0.00/0.17	YES
0.00/0.18	We consider the system theBenchmark.
0.00/0.18	
0.00/0.18	  Alphabet:
0.00/0.18	
0.00/0.18	    rec : [] --> N -> a -> (N -> a -> a) -> a 
0.00/0.18	    s : [] --> N -> N 
0.00/0.18	    z : [] --> N 
0.00/0.18	
0.00/0.18	  Rules:
0.00/0.18	
0.00/0.18	    rec z x (/\y.f y) => x 
0.00/0.18	    rec (s x) y (/\u.f u) => f x (rec x y (/\v.f v)) 
0.00/0.18	
0.00/0.18	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.
0.00/0.18	
0.00/0.18	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:
0.00/0.18	
0.00/0.18	  Alphabet:
0.00/0.18	
0.00/0.18	    rec : [N * a * N -> a -> a] --> a 
0.00/0.18	    s : [N] --> N 
0.00/0.18	    z : [] --> N 
0.00/0.18	    ~AP1 : [N -> a -> a * N] --> a -> a 
0.00/0.18	
0.00/0.18	  Rules:
0.00/0.18	
0.00/0.18	    rec(z, X, /\x.~AP1(F, x)) => X 
0.00/0.18	    rec(s(X), Y, /\x.~AP1(F, x)) => ~AP1(F, X) rec(X, Y, /\y.~AP1(F, y)) 
0.00/0.18	    ~AP1(F, X) => F X 
0.00/0.18	
0.00/0.18	Symbol ~AP1 is an encoding for application that is only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing it.  This gives:
0.00/0.18	
0.00/0.18	  Alphabet:
0.00/0.18	
0.00/0.18	    rec : [N * a * N -> a -> a] --> a 
0.00/0.18	    s : [N] --> N 
0.00/0.18	    z : [] --> N 
0.00/0.18	
0.00/0.18	  Rules:
0.00/0.18	
0.00/0.18	    rec(z, X, /\x.F(x)) => X 
0.00/0.18	    rec(s(X), Y, /\x.F(x)) => F(X) rec(X, Y, /\y.F(y)) 
0.00/0.18	
0.00/0.18	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.18	
0.00/0.18	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.18	
0.00/0.18	  rec(z, X, /\x.F(x)) >? X 
0.00/0.18	  rec(s(X), Y, /\x.F(x)) >? F(X) rec(X, Y, /\y.F(y)) 
0.00/0.18	
0.00/0.18	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.18	
0.00/0.18	We choose Lex = {} and Mul = {@_{o -> o}, rec, s, z}, and the following precedence: rec > @_{o -> o} > s > z
0.00/0.18	
0.00/0.18	With these choices, we have:
0.00/0.18	
0.00/0.18	  1] rec(z, X, /\x.F(x)) > X  because [2], by definition 
0.00/0.18	  2] rec*(z, X, /\x.F(x)) >= X  because [3], by (Select) 
0.00/0.18	  3] X >= X  by (Meta) 
0.00/0.18	
0.00/0.18	  4] rec(s(X), Y, /\x.F(x)) >= @_{o -> o}(F(X), rec(X, Y, /\x.F(x)))  because [5], by (Star) 
0.00/0.18	  5] rec*(s(X), Y, /\x.F(x)) >= @_{o -> o}(F(X), rec(X, Y, /\x.F(x)))  because rec > @_{o -> o}, [6] and [12], by (Copy) 
0.00/0.18	  6] rec*(s(X), Y, /\x.F(x)) >= F(X)  because [7], by (Select) 
0.00/0.18	  7] F(rec*(s(X), Y, /\x.F(x))) >= F(X)  because [8], by (Meta) 
0.00/0.18	  8] rec*(s(X), Y, /\x.F(x)) >= X  because [9], by (Select) 
0.00/0.18	  9] s(X) >= X  because [10], by (Star) 
0.00/0.18	  10] s*(X) >= X  because [11], by (Select) 
0.00/0.18	  11] X >= X  by (Meta) 
