0.00/0.40	YES
0.00/0.41	We consider the system theBenchmark.
0.00/0.41	
0.00/0.41	  Alphabet:
0.00/0.41	
0.00/0.41	    0 : [] --> c 
0.00/0.41	    add : [] --> a -> c -> c 
0.00/0.41	    cons : [a * b] --> b 
0.00/0.41	    fold : [a -> c -> c * c] --> b -> c 
0.00/0.41	    mul : [] --> a -> c -> c 
0.00/0.41	    nil : [] --> b 
0.00/0.41	    plus : [c * c] --> c 
0.00/0.41	    prod : [] --> b -> c 
0.00/0.41	    s : [c] --> c 
0.00/0.41	    sum : [] --> b -> c 
0.00/0.41	    times : [c * c] --> c 
0.00/0.41	
0.00/0.41	  Rules:
0.00/0.41	
0.00/0.41	    fold(f, x) nil => x 
0.00/0.41	    fold(f, x) cons(y, z) => f y (fold(f, x) z) 
0.00/0.41	    plus(0, x) => x 
0.00/0.41	    plus(s(x), y) => s(plus(x, y)) 
0.00/0.41	    times(0, x) => 0 
0.00/0.41	    times(s(x), y) => plus(times(x, y), y) 
0.00/0.41	    sum => fold(add, 0) 
0.00/0.41	    prod => fold(mul, s(0)) 
0.00/0.41	
0.00/0.41	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.41	
0.00/0.41	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.41	
0.00/0.41	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.41	
0.00/0.41	  fold(F, X) nil >? X 
0.00/0.41	  fold(F, X) cons(Y, Z) >? F Y (fold(F, X) Z) 
0.00/0.41	  plus(0, X) >? X 
0.00/0.41	  plus(s(X), Y) >? s(plus(X, Y)) 
0.00/0.41	  times(0, X) >? 0 
0.00/0.41	  times(s(X), Y) >? plus(times(X, Y), Y) 
0.00/0.41	  sum >? fold(add, 0) 
0.00/0.41	  prod >? fold(mul, s(0)) 
0.00/0.41	
0.00/0.41	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.41	
0.00/0.41	Argument functions:
0.00/0.41	
0.00/0.41	  [[0]] = _|_ 
0.00/0.41	  [[add]] = _|_ 
0.00/0.41	  [[mul]] = _|_ 
0.00/0.41	
0.00/0.41	We choose Lex = {} and Mul = {@_{o -> o -> o}, @_{o -> o}, cons, fold, nil, plus, prod, s, sum, times}, and the following precedence: nil > prod > sum > fold > @_{o -> o -> o} > @_{o -> o} > cons > times > plus > s
0.00/0.41	
0.00/0.41	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
0.00/0.41	
0.00/0.41	  @_{o -> o}(fold(F, X), nil) > X 
0.00/0.41	  @_{o -> o}(fold(F, X), cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z)) 
0.00/0.41	  plus(_|_, X) >= X 
0.00/0.41	  plus(s(X), Y) >= s(plus(X, Y)) 
0.00/0.41	  times(_|_, X) >= _|_ 
0.00/0.41	  times(s(X), Y) > plus(times(X, Y), Y) 
0.00/0.41	  sum >= fold(_|_, _|_) 
0.00/0.41	  prod >= fold(_|_, s(_|_)) 
0.00/0.41	
0.00/0.41	With these choices, we have:
0.00/0.41	
0.00/0.41	  1] @_{o -> o}(fold(F, X), nil) > X  because [2], by definition 
0.00/0.41	  2] @_{o -> o}*(fold(F, X), nil) >= X  because [3], by (Select) 
0.00/0.41	  3] fold(F, X) @_{o -> o}*(fold(F, X), nil) >= X  because [4] 
0.00/0.41	  4] fold*(F, X, @_{o -> o}*(fold(F, X), nil)) >= X  because [5], by (Select) 
0.00/0.41	  5] X >= X  by (Meta) 
0.00/0.41	
0.00/0.41	  6] @_{o -> o}(fold(F, X), cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [7], by definition 
0.00/0.41	  7] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [8], by (Select) 
