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 > sum > foldl > @_{o -> o} = yap > plus > nil 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 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], [19] and [29], 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 (F-Abs) 11] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z, u) >= yap(F(z), u) because foldl > yap, [12] and [17], by (Copy) 12] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z, u) >= F(z) because [13], by (Select) 13] /\x.yap(F(foldl*(/\y./\v.yap(F(y), v), X, cons(Y, Z), z, u)), x) >= F(z) because [14], by (Eta)[Kop13:2] 14] F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z, u)) >= F(z) because [15], by (Meta) 15] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z, u) >= z because [16], by (Select) 16] z >= z by (Var) 17] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z, u) >= u because [18], by (Select) 18] u >= u by (Var) 19] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= yap(F(X), Y) because foldl > yap, [20] and [25], by (Copy) 20] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= F(X) because [21], by (Select) 21] /\x.yap(F(foldl*(/\y./\v.yap(F(y), v), X, cons(Y, Z))), x) >= F(X) because [22], by (Eta)[Kop13:2] 22] F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))) >= F(X) because [23], by (Meta) 23] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= X because [24], by (Select) 24] X >= X by (Meta) 25] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= Y because [26], by (Select) 26] cons(Y, Z) >= Y because [27], by (Star) 27] cons*(Y, Z) >= Y because [28], by (Select) 28] Y >= Y by (Meta) 29] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= Z because [30], by (Select) 30] cons(Y, Z) >= Z because [7], by (Star) 31] sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because [32], by (Star) 32] sum*(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because sum > foldl, [33], [41] and [42], by (Copy) 33] sum*(X) >= /\y./\z.yap(plus(y), z) because [34], by (F-Abs) 34] sum*(X, x) >= /\z.yap(plus(x), z) because [35], by (F-Abs) 35] sum*(X, x, y) >= yap(plus(x), y) because sum > yap, [36] and [39], by (Copy) 36] sum*(X, x, y) >= plus(x) because sum > plus and [37], by (Copy) 37] sum*(X, x, y) >= x because [38], by (Select) 38] x >= x by (Var) 39] sum*(X, x, y) >= y because [40], by (Select) 40] y >= y by (Var) 41] sum*(X) >= _|_ by (Bot) 42] sum*(X) >= X because [43], by (Select) 43] X >= X by (Meta) 44] yap(F, X) >= @_{o -> o}(F, X) because yap = @_{o -> o}, yap in Mul, [45] and [46], by (Fun) 45] F >= F by (Meta) 46] 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) 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, plus, sum, yap}, and the following precedence: sum > cons > @_{o -> o} = yap > foldl > plus 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) yap(F, X) >= @_{o -> o}(F, 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], [14] 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 (F-Abs) 7] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z) >= /\x.yap(F(z), x) because [8], by (Select) 8] /\x.yap(F(foldl*(/\y./\v.yap(F(y), v), X, cons(Y, Z), z)), x) >= /\x.yap(F(z), x) because [9], by (Abs) 9] yap(F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z)), u) >= yap(F(z), u) because yap in Mul, [10] and [13], by (Fun) 10] F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z)) >= F(z) because [11], by (Meta) 11] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z), z) >= z because [12], by (Select) 12] z >= z by (Var) 13] u >= u by (Var) 14] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= yap(F(X), Y) because [15], by (Select) 15] 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, [16] and [19], by (Fun) 16] F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))) >= F(X) because [17], by (Meta) 17] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= X because [18], by (Select) 18] X >= X by (Meta) 19] foldl*(/\x./\y.yap(F(x), y), 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*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= Z