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.
0.98/1.02	
0.98/1.02	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):
0.98/1.02	
0.98/1.02	  Dependency Pairs P_0:
0.98/1.02	
0.98/1.02	    0] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
0.98/1.02	    1] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	    2] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
0.98/1.02	    3] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	    4] !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(~AP1(g(X), Z), !facplus(Y, a))   
0.98/1.02	    5] !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> ~AP1#(g(X), Z)   
0.98/1.02	    6] map#(F, cons(X, Y)) =#> F(X)   
0.98/1.02	    7] map#(F, cons(X, Y)) =#> map#(F, Y)   
0.98/1.02	    8] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
0.98/1.02	    9] filter#(F, cons(X, Y)) =#> F(X)   
0.98/1.02	    10] filter2#(true, F, X, Y) =#> filter#(F, Y)   
0.98/1.02	    11] filter2#(false, F, X, Y) =#> filter#(F, Y)   
0.98/1.02	    12] ~AP1#(F, X) =#> F(X)   
0.98/1.02	
0.98/1.02	  Rules R_0:
0.98/1.02	
0.98/1.02	    !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) 
0.98/1.02	    !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
0.98/1.02	    !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	    map(F, nil) => nil 
0.98/1.02	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.98/1.02	    filter(F, nil) => nil 
0.98/1.02	    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
0.98/1.02	    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
0.98/1.02	    filter2(false, F, X, Y) => filter(F, Y) 
0.98/1.02	    ~AP1(F, X) => F X 
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_0, R_0, minimal, formative).
0.98/1.02	
0.98/1.02	The formative rules of (P_0, R_0) are R_1 ::=
0.98/1.02	
0.98/1.02	  !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.98/1.02	  filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
0.98/1.02	  filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
0.98/1.02	  filter2(false, F, X, Y) => filter(F, Y) 
0.98/1.02	  ~AP1(F, X) => F X 
0.98/1.02	
0.98/1.02	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).
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_0, R_1, minimal, formative).
0.98/1.02	
0.98/1.02	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:
0.98/1.02	
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
0.98/1.02	  !faccolon#(X, !facplus(Y, ~AP1(F, Z))) >? !faccolon#(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	  !faccolon#(X, !facplus(Y, ~AP1(F, Z))) >? ~AP1#(g(X), Z) 
0.98/1.02	  map#(F, cons(X, Y)) >? F(X) 
0.98/1.02	  map#(F, cons(X, Y)) >? map#(F, Y) 
0.98/1.02	  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
0.98/1.02	  filter#(F, cons(X, Y)) >? F(X) 
0.98/1.02	  filter2#(true, F, X, Y) >? filter#(F, Y) 
0.98/1.02	  filter2#(false, F, X, Y) >? filter#(F, Y) 
0.98/1.02	  ~AP1#(F, X) >? F(X) 
0.98/1.02	  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
0.98/1.02	  filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) 
0.98/1.02	  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
0.98/1.02	  filter2(false, F, X, Y) >= filter(F, Y) 
0.98/1.02	  ~AP1(F, X) >= F X 
0.98/1.02	
0.98/1.02	We apply [Kop12, Thm. 6.75] and use the following argument functions:
0.98/1.02	
0.98/1.02	  pi( ~AP1#(F, X) ) = #argfun-~AP1##(F X) 
0.98/1.02	
0.98/1.02	We orient these requirements with a polynomial interpretation in the natural numbers.
