We consider the system 434. Alphabet: 0 : [] --> nat cons : [nat * natlist] --> natlist foldl : [nat -> nat -> nat * nat * natlist] --> nat nil : [] --> natlist plus : [nat * nat] --> nat sum : [natlist] --> nat Rules: foldl(/\x./\y.X(x, y), Y, nil) => Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) => foldl(/\z./\u.X(z, u), X(Y, Z), U) sum(X) => foldl(/\x./\y.plus(x, y), 0, 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.X(x, y), Y, nil) >? Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >? foldl(/\z./\u.X(z, u), X(Y, Z), U) sum(X) >? foldl(/\x./\y.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_2, x_1) We choose Lex = {foldl} and Mul = {cons, nil, plus, sum}, and the following precedence: cons > nil > sum > 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.X(x, y), Y, nil) > Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) sum(X) >= foldl(/\x./\y.plus(x, y), _|_, X) With these choices, we have: 1] foldl(/\x./\y.X(x, y), Y, nil) > Y because [2], by definition 2] foldl*(/\x./\y.X(x, y), Y, nil) >= Y because [3], by (Select) 3] Y >= Y by (Meta) 4] foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) because [5], by (Star) 5] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) because [6], [9], [15] and [23], by (Stat) 6] cons(Z, U) > U because [7], by definition 7] cons*(Z, U) >= U because [8], by (Select) 8] U >= U by (Meta) 9] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= /\x./\y.X(x, y) because [10], by (Select) 10] /\x./\z.X(x, z) >= /\x./\z.X(x, z) because [11], by (Abs) 11] /\z.X(y, z) >= /\z.X(y, z) because [12], by (Abs) 12] X(y, x) >= X(y, x) because [13] and [14], by (Meta) 13] y >= y by (Var) 14] x >= x by (Var) 15] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= X(Y, Z) because [16], by (Select) 16] X(foldl*(/\z./\u.X(z, u), Y, cons(Z, U)), foldl*(/\v./\w.X(v, w), Y, cons(Z, U))) >= X(Y, Z) because [17] and [19], by (Meta) 17] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= Y because [18], by (Select) 18] Y >= Y by (Meta) 19] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= Z because [20], by (Select) 20] cons(Z, U) >= Z because [21], by (Star) 21] cons*(Z, U) >= Z because [22], by (Select) 22] Z >= Z by (Meta) 23] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= U because [24], by (Select) 24] cons(Z, U) >= U because [7], by (Star) 25] sum(X) >= foldl(/\x./\y.plus(x, y), _|_, X) because [26], by (Star) 26] sum*(X) >= foldl(/\x./\y.plus(x, y), _|_, X) because sum > foldl, [27], [34] and [35], by (Copy) 27] sum*(X) >= /\y./\z.plus(y, z) because [28], by (F-Abs) 28] sum*(X, x) >= /\z.plus(x, z) because [29], by (F-Abs) 29] sum*(X, x, y) >= plus(x, y) because sum > plus, [30] and [32], by (Copy) 30] sum*(X, x, y) >= x because [31], by (Select) 31] x >= x by (Var) 32] sum*(X, x, y) >= y because [33], by (Select) 33] y >= y by (Var) 34] sum*(X) >= _|_ by (Bot) 35] sum*(X) >= X because [36], by (Select) 36] X >= X by (Meta) We can thus remove the following rules: foldl(/\x./\y.X(x, y), Y, nil) => Y 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.X(x, y), Y, cons(Z, U)) >? foldl(/\z./\u.X(z, u), X(Y, Z), U) sum(X) >? foldl(/\x./\y.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_2, x_1) We choose Lex = {foldl} and Mul = {cons, plus, sum}, and the following precedence: cons > sum > 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.