because [24], by (Select) 24] yap(F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))), foldl*(/\v./\w.yap(F(v), w), X, cons(Y, Z))) >= Z because [25], by (Star) 25] yap*(F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))), foldl*(/\v./\w.yap(F(v), w), X, cons(Y, Z))) >= Z because [26], by (Select) 26] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= Z because [27], by (Select) 27] yap(F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))), foldl*(/\v./\w.yap(F(v), w), X, cons(Y, Z))) >= Z because [28], by (Star) 28] yap*(F(foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z))), foldl*(/\v./\w.yap(F(v), w), X, cons(Y, Z))) >= Z because [29], by (Select) 29] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= Z because [30], by (Select) 30] cons(Y, Z) >= Z because [4], by (Star) 31] sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because [32], by (Star) 32] sum*(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because sum > foldl, [33], [41] and [42], by (Copy) 33] sum*(X) >= /\y./\z.yap(plus(y), z) because [34], by (F-Abs) 34] sum*(X, x) >= /\z.yap(plus(x), z) because [35], by (F-Abs) 35] sum*(X, x, y) >= yap(plus(x), y) because sum > yap, [36] and [39], by (Copy) 36] sum*(X, x, y) >= plus(x) because sum > plus and [37], by (Copy) 37] sum*(X, x, y) >= x because [38], by (Select) 38] x >= x by (Var) 39] sum*(X, x, y) >= y because [40], by (Select) 40] y >= y by (Var) 41] sum*(X) >= _|_ by (Bot) 42] sum*(X) >= X because [43], by (Select) 43] X >= X by (Meta) 44] yap(F, X) >= @_{o -> o}(F, X) because yap = @_{o -> o}, yap in Mul, [45] and [46], by (Fun) 45] F >= F by (Meta) 46] 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) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, foldl, plus, sum, yap}, and the following precedence: sum > yap > @_{o -> o} > foldl > plus Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) yap(F, X) > @_{o -> o}(F, X) With these choices, we have: 1] sum(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because [2], by (Star) 2] sum*(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because sum > foldl, [3], [11] and [12], by (Copy) 3] sum*(X) >= /\y./\z.yap(plus(y), z) because [4], by (F-Abs) 4] sum*(X, x) >= /\z.yap(plus(x), z) because [5], by (F-Abs) 5] sum*(X, x, y) >= yap(plus(x), y) because sum > yap, [6] and [9], by (Copy) 6] sum*(X, x, y) >= plus(x) because sum > plus and [7], by (Copy) 7] sum*(X, x, y) >= x because [8], by (Select) 8] x >= x by (Var) 9] sum*(X, x, y) >= y because [10], by (Select) 10] y >= y by (Var) 11] sum*(X) >= _|_ by (Bot) 12] sum*(X) >= X because [13], by (Select) 13] X >= X by (Meta) 14] yap(F, X) > @_{o -> o}(F, X) because [15], by definition 15] yap*(F, X) >= @_{o -> o}(F, X) because yap > @_{o -> o}, [16] and [18], by (Copy) 16] yap*(F, X) >= F because [17], by (Select) 17] F >= F by (Meta) 18] yap*(F, X) >= X because [19], by (Select) 19] 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]): 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]] = _|_ We choose Lex = {} and Mul = {foldl, plus, sum, yap}, and the following precedence: sum > plus > foldl > yap Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: sum(X) > foldl(/\x./\y.yap(plus(x), y), _|_, X) With these choices, we have: 1] sum(X) > foldl(/\x./\y.yap(plus(x), y), _|_, X) because [2], by definition 2] sum*(X) >= foldl(/\x./\y.yap(plus(x), y), _|_, X) because sum > foldl, [3], [11] and [12], by (Copy) 3] sum*(X) >= /\y./\z.yap(plus(y), z) because [4], by (F-Abs) 4] sum*(X, x) >= /\z.yap(plus(x), z) because [5], by (F-Abs) 5] sum*(X, x, y) >= yap(plus(x), y) because sum > yap, [6] and [9], by (Copy) 6] sum*(X, x, y) >= plus(x) because sum > plus and [7], by (Copy) 7] sum*(X, x, y) >= x because [8], by (Select) 8] x >= x by (Var) 9] sum*(X, x, y) >= y because [10], by (Select) 10] y >= y by (Var) 11] sum*(X) >= _|_ by (Bot) 12] sum*(X) >= X because [13], by (Select) 13] X >= X by (Meta) 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.