0.98/1.02	
0.98/1.02	The following interpretation satisfies the requirements:
0.98/1.02	
0.98/1.02	  !faccolon = \y0y1.0 
0.98/1.02	  !faccolon# = \y0y1.3 + y1 
0.98/1.02	  !facplus = \y0y1.2y1 
0.98/1.02	  #argfun-~AP1## = \y0.y0 
0.98/1.02	  a = 0 
0.98/1.02	  cons = \y0y1.2 + y0 + y1 
0.98/1.02	  false = 3 
0.98/1.02	  filter = \G0y1.y1 + 2G0(y1) + 3y1G0(y1) 
0.98/1.02	  filter2 = \y0G1y2y3.2 + y2 + y3 + 2G1(y3) + 3y3G1(y3) 
0.98/1.02	  filter2# = \y0G1y2y3.1 + 2y3 + G1(0) + y3G1(y3) 
0.98/1.02	  filter# = \G0y1.2y1 + G0(0) + y1G0(y1) 
0.98/1.02	  g = \y0y1.0 
0.98/1.02	  map = \G0y1.y1 + G0(y1) + 2y1G0(y1) 
0.98/1.02	  map# = \G0y1.3 + G0(y1) 
0.98/1.02	  true = 3 
0.98/1.02	  ~AP1 = \G0y1.2y1 + G0(y1) 
0.98/1.02	  ~AP1# = \G0y1.0 
0.98/1.02	
0.98/1.02	Using this interpretation, the requirements translate to:
0.98/1.02	
0.98/1.02	  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 3 + x2 >= 3 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
0.98/1.02	  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 3 + x2 >= 3 + x2 = [[!faccolon#(_x1, _x2)]] 
0.98/1.02	  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 3 + x2 >= 3 + x2 = [[!faccolon#(_x0, _x2)]] 
0.98/1.02	  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 3 + x2 >= 3 + x2 = [[!faccolon#(_x1, _x2)]] 
0.98/1.02	  [[!faccolon#(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 3 + 4x3 + 2F2(x3) >= 3 = [[!faccolon#(~AP1(g(_x0), _x3), !facplus(_x1, a))]] 
0.98/1.02	  [[!faccolon#(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 3 + 4x3 + 2F2(x3) > x3 = [[#argfun-~AP1##(g(_x0) _x3)]] 
0.98/1.02	  [[map#(_F0, cons(_x1, _x2))]] = 3 + F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
0.98/1.02	  [[map#(_F0, cons(_x1, _x2))]] = 3 + F0(2 + x1 + x2) >= 3 + F0(x2) = [[map#(_F0, _x2)]] 
0.98/1.02	  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > 1 + 2x2 + F0(0) + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
0.98/1.02	  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
0.98/1.02	  [[filter2#(true, _F0, _x1, _x2)]] = 1 + 2x2 + F0(0) + x2F0(x2) > 2x2 + F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] 
0.98/1.02	  [[filter2#(false, _F0, _x1, _x2)]] = 1 + 2x2 + F0(0) + x2F0(x2) > 2x2 + F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] 
0.98/1.02	  [[#argfun-~AP1##(_F0 _x1)]] = max(x1, F0(x1)) >= F0(x1) = [[_F0(_x1)]] 
0.98/1.02	  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
0.98/1.02	  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
0.98/1.02	  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(~AP1(g(_x0), _x3), !facplus(_x1, a))]] 
0.98/1.02	  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 5F0(2 + x1 + x2) >= 2 + x2 + F0(x2) + 2x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
0.98/1.02	  [[filter(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 3x1F0(2 + x1 + x2) + 3x2F0(2 + x1 + x2) + 8F0(2 + x1 + x2) >= 2 + x1 + x2 + 2F0(x2) + 3x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
0.98/1.02	  [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + x2 + 2F0(x2) + 3x2F0(x2) >= 2 + x1 + x2 + 2F0(x2) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
0.98/1.02	  [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + x2 + 2F0(x2) + 3x2F0(x2) >= x2 + 2F0(x2) + 3x2F0(x2) = [[filter(_F0, _x2)]] 
0.98/1.02	  [[~AP1(_F0, _x1)]] = 2x1 + F0(x1) >= max(x1, F0(x1)) = [[_F0 _x1]] 
0.98/1.02	
0.98/1.02	By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of:
0.98/1.02	
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	  !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(~AP1(g(X), Z), !facplus(Y, a))   
0.98/1.02	  map#(F, cons(X, Y)) =#> map#(F, Y)   
0.98/1.02	  ~AP1#(F, X) =#> F(X)   
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_1, R_1, minimal, formative).