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) sum(X) > foldl(/\x./\y.plus(x, y), _|_, X) With these choices, we have: 1] foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) because [2], by (Star) 2] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) because [3], [6], [12] and [20], by (Stat) 3] cons(Z, U) > U because [4], by definition 4] cons*(Z, U) >= U because [5], by (Select) 5] U >= U by (Meta) 6] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= /\x./\y.X(x, y) because [7], by (Select) 7] /\x./\z.X(x, z) >= /\x./\z.X(x, z) because [8], by (Abs) 8] /\z.X(y, z) >= /\z.X(y, z) because [9], by (Abs) 9] X(y, x) >= X(y, x) because [10] and [11], by (Meta) 10] y >= y by (Var) 11] x >= x by (Var) 12] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= X(Y, Z) because [13], by (Select) 13] X(foldl*(/\z./\u.X(z, u), Y, cons(Z, U)), foldl*(/\v./\w.X(v, w), Y, cons(Z, U))) >= X(Y, Z) because [14] and [16], by (Meta) 14] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= Y because [15], by (Select) 15] Y >= Y by (Meta) 16] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= Z because [17], by (Select) 17] cons(Z, U) >= Z because [18], by (Star) 18] cons*(Z, U) >= Z because [19], by (Select) 19] Z >= Z by (Meta) 20] foldl*(/\z./\u.X(z, u), Y, cons(Z, U)) >= U because [21], by (Select) 21] cons(Z, U) >= U because [4], by (Star) 22] sum(X) > foldl(/\x./\y.plus(x, y), _|_, X) because [23], by definition 23] sum*(X) >= foldl(/\x./\y.plus(x, y), _|_, X) because sum > foldl, [24], [31] and [32], by (Copy) 24] sum*(X) >= /\y./\z.plus(y, z) because [25], by (F-Abs) 25] sum*(X, x) >= /\z.plus(x, z) because [26], by (F-Abs) 26] sum*(X, x, y) >= plus(x, y) because sum > plus, [27] and [29], by (Copy) 27] sum*(X, x, y) >= x because [28], by (Select) 28] x >= x by (Var) 29] sum*(X, x, y) >= y because [30], by (Select) 30] y >= y by (Var) 31] sum*(X) >= _|_ by (Bot) 32] sum*(X) >= X because [33], by (Select) 33] X >= X by (Meta) We can thus remove the following rules: sum(X) => foldl(/\x./\y.plus(x, y), 0, 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.X(x, y), Y, cons(Z, U)) >? foldl(/\z./\u.X(z, u), X(Y, Z), U) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_2, x_1) We choose Lex = {foldl} and Mul = {cons}, and the following precedence: cons > foldl Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: foldl(/\x./\y.X(x, y), Y, cons(Z, U)) > foldl(/\x./\y.X(x, y), X(Y, Z), U) With these choices, we have: 1] foldl(/\x./\y.X(x, y), Y, cons(Z, U)) > foldl(/\x./\y.X(x, y), X(Y, Z), U) because [2], by definition 2] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) because [3], [6], [14] and [22], by (Stat) 3] cons(Z, U) > U because [4], by definition 4] cons*(Z, U) >= U because [5], by (Select) 5] U >= U by (Meta) 6] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= /\x./\y.X(x, y) because [7], by (F-Abs) 7] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z) >= /\x.X(z, x) because [8], by (F-Abs) 8] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u) >= X(z, u) because [9], by (Select) 9] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u), foldl*(/\v./\w.X(v, w), Y, cons(Z, U), z, u)) >= X(z, u) because [10] and [12], by (Meta) 10] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u) >= z because [11], by (Select) 11] z >= z by (Var) 12] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u) >= u because [13], by (Select) 13] u >= u by (Var) 14] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Y, Z) because [15], by (Select) 15] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U)), foldl*(/\v./\w.X(v, w), Y, cons(Z, U))) >= X(Y, Z) because [16] and [18], by (Meta) 16] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Y because [17], by (Select) 17] Y >= Y by (Meta) 18] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z because [19], by (Select) 19] cons(Z, U) >= Z because [20], by (Star) 20] cons*(Z, U) >= Z because [21], by (Select) 21] Z >= Z by (Meta) 22] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= U because [23], by (Select) 23] cons(Z, U) >= U because [4], by (Star) We can thus remove the following rules: foldl(/\x./\y.X(x, y), Y, cons(Z, U)) => foldl(/\z./\u.X(z, u), X(Y, Z), U) 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.