We consider the system h15. Alphabet: 0 : [] --> nat cons : [nat * natlist] --> natlist foldl : [nat -> nat -> nat * nat * natlist] --> nat nil : [] --> natlist plus : [] --> nat -> nat -> nat sum : [natlist] --> nat xap : [nat -> nat -> nat * nat] --> nat -> nat yap : [nat -> nat * nat] --> nat Rules: foldl(/\x./\y.yap(xap(f, x), y), z, nil) => z foldl(/\x./\y.yap(xap(f, x), y), z, cons(u, v)) => foldl(/\w./\x'.yap(xap(f, w), x'), yap(xap(f, z), u), v) sum(x) => foldl(/\y./\z.yap(xap(plus, y), z), 0, x) xap(f, x) => f x yap(f, x) => f x This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). Symbol xap 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: Alphabet: 0 : [] --> nat cons : [nat * natlist] --> natlist foldl : [nat -> nat -> nat * nat * natlist] --> nat nil : [] --> natlist plus : [nat] --> nat -> nat sum : [natlist] --> nat yap : [nat -> nat * nat] --> nat Rules: foldl(/\x./\y.yap(F(x), y), X, nil) => X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) => foldl(/\z./\u.yap(F(z), u), yap(F(X), Y), Z) sum(X) => foldl(/\x./\y.yap(plus(x), y), 0, X) yap(F, X) => F X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): foldl(/\x./\y.yap(F(x), y), X, nil) >? X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >? foldl(/\z./\u.yap(F(z), u), yap(F(X), Y), Z) sum(X) >? foldl(/\x./\y.yap(plus(x), y), 0, X) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_1, x_2) We choose Lex = {foldl} and Mul = {@_{o -> o}, cons, nil, plus, sum, yap}, and the following precedence: cons > nil > sum > plus > yap > @_{o -> o} > foldl Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: foldl(/\x./\y.yap(F(x), y), X, nil) >= X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) yap(F, X) > @_{o -> o}(F, X) With these choices, we have: 1] foldl(/\x./\y.yap(F(x), y), X, nil) >= X because [2], by (Star) 2] foldl*(/\x./\y.yap(F(x), y), X, nil) >= X because [3], by (Select) 3] X >= X by (Meta) 4] foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [5], by (Star) 5] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [6], [9], [17] and [26], by (Stat) 6] cons(Y, Z) > Z because [7], by definition 7] cons*(Y, Z) >= Z because [8], by (Select) 8] Z >= Z by (Meta) 9] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= /\x./\y.yap(F(x), y) because [10], by (F-Abs) 10] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z) >= /\x.yap(F(z), x) because [11], by (Select) 11] /\x.yap(F(foldl*(/\y./\v.yap(F(y), v), X, cons(Y, Z), z)), x) >= /\x.yap(F(z), x) because [12], by (Abs) 12] yap(F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z)), u) >= yap(F(z), u) because yap in Mul, [13] and [16], by (Fun) 13] F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z)) >= F(z) because [14], by (Meta) 14] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z) >= z because [15], by (Select) 15] z >= z by (Var) 16] u >= u by (Var) 17] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= yap(F(X), Y) because [18], by (Select) 18] yap(F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))), foldl*(/\v./\w.yap(F(v), w), X, cons(Y, Z))) >= yap(F(X), Y) because yap in Mul, [19] and [22], by (Fun) 19] F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))) >= F(X) because [20], by (Meta) 20] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= X because [21], by (Select) 21] X >= X by (Meta) 22] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= Y because [23], by (Select) 23] cons(Y, Z) >= Y because [24], by (Star) 24] cons*(Y, Z) >= Y because [25], by (Select) 25] Y >= Y by (Meta) 26] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= Z because [27], by (Select) 27] cons(Y, Z) >= Z because [7], by (Star) 28] sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because [29], by (Star) 29] sum*(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because sum > foldl, [30], [38] and [39], by (Copy) 30] sum*(X) >= /\y./\z.yap(plus(y), z) because [31], by (F-Abs) 31] sum*(X, x) >= /\z.yap(plus(x), z) because [32], by (F-Abs) 32] sum*(X, x, y) >= yap(plus(x), y) because sum > yap, [33] and [36], by (Copy) 33] sum*(X, x, y) >= plus(x) because sum > plus and [34], by (Copy) 34] sum*(X, x, y) >= x because [35], by (Select) 35] x >= x by (Var) 36] sum*(X, x, y) >= y because [37], by (Select) 37] y >= y by (Var) 38] sum*(X) >= _|_ by (Bot) 39] sum*(X) >= X because [40], by (Select) 40] X >= X by (Meta) 41] yap(F, X) > @_{o -> o}(F, X) because [42], by definition 42] yap*(F, X) >= @_{o -> o}(F, X) because yap > @_{o -> o}, [43] and [45], by (Copy) 43] yap*(F, X) >= F because [44], by (Select) 44] F >= F by (Meta) 45] yap*(F, X) >= X because [46], by (Select) 46] X >= X by (Meta) We can thus remove the following rules: yap(F, X) => F X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): foldl(/\x./\y.yap(F(x), y), X, nil) >? X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >? foldl(/\z./\u.yap(F(z), u), yap(F(X), Y), Z) sum(X) >? foldl(/\x./\y.yap(plus(x), y), 0, X) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_1, x_2) We choose Lex = {foldl} and Mul = {cons, nil, plus, sum, yap}, and the following precedence: cons > nil > sum > plus > foldl > yap Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: foldl(/\x./\y.yap(F(x), y), X, nil) > X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) With these choices, we have: 1] foldl(/\x./\y.yap(F(x), y), X, nil) > X because [2], by definition 2] foldl*(/\x./\y.yap(F(x), y), X, nil) >= X because [3], by (Select) 3] X >= X by (Meta) 4] foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [5], by (Star) 5] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [6], [9], [16] and [26], by (Stat) 6] cons(Y, Z) > Z because [7], by definition 