We consider the system 468. Alphabet: 0 : [] --> nat add : [nat * nat] --> nat rec : [nat -> nat -> nat * nat * nat] --> nat s : [nat] --> nat Rules: rec(/\x./\y.X(x, y), Y, 0) => Y rec(/\x./\y.X(x, y), Y, s(Z)) => X(Z, rec(/\z./\u.X(z, u), Y, Z)) add(X, Y) => rec(/\x./\y.s(y), X, Y) add(X, 0) => X add(X, s(Y)) => s(add(X, 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]): rec(/\x./\y.X(x, y), Y, 0) >? Y rec(/\x./\y.X(x, y), Y, s(Z)) >? X(Z, rec(/\z./\u.X(z, u), Y, Z)) add(X, Y) >? rec(/\x./\y.s(y), X, Y) add(X, 0) >? X add(X, s(Y)) >? s(add(X, Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[add(x_1, x_2)]] = add(x_2, x_1) We choose Lex = {add} and Mul = {0, rec, s}, and the following precedence: add > 0 > s > rec Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: rec(/\x./\y.X(x, y), Y, 0) > Y rec(/\x./\y.X(x, y), Y, s(Z)) > X(Z, rec(/\x./\y.X(x, y), Y, Z)) add(X, Y) > rec(/\x./\y.s(y), X, Y) add(X, 0) >= X add(X, s(Y)) >= s(add(X, Y)) With these choices, we have: 1] rec(/\x./\y.X(x, y), Y, 0) > Y because [2], by definition 2] rec*(/\x./\y.X(x, y), Y, 0) >= Y because [3], by (Select) 3] Y >= Y by (Meta) 4] rec(/\x./\y.X(x, y), Y, s(Z)) > X(Z, rec(/\x./\y.X(x, y), Y, Z)) because [5], by definition 5] rec*(/\x./\y.X(x, y), Y, s(Z)) >= X(Z, rec(/\x./\y.X(x, y), Y, Z)) because [6], by (Select) 6] X(rec*(/\x./\y.X(x, y), Y, s(Z)), rec*(/\z./\u.X(z, u), Y, s(Z))) >= X(Z, rec(/\x./\y.X(x, y), Y, Z)) because [7] and [11], by (Meta) 7] rec*(/\x./\y.X(x, y), Y, s(Z)) >= Z because [8], by (Select) 8] s(Z) >= Z because [9], by (Star) 9] s*(Z) >= Z because [10], by (Select) 10] Z >= Z by (Meta) 11] rec*(/\x./\y.X(x, y), Y, s(Z)) >= rec(/\x./\y.X(x, y), Y, Z) because rec in Mul, [12], [17] and [18], by (Stat) 12] /\x./\z.X(x, z) >= /\x./\z.X(x, z) because [13], by (Abs) 13] /\z.X(y, z) >= /\z.X(y, z) because [14], by (Abs) 14] X(y, x) >= X(y, x) because [15] and [16], by (Meta) 15] y >= y by (Var) 16] x >= x by (Var) 17] Y >= Y by (Meta) 18] s(Z) > Z because [19], by definition 19] s*(Z) >= Z because [10], by (Select) 20] add(X, Y) > rec(/\x./\y.s(y), X, Y) because [21], by definition 21] add*(X, Y) >= rec(/\x./\y.s(y), X, Y) because add > rec, [22], [27] and [29], by (Copy) 22] add*(X, Y) >= /\y./\z.s(z) because [23], by (F-Abs) 23] add*(X, Y, x) >= /\z.s(z) because [24], by (F-Abs) 24] add*(X, Y, x, y) >= s(y) because add > s and [25], by (Copy) 25] add*(X, Y, x, y) >= y because [26], by (Select) 26] y >= y by (Var) 27] add*(X, Y) >= X because [28], by (Select) 28] X >= X by (Meta) 29] add*(X, Y) >= Y because [30], by (Select) 30] Y >= Y by (Meta) 31] add(X, 0) >= X because [32], by (Star) 32] add*(X, 0) >= X because [33], by (Select) 33] X >= X by (Meta) 34] add(X, s(Y)) >= s(add(X, Y)) because [35], by (Star) 35] add*(X, s(Y)) >= s(add(X, Y)) because add > s and [36], by (Copy) 36] add*(X, s(Y)) >= add(X, Y) because [37], [40] and [42], by (Stat) 37] s(Y) > Y because [38], by definition 38] s*(Y) >= Y because [39], by (Select) 39] Y >= Y by (Meta) 40] add*(X, s(Y)) >= X because [41], by (Select) 41] X >= X by (Meta) 42] add*(X, s(Y)) >= Y because [43], by (Select) 43] s(Y) >= Y because [38], by (Star) We can thus remove the following rules: rec(/\x./\y.X(x, y), Y, 0) => Y rec(/\x./\y.X(x, y), Y, s(Z)) => X(Z, rec(/\z./\u.X(z, u), Y, Z)) add(X, Y) => rec(/\x./\y.s(y), X, 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]): add(X, 0) >? X add(X, s(Y)) >? s(add(X, Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {0, add, s}, and the following precedence: add > 0 > s With these choices, we have: 1] add(X, 0) > X because [2], by definition 2] add*(X, 0) >= X because [3], by (Select) 3] X >= X by (Meta) 4] add(X, s(Y)) >= s(add(X, Y)) because [5], by (Star) 5] add*(X, s(Y)) >= s(add(X, Y)) because add > s and [6], by (Copy) 6] add*(X, s(Y)) >= add(X, Y) because add in Mul, [7] and [8], by (Stat) 7] X >= X by (Meta) 8] s(Y) > Y because [9], by definition 9] s*(Y) >= Y because [10], by (Select) 10] Y >= Y by (Meta) We can thus remove the following rules: add(X, 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]): add(X, s(Y)) >? s(add(X, Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {add, s}, and the following precedence: add > s With these choices, we have: 1] add(X, s(Y)) > s(add(X, Y)) because [2], by definition 2] add*(X, s(Y)) >= s(add(X, Y)) because add > s and [3], by (Copy) 3] add*(X, s(Y)) >= add(X, Y) because add in Mul, [4] and [5], by (Stat) 4] X >= X by (Meta) 5] s(Y) > Y because [6], by definition 6] s*(Y) >= Y because [7], by (Select) 7] Y >= Y by (Meta) We can thus remove the following rules: add(X, s(Y)) => s(add(X, Y)) 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.