0.98/1.02	
0.98/1.02	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.98/1.02	
0.98/1.02	    * 0 : 0, 1, 2, 3, 4 
0.98/1.02	    * 1 : 0, 1, 2, 3, 4 
0.98/1.02	    * 2 : 0, 1, 2, 3, 4 
0.98/1.02	    * 3 : 0, 1, 2, 3, 4 
0.98/1.02	    * 4 : 0, 1, 2, 3 
0.98/1.02	    * 5 : 5 
0.98/1.02	    * 6 : 0, 1, 2, 3, 4, 5, 6 
0.98/1.02	
0.98/1.02	This graph has the following strongly connected components:
0.98/1.02	
0.98/1.02	  P_2:
0.98/1.02	
0.98/1.02	    !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
0.98/1.02	    !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	    !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
0.98/1.02	    !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	    !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(~AP1(g(X), Z), !facplus(Y, a))   
0.98/1.02	
0.98/1.02	  P_3:
0.98/1.02	
0.98/1.02	    map#(F, cons(X, Y)) =#> map#(F, Y)   
0.98/1.02	
0.98/1.02	  P_4:
0.98/1.02	
0.98/1.02	    ~AP1#(F, X) =#> F(X)   
0.98/1.02	
0.98/1.02	By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_1, m, f) by (P_2, R_1, m, f), (P_3, R_1, m, f) and (P_4, R_1, m, f).
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if each of (P_2, R_1, minimal, formative), (P_3, R_1, minimal, formative) and (P_4, R_1, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_4, R_1, minimal, formative).
0.98/1.02	
0.98/1.02	This combination (P_4, R_1) has no formative rules!  We will name the empty set of rules:R_2.
0.98/1.02	
0.98/1.02	By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_4, R_1, minimal, formative) by (P_4, R_2, minimal, formative).
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if each of (P_2, R_1, minimal, formative), (P_3, R_1, minimal, formative) and (P_4, R_2, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_4, R_2, minimal, formative).
0.98/1.02	
0.98/1.02	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:
0.98/1.02	
0.98/1.02	  ~AP1#(F, X) >? F(X) 
0.98/1.02	
0.98/1.02	We apply [Kop12, Thm. 6.75] and use the following argument functions:
0.98/1.02	
0.98/1.02	  pi( ~AP1#(F, X) ) = #argfun-~AP1##(F X) 
0.98/1.02	
0.98/1.02	We orient these requirements with a polynomial interpretation in the natural numbers.
0.98/1.02	
0.98/1.02	The following interpretation satisfies the requirements:
0.98/1.02	
0.98/1.02	  #argfun-~AP1## = \y0.1 + y0 
0.98/1.02	  ~AP1# = \G0y1.0 
0.98/1.02	
0.98/1.02	Using this interpretation, the requirements translate to:
0.98/1.02	
0.98/1.02	  [[#argfun-~AP1##(_F0 _x1)]] = 1 + max(x1, F0(x1)) > F0(x1) = [[_F0(_x1)]] 
0.98/1.02	
0.98/1.02	By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_2) by ({}, R_2).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if each of (P_2, R_1, minimal, formative) and (P_3, R_1, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_3, R_1, minimal, formative).
0.98/1.02	
0.98/1.02	We apply the subterm criterion with the following projection function:
0.98/1.02	
0.98/1.02	  nu(map#) = 2 
0.98/1.02	
0.98/1.02	Thus, we can orient the dependency pairs as follows:
0.98/1.02	
0.98/1.02	  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 
0.98/1.02	
0.98/1.02	By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_1, minimal, f) by ({}, R_1, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_2, R_1, minimal, formative).