7] cons*(Y, Z) >= Z because [8], by (Select) 8] Z >= Z by (Meta) 9] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= /\x./\y.yap(F(x), y) because [10], by (Select) 10] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [11], by (Abs) 11] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [12], by (Abs) 12] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [13] and [15], by (Fun) 13] F(y) >= F(y) because [14], by (Meta) 14] y >= y by (Var) 15] x >= x by (Var) 16] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= yap(F(X), Y) because foldl > yap, [17] and [22], by (Copy) 17] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= F(X) because [18], by (Select) 18] /\z.yap(F(foldl*(/\u./\v.yap(F(u), v), X, cons(Y, Z))), z) >= F(X) because [19], by (Eta)[Kop13:2] 19] F(foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z))) >= F(X) because [20], by (Meta) 20] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= X because [21], by (Select) 21] X >= X by (Meta) 22] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= Y because [23], by (Select) 23] cons(Y, Z) >= Y because [24], by (Star) 24] cons*(Y, Z) >= Y because [25], by (Select) 25] Y >= Y by (Meta) 26] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= Z because [27], by (Select) 27] cons(Y, Z) >= Z because [7], by (Star) 28] sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because [29], by (Star) 29] sum*(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because sum > foldl, [30], [38] and [39], by (Copy) 30] sum*(X) >= /\y./\z.yap(plus(y), z) because [31], by (F-Abs) 31] sum*(X, x) >= /\z.yap(plus(x), z) because [32], by (F-Abs) 32] sum*(X, x, y) >= yap(plus(x), y) because sum > yap, [33] and [36], by (Copy) 33] sum*(X, x, y) >= plus(x) because sum > plus and [34], by (Copy) 34] sum*(X, x, y) >= x because [35], by (Select) 35] x >= x by (Var) 36] sum*(X, x, y) >= y because [37], by (Select) 37] y >= y by (Var) 38] sum*(X) >= _|_ by (Bot) 39] sum*(X) >= X because [40], by (Select) 40] X >= X by (Meta) We can thus remove the following rules: foldl(/\x./\y.yap(F(x), y), X, nil) => X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >? foldl(/\z./\u.yap(F(z), u), yap(F(X), Y), Z) sum(X) >? foldl(/\x./\y.yap(plus(x), y), 0, X) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_1, x_2) We choose Lex = {foldl} and Mul = {cons, plus, sum, yap}, and the following precedence: sum > plus > foldl > yap > cons Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) > foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) With these choices, we have: 1] foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) > foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [2], by definition 2] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [3], [6], [13] and [23], by (Stat) 3] cons(Y, Z) > Z because [4], by definition 4] cons*(Y, Z) >= Z because [5], by (Select) 5] Z >= Z by (Meta) 6] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= /\x./\y.yap(F(x), y) because [7], by (Select) 7] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [8], by (Abs) 8] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [9], by (Abs) 9] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [10] and [12], by (Fun) 10] F(y) >= F(y) because [11], by (Meta) 11] y >= y by (Var) 12] x >= x by (Var) 13] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= yap(F(X), Y) because foldl > yap, [14] and [19], by (Copy) 14] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= F(X) because [15], by (Select) 15] /\z.yap(F(foldl*(/\u./\v.yap(F(u), v), X, cons(Y, Z))), z) >= F(X) because [16], by (Eta)[Kop13:2] 16] F(foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z))) >= F(X) because [17], by (Meta) 17] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= X because [18], by (Select) 18] X >= X by (Meta) 19] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= Y because [20], by (Select) 20] cons(Y, Z) >= Y because [21], by (Star) 21] cons*(Y, Z) >= Y because [22], by (Select) 22] Y >= Y by (Meta) 23] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= Z because [24], by (Select) 24] cons(Y, Z) >= Z because [4], by (Star) 25] sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because [26], by (Star) 26] sum*(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because sum > foldl, [27], [35] and [36], by (Copy) 27] sum*(X) >= /\y./\z.yap(plus(y), z) because [28], by (F-Abs) 28] sum*(X, x) >= /\z.yap(plus(x), z) because [29], by (F-Abs) 29] sum*(X, x, y) >= yap(plus(x), y) because sum > yap, [30] and [33], by (Copy) 30] sum*(X, x, y) >= plus(x) because sum > plus and [31], by (Copy) 31] sum*(X, x, y) >= x because [32], by (Select) 32] x >= x by (Var) 33] sum*(X, x, y) >= y because [34], by (Select) 34] y >= y by (Var) 35] sum*(X) >= _|_ by (Bot) 36] sum*(X) >= X because [37], by (Select) 37] X >= X by (Meta) We can thus remove the following rules: foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) => foldl(/\z./\u.yap(F(z), u), yap(F(X), Y), Z) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): sum(X) >? foldl(/\x./\y.yap(plus(x), y), 0, X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 foldl = \G0y1y2.y1 + y2 + G0(0,0) plus = \y0y1.y0 sum = \y0.3 + 3y0 yap = \G0y1.y1 + G0(0) Using this interpretation, the requirements translate to: [[sum(_x0)]] = 3 + 3x0 > x0 = [[foldl(/\x./\y.yap(plus(x), y), 0, _x0)]] We can thus remove the following rules: sum(X) => foldl(/\x./\y.yap(plus(x), y), 0, X) 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. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [Kop13:2] C. Kop. StarHorpo with an Eta-Rule. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/etahorpo.pdf, 2013.