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 : [] --> nat 
0.00/0.41	    cons : [nat * list] --> list 
0.00/0.41	    foldl : [nat -> nat -> nat * nat * list] --> nat 
0.00/0.41	    nil : [] --> list 
0.00/0.41	    plus : [nat * nat] --> nat 
0.00/0.41	    plusc : [] --> nat -> nat -> nat 
0.00/0.41	    sum : [list] --> nat 
0.00/0.41	
0.00/0.41	  Rules:
0.00/0.41	
0.00/0.41	    foldl(f, x, nil) => x 
0.00/0.41	    foldl(f, x, cons(y, z)) => foldl(f, f x y, z) 
0.00/0.41	    plusc => /\x./\y.plus(x, y) 
0.00/0.41	    sum(x) => foldl(plusc, 0, x) 
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	  foldl(F, X, nil) >? X 
0.00/0.41	  foldl(F, X, cons(Y, Z)) >? foldl(F, F X Y, Z) 
0.00/0.41	  plusc >? /\x./\y.plus(x, y) 
0.00/0.41	  sum(X) >? foldl(plusc, 0, X) 
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	  [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_1, x_2) 
0.00/0.41	
0.00/0.41	We choose Lex = {foldl} and Mul = {@_{o -> o -> o}, @_{o -> o}, cons, nil, plus, plusc, sum}, and the following precedence: nil > plusc = sum > foldl > @_{o -> o -> o} > @_{o -> o} > cons > plus
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	  foldl(F, X, nil) > X 
0.00/0.41	  foldl(F, X, cons(Y, Z)) >= foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z) 
0.00/0.41	  plusc >= /\x./\y.plus(x, y) 
0.00/0.41	  sum(X) >= foldl(plusc, _|_, X) 
0.00/0.41	
0.00/0.41	With these choices, we have:
0.00/0.41	
0.00/0.41	  1] foldl(F, X, nil) > X  because [2], by definition 
0.00/0.41	  2] foldl*(F, X, nil) >= X  because [3], by (Select) 
0.00/0.41	  3] X >= X  by (Meta) 
0.00/0.41	
0.00/0.41	  4] foldl(F, X, cons(Y, Z)) >= foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z)  because [5], by (Star) 
0.00/0.41	  5] foldl*(F, X, cons(Y, Z)) >= foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z)  because [6], [9], [11] and [19], by (Stat) 
0.00/0.41	  6] cons(Y, Z) > Z  because [7], by definition 
0.00/0.41	  7] cons*(Y, Z) >= Z  because [8], by (Select) 
0.00/0.41	  8] Z >= Z  by (Meta) 
0.00/0.41	  9] foldl*(F, X, cons(Y, Z)) >= F  because [10], by (Select) 
0.00/0.41	  10] F >= F  by (Meta) 
0.00/0.41	  11] foldl*(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, X), Y)  because foldl > @_{o -> o}, [12] and [15], by (Copy) 
0.00/0.41	  12] foldl*(F, X, cons(Y, Z)) >= @_{o -> o -> o}(F, X)  because foldl > @_{o -> o -> o}, [9] and [13], by (Copy) 
0.00/0.41	  13] foldl*(F, X, cons(Y, Z)) >= X  because [14], by (Select) 
0.00/0.41	  14] X >= X  by (Meta) 
0.00/0.41	  15] foldl*(F, X, cons(Y, Z)) >= Y  because [16], by (Select) 
0.00/0.41	  16] cons(Y, Z) >= Y  because [17], by (Star) 
0.00/0.41	  17] cons*(Y, Z) >= Y  because [18], by (Select) 
0.00/0.41	  18] Y >= Y  by (Meta) 
0.00/0.41	  19] foldl*(F, X, cons(Y, Z)) >= Z  because [20], by (Select) 
0.00/0.41	  20] cons(Y, Z) >= Z  because [7], by (Star) 
0.00/0.41	
0.00/0.41	  21] plusc >= /\x./\y.plus(x, y)  because [22], by (Star) 
0.00/0.41	  22] plusc* >= /\y./\z.plus(y, z)  because [23], by (F-Abs) 
0.00/0.41	  23] plusc*(x) >= /\z.plus(x, z)  because [24], by (F-Abs) 
0.00/0.41	  24] plusc*(x, y) >= plus(x, y)  because plusc > plus, [25] and [27], by (Copy) 
0.00/0.41	  25] plusc*(x, y) >= x  because [26], by (Select) 
0.00/0.41	  26] x >= x  by (Var) 
0.00/0.41	  27] plusc*(x, y) >= y  because [28], by (Select) 
0.00/0.41	  28] y >= y  by (Var) 
0.00/0.41	
0.00/0.41	  29] sum(X) >= foldl(plusc, _|_, X)  because [30], by (Star) 
0.00/0.41	  30] sum*(X) >= foldl(plusc, _|_, X)  because sum > foldl, [31], [32] and [33], by (Copy) 
0.00/0.41	  31] sum*(X) >= plusc  because sum = plusc and sum in Mul, by (Stat) 
0.00/0.41	  32] sum*(X) >= _|_  by (Bot) 
0.00/0.41	  33] sum*(X) >= X  because [34], by (Select) 
0.00/0.41	  34] X >= X  by (Meta) 
0.00/0.41	
0.00/0.41	We can thus remove the following rules:
0.00/0.41	
0.00/0.41	  foldl(F, X, nil) => X 