0.00/0.41	  8] fold(F, X) @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because [9] 
0.00/0.41	  9] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o}(@_{o -> o -> o}(F, Y), @_{o -> o}(fold(F, X), Z))  because fold > @_{o -> o}, [10] and [18], by (Copy) 
0.00/0.41	  10] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o -> o}(F, Y)  because fold > @_{o -> o -> o}, [11] and [13], by (Copy) 
0.00/0.41	  11] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= F  because [12], by (Select) 
0.00/0.41	  12] F >= F  by (Meta) 
0.00/0.41	  13] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= Y  because [14], by (Select) 
0.00/0.41	  14] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= Y  because [15], by (Select) 
0.00/0.41	  15] cons(Y, Z) >= Y  because [16], by (Star) 
0.00/0.41	  16] cons*(Y, Z) >= Y  because [17], by (Select) 
0.00/0.41	  17] Y >= Y  by (Meta) 
0.00/0.41	  18] fold*(F, X, @_{o -> o}*(fold(F, X), cons(Y, Z))) >= @_{o -> o}(fold(F, X), Z)  because [19], by (Select) 
0.00/0.41	  19] @_{o -> o}*(fold(F, X), cons(Y, Z)) >= @_{o -> o}(fold(F, X), Z)  because @_{o -> o} in Mul, [20] and [23], by (Stat) 
0.00/0.41	  20] fold(F, X) >= fold(F, X)  because fold in Mul, [21] and [22], by (Fun) 
0.00/0.41	  21] F >= F  by (Meta) 
0.00/0.41	  22] X >= X  by (Meta) 
0.00/0.41	  23] cons(Y, Z) > Z  because [24], by definition 
0.00/0.41	  24] cons*(Y, Z) >= Z  because [25], by (Select) 
0.00/0.41	  25] Z >= Z  by (Meta) 
0.00/0.41	
0.00/0.41	  26] plus(_|_, X) >= X  because [27], by (Star) 
0.00/0.41	  27] plus*(_|_, X) >= X  because [28], by (Select) 
0.00/0.41	  28] X >= X  by (Meta) 
0.00/0.41	
0.00/0.41	  29] plus(s(X), Y) >= s(plus(X, Y))  because [30], by (Star) 
0.00/0.41	  30] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [31], by (Copy) 
0.00/0.41	  31] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [32] and [35], by (Stat) 
0.00/0.41	  32] s(X) > X  because [33], by definition 
0.00/0.41	  33] s*(X) >= X  because [34], by (Select) 
0.00/0.41	  34] X >= X  by (Meta) 
0.00/0.41	  35] Y >= Y  by (Meta) 
0.00/0.41	
0.00/0.41	  36] times(_|_, X) >= _|_  by (Bot) 
0.00/0.41	
0.00/0.41	  37] times(s(X), Y) > plus(times(X, Y), Y)  because [38], by definition 
0.00/0.41	  38] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [39] and [44], by (Copy) 
0.00/0.41	  39] times*(s(X), Y) >= times(X, Y)  because times in Mul, [40] and [43], by (Stat) 
0.00/0.41	  40] s(X) > X  because [41], by definition 
0.00/0.41	  41] s*(X) >= X  because [42], by (Select) 
0.00/0.41	  42] X >= X  by (Meta) 
0.00/0.41	  43] Y >= Y  by (Meta) 
0.00/0.41	  44] times*(s(X), Y) >= Y  because [43], by (Select) 
0.00/0.41	
0.00/0.41	  45] sum >= fold(_|_, _|_)  because [46], by (Star) 
0.00/0.41	  46] sum* >= fold(_|_, _|_)  because sum > fold, [47] and [48], by (Copy) 
0.00/0.41	  47] sum* >= _|_  by (Bot) 