0.98/1.02	
0.98/1.02	The formative rules of (P_2, R_1) are R_3 ::=
0.98/1.02	
0.98/1.02	  !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	  ~AP1(F, X) => F X 
0.98/1.02	
0.98/1.02	By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_1, minimal, formative) by (P_2, R_3, minimal, formative).
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if (P_2, R_3, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_2, R_3, minimal, formative).
0.98/1.02	
0.98/1.02	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:
0.98/1.02	
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) 
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) 
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) 
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) 
0.98/1.02	  !faccolon#(X, !facplus(Y, ~AP1(F, Z))) >? !faccolon#(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	  !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) 
0.98/1.02	  !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(~AP1(g(X), Z), !facplus(Y, a)) 
0.98/1.02	  ~AP1(F, X) >= F X 
0.98/1.02	
0.98/1.02	We orient these requirements with a polynomial interpretation in the natural numbers.
0.98/1.02	
0.98/1.02	The following interpretation satisfies the requirements:
0.98/1.02	
0.98/1.02	  !faccolon = \y0y1.0 
0.98/1.02	  !faccolon# = \y0y1.y1 
0.98/1.02	  !facplus = \y0y1.y1 
0.98/1.02	  a = 0 
0.98/1.02	  g = \y0y1.0 
0.98/1.02	  ~AP1 = \G0y1.2 + G0(y1) 
0.98/1.02	
0.98/1.02	Using this interpretation, the requirements translate to:
0.98/1.02	
0.98/1.02	  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= 0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] 
0.98/1.02	  [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] 
0.98/1.02	  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x0, _x2)]] 
0.98/1.02	  [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] 
0.98/1.02	  [[!faccolon#(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 2 + F2(x3) > 0 = [[!faccolon#(~AP1(g(_x0), _x3), !facplus(_x1, a))]] 
0.98/1.02	  [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] 
0.98/1.02	  [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] 
0.98/1.02	  [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(~AP1(g(_x0), _x3), !facplus(_x1, a))]] 
0.98/1.02	  [[~AP1(_F0, _x1)]] = 2 + F0(x1) >= F0(x1) = [[_F0 _x1]] 
0.98/1.02	
0.98/1.02	By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_3, minimal, formative) by (P_5, R_3, minimal, formative), where P_5 consists of:
0.98/1.02	
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z))   
0.98/1.02	  !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z)   
0.98/1.02	  !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z)   
0.98/1.02	
0.98/1.02	Thus, the original system is terminating if (P_5, R_3, minimal, formative) is finite.
0.98/1.02	
0.98/1.02	We consider the dependency pair problem (P_5, R_3, minimal, formative).
0.98/1.02	
0.98/1.02	We apply the subterm criterion with the following projection function:
0.98/1.02	
0.98/1.02	  nu(!faccolon#) = 1 
0.98/1.02	
0.98/1.02	Thus, we can orient the dependency pairs as follows:
0.98/1.02	
0.98/1.02	  nu(!faccolon#(!faccolon(X, Y), Z)) = !faccolon(X, Y) |> X = nu(!faccolon#(X, !faccolon(Y, Z))) 
0.98/1.02	  nu(!faccolon#(!faccolon(X, Y), Z)) = !faccolon(X, Y) |> Y = nu(!faccolon#(Y, Z)) 
0.98/1.02	  nu(!faccolon#(!facplus(X, Y), Z)) = !facplus(X, Y) |> X = nu(!faccolon#(X, Z)) 
0.98/1.02	  nu(!faccolon#(!facplus(X, Y), Z)) = !facplus(X, Y) |> Y = nu(!faccolon#(Y, Z)) 
0.98/1.02	
0.98/1.02	By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_3, minimal, f) by ({}, R_3, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
0.98/1.02	
0.98/1.02	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.98/1.02	
0.98/1.02	
0.98/1.02	+++ Citations +++
0.98/1.02	
0.98/1.02	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
0.98/1.02	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.98/1.02	EOF