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	  foldl(F, X, cons(Y, Z)) >? foldl(F, F X Y, Z) 
0.00/0.41	  plusc >? /\x./\y.plus(x, y) 
0.00/0.41	  sum(X) >? foldl(plusc, 0, X) 
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	  [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_2, x_1) 
0.00/0.41	
0.00/0.41	We choose Lex = {foldl} and Mul = {@_{o -> o -> o}, @_{o -> o}, cons, plus, plusc, sum}, and the following precedence: cons > sum > foldl > @_{o -> o -> o} > @_{o -> o} > plusc > plus
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	  foldl(F, X, cons(Y, Z)) > foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z) 
0.00/0.41	  plusc >= /\x./\y.plus(x, y) 
0.00/0.41	  sum(X) >= foldl(plusc, _|_, X) 
0.00/0.41	
0.00/0.41	With these choices, we have:
0.00/0.41	
0.00/0.41	  1] foldl(F, X, cons(Y, Z)) > foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z)  because [2], by definition 
0.00/0.41	  2] foldl*(F, X, cons(Y, Z)) >= foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z)  because [3], [6], [8] and [16], by (Stat) 
0.00/0.41	  3] cons(Y, Z) > Z  because [4], by definition 
0.00/0.41	  4] cons*(Y, Z) >= Z  because [5], by (Select) 
0.00/0.41	  5] Z >= Z  by (Meta) 
0.00/0.41	  6] foldl*(F, X, cons(Y, Z)) >= F  because [7], by (Select) 
0.00/0.41	  7] F >= F  by (Meta) 
0.00/0.41	  8] foldl*(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, X), Y)  because foldl > @_{o -> o}, [9] and [12], by (Copy) 
0.00/0.41	  9] foldl*(F, X, cons(Y, Z)) >= @_{o -> o -> o}(F, X)  because foldl > @_{o -> o -> o}, [6] and [10], by (Copy) 
0.00/0.41	  10] foldl*(F, X, cons(Y, Z)) >= X  because [11], by (Select) 
0.00/0.41	  11] X >= X  by (Meta) 
0.00/0.41	  12] foldl*(F, X, cons(Y, Z)) >= Y  because [13], by (Select) 
0.00/0.41	  13] cons(Y, Z) >= Y  because [14], by (Star) 
0.00/0.41	  14] cons*(Y, Z) >= Y  because [15], by (Select) 
0.00/0.41	  15] Y >= Y  by (Meta) 
0.00/0.41	  16] foldl*(F, X, cons(Y, Z)) >= Z  because [17], by (Select) 
0.00/0.41	  17] cons(Y, Z) >= Z  because [4], by (Star) 
0.00/0.41	
0.00/0.41	  18] plusc >= /\x./\y.plus(x, y)  because [19], by (Star) 
0.00/0.41	  19] plusc* >= /\y./\z.plus(y, z)  because [20], by (F-Abs) 
0.00/0.41	  20] plusc*(x) >= /\z.plus(x, z)  because [21], by (F-Abs) 
0.00/0.41	  21] plusc*(x, y) >= plus(x, y)  because plusc > plus, [22] and [24], by (Copy) 
0.00/0.41	  22] plusc*(x, y) >= x  because [23], by (Select) 
0.00/0.41	  23] x >= x  by (Var) 
0.00/0.41	  24] plusc*(x, y) >= y  because [25], by (Select) 
0.00/0.41	  25] y >= y  by (Var) 
0.00/0.41	
0.00/0.41	  26] sum(X) >= foldl(plusc, _|_, X)  because [27], by (Star) 
0.00/0.41	  27] sum*(X) >= foldl(plusc, _|_, X)  because sum > foldl, [28], [29] and [30], by (Copy) 
0.00/0.41	  28] sum*(X) >= plusc  because sum > plusc, by (Copy) 
0.00/0.41	  29] sum*(X) >= _|_  by (Bot) 
0.00/0.41	  30] sum*(X) >= X  because [31], by (Select) 
0.00/0.41	  31] X >= X  by (Meta) 
0.00/0.41	
0.00/0.41	We can thus remove the following rules:
0.00/0.41	
0.00/0.41	  foldl(F, X, cons(Y, Z)) => foldl(F, F X Y, Z) 
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	  plusc >? /\x./\y.plus(x, y) 
0.00/0.41	  sum(X) >? foldl(plusc, 0, X) 
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	  foldl = \G0y1y2.y1 + y2 + G0(0,0) 
0.00/0.41	  plus = \y0y1.y0 + y1 
0.00/0.41	  plusc = \y0y1.1 + y0 + y1 
0.00/0.41	  sum = \y0.3 + 3y0 
0.00/0.41	
0.00/0.41	Using this interpretation, the requirements translate to:
0.00/0.41	
0.00/0.41	  [[plusc]] = \y0y1.1 + y0 + y1 > \y0y1.y0 + y1 = [[/\x./\y.plus(x, y)]] 
0.00/0.41	  [[sum(_x0)]] = 3 + 3x0 > 1 + x0 = [[foldl(plusc, 0, _x0)]] 
0.00/0.41	
0.00/0.41	We can thus remove the following rules:
0.00/0.41	
0.00/0.41	  plusc => /\x./\y.plus(x, y) 
0.00/0.41	  sum(X) => foldl(plusc, 0, X) 
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.
0.00/0.41	EOF