0.00/0.41	  48] sum* >= _|_  by (Bot) 
0.00/0.41	
0.00/0.41	  49] prod >= fold(_|_, s(_|_))  because [50], by (Star) 
0.00/0.41	  50] prod* >= fold(_|_, s(_|_))  because prod > fold, [51] and [52], by (Copy) 
0.00/0.41	  51] prod* >= _|_  by (Bot) 
0.00/0.41	  52] prod* >= s(_|_)  because prod > s and [53], by (Copy) 
0.00/0.41	  53] prod* >= _|_  by (Bot) 
0.00/0.41	
0.00/0.41	We can thus remove the following rules:
0.00/0.41	
0.00/0.41	  fold(F, X) nil => X 
0.00/0.41	  fold(F, X) cons(Y, Z) => F Y (fold(F, X) Z) 
0.00/0.41	  times(s(X), Y) => plus(times(X, Y), Y) 
0.00/0.41	
0.00/0.41	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.41	
0.00/0.41	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.41	
0.00/0.41	  plus(0, X) >? X 
0.00/0.41	  plus(s(X), Y) >? s(plus(X, Y)) 
0.00/0.41	  times(0, X) >? 0 
0.00/0.41	  sum >? fold(add, 0) 
0.00/0.41	  prod >? fold(mul, s(0)) 
0.00/0.41	
0.00/0.41	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.41	
0.00/0.41	The following interpretation satisfies the requirements:
0.00/0.41	
0.00/0.41	  0 = 0 
0.00/0.41	  add = \y0y1.0 
0.00/0.41	  fold = \G0y1y2.y1 + G0(0,0) 
0.00/0.41	  mul = \y0y1.0 
0.00/0.41	  plus = \y0y1.3 + y1 + 3y0 
0.00/0.41	  prod = \y0.3 + 3y0 
0.00/0.41	  s = \y0.y0 
0.00/0.41	  sum = \y0.3 + 3y0 
0.00/0.41	  times = \y0y1.3 + y1 + 3y0 
0.00/0.41	
0.00/0.41	Using this interpretation, the requirements translate to:
0.00/0.41	
0.00/0.41	  [[plus(0, _x0)]] = 3 + x0 > x0 = [[_x0]] 
0.00/0.41	  [[plus(s(_x0), _x1)]] = 3 + x1 + 3x0 >= 3 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
0.00/0.41	  [[times(0, _x0)]] = 3 + x0 > 0 = [[0]] 
0.00/0.41	  [[sum]] = \y0.3 + 3y0 > \y0.0 = [[fold(add, 0)]] 
0.00/0.41	  [[prod]] = \y0.3 + 3y0 > \y0.0 = [[fold(mul, s(0))]] 
0.00/0.41	
0.00/0.41	We can thus remove the following rules:
0.00/0.41	
0.00/0.41	  plus(0, X) => X 
0.00/0.41	  times(0, X) => 0 
0.00/0.41	  sum => fold(add, 0) 
0.00/0.41	  prod => fold(mul, s(0)) 
0.00/0.41	
0.00/0.41	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.41	
0.00/0.41	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.41	
0.00/0.41	  plus(s(X), Y) >? s(plus(X, Y)) 
0.00/0.41	
0.00/0.41	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.41	
0.00/0.41	The following interpretation satisfies the requirements:
0.00/0.41	
0.00/0.41	  plus = \y0y1.y1 + 3y0 
0.00/0.41	  s = \y0.1 + y0 
0.00/0.41	
0.00/0.41	Using this interpretation, the requirements translate to:
0.00/0.41	
0.00/0.41	  [[plus(s(_x0), _x1)]] = 3 + x1 + 3x0 > 1 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
0.00/0.41	
0.00/0.41	We can thus remove the following rules:
0.00/0.41	
0.00/0.41	  plus(s(X), Y) => s(plus(X, Y)) 
0.00/0.41	
0.00/0.41	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.
0.00/0.41	
0.00/0.41	
0.00/0.41	+++ Citations +++
0.00/0.41	
0.00/0.41	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.42	EOF