0.00/0.18	  12] rec*(s(X), Y, /\x.F(x)) >= rec(X, Y, /\x.F(x))  because rec in Mul, [13], [15] and [16], by (Stat) 
0.00/0.18	  13] s(X) > X  because [14], by definition 
0.00/0.18	  14] s*(X) >= X  because [11], by (Select) 
0.00/0.18	  15] Y >= Y  by (Meta) 
0.00/0.18	  16] /\y.F(y) >= /\y.F(y)  because [17], by (Abs) 
0.00/0.18	  17] F(x) >= F(x)  because [18], by (Meta) 
0.00/0.18	  18] x >= x  by (Var) 
0.00/0.18	
0.00/0.18	We can thus remove the following rules:
0.00/0.18	
0.00/0.18	  rec(z, X, /\x.F(x)) => X 
0.00/0.18	
0.00/0.18	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.18	
0.00/0.18	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.18	
0.00/0.18	  rec(s(X), Y, /\x.F(x)) >? F(X) rec(X, Y, /\y.F(y)) 
0.00/0.18	
0.00/0.18	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.18	
0.00/0.18	We choose Lex = {rec} and Mul = {@_{o -> o}, s}, and the following precedence: rec > @_{o -> o} > s
0.00/0.18	
0.00/0.18	With these choices, we have:
0.00/0.18	
0.00/0.18	  1] rec(s(X), Y, /\x.F(x)) > @_{o -> o}(F(X), rec(X, Y, /\x.F(x)))  because [2], by definition 
0.00/0.18	  2] rec*(s(X), Y, /\x.F(x)) >= @_{o -> o}(F(X), rec(X, Y, /\x.F(x)))  because rec > @_{o -> o}, [3] and [9], by (Copy) 
0.00/0.18	  3] rec*(s(X), Y, /\x.F(x)) >= F(X)  because [4], by (Select) 
0.00/0.18	  4] F(rec*(s(X), Y, /\x.F(x))) >= F(X)  because [5], by (Meta) 
0.00/0.18	  5] rec*(s(X), Y, /\x.F(x)) >= X  because [6], by (Select) 
0.00/0.18	  6] s(X) >= X  because [7], by (Star) 
0.00/0.18	  7] s*(X) >= X  because [8], by (Select) 
0.00/0.18	  8] X >= X  by (Meta) 
0.00/0.18	  9] rec*(s(X), Y, /\x.F(x)) >= rec(X, Y, /\x.F(x))  because [10], [5], [12] and [14], by (Stat) 
0.00/0.18	  10] s(X) > X  because [11], by definition 
0.00/0.18	  11] s*(X) >= X  because [8], by (Select) 
0.00/0.18	  12] rec*(s(X), Y, /\x.F(x)) >= Y  because [13], by (Select) 
0.00/0.18	  13] Y >= Y  by (Meta) 
0.00/0.18	  14] rec*(s(X), Y, /\x.F(x)) >= /\x.F(x)  because [15], by (F-Abs) 
0.00/0.18	  15] rec*(s(X), Y, /\x.F(x), y) >= F(y)  because [16], by (Select) 
0.00/0.18	  16] F(rec*(s(X), Y, /\x.F(x), y)) >= F(y)  because [17], by (Meta) 
0.00/0.18	  17] rec*(s(X), Y, /\x.F(x), y) >= y  because [18], by (Select) 
0.00/0.18	  18] y >= y  by (Var) 
0.00/0.18	
0.00/0.18	We can thus remove the following rules:
0.00/0.18	
0.00/0.18	  rec(s(X), Y, /\x.F(x)) => F(X) rec(X, Y, /\y.F(y)) 
0.00/0.18	
0.00/0.18	All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.
0.00/0.18	
0.00/0.18	
0.00/0.18	+++ Citations +++
0.00/0.18	
0.00/0.18	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
0.00/0.18	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.18	EOF