0.00/0.22 YES 0.00/0.24 We consider the system theBenchmark. 0.00/0.24 0.00/0.24 Alphabet: 0.00/0.24 0.00/0.24 0 : [] --> nat 0.00/0.24 add : [] --> nat -> nat -> nat 0.00/0.24 rec : [] --> (nat -> nat -> nat) -> nat -> nat -> nat 0.00/0.24 s : [] --> nat -> nat 0.00/0.24 0.00/0.24 Rules: 0.00/0.24 0.00/0.24 rec (/\x./\y.f x y) z 0 => z 0.00/0.24 rec (/\x./\y.f x y) z (s u) => f u (rec (/\v./\w.f v w) z u) 0.00/0.24 add x y => rec (/\z./\u.s u) x y 0.00/0.24 add x 0 => x 0.00/0.24 add x (s y) => s (add x y) 0.00/0.24 0.00/0.24 Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. 0.00/0.24 0.00/0.24 We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: 0.00/0.24 0.00/0.24 Alphabet: 0.00/0.24 0.00/0.24 0 : [] --> nat 0.00/0.24 add : [nat * nat] --> nat 0.00/0.24 rec : [nat -> nat -> nat * nat * nat] --> nat 0.00/0.24 s : [nat] --> nat 0.00/0.24 ~AP1 : [nat -> nat -> nat * nat] --> nat -> nat 0.00/0.24 0.00/0.24 Rules: 0.00/0.24 0.00/0.24 rec(/\x./\y.~AP1(F, x) y, X, 0) => X 0.00/0.24 rec(/\x./\y.~AP1(F, x) y, X, s(Y)) => ~AP1(F, Y) rec(/\z./\u.~AP1(F, z) u, X, Y) 0.00/0.24 add(X, Y) => rec(/\x./\y.s(y), X, Y) 0.00/0.24 add(X, 0) => X 0.00/0.24 add(X, s(Y)) => s(add(X, Y)) 0.00/0.24 rec(/\x./\y.add(x, y), X, 0) => X 0.00/0.24 rec(/\x./\y.rec(F, x, y), X, 0) => X 0.00/0.24 rec(/\x./\y.add(x, y), X, s(Y)) => add(Y, rec(/\z./\u.add(z, u), X, Y)) 0.00/0.24 rec(/\x./\y.rec(F, x, y), X, s(Y)) => rec(F, Y, rec(/\z./\u.rec(F, z, u), X, Y)) 0.00/0.24 ~AP1(F, X) => F X 0.00/0.24 0.00/0.24 Symbol ~AP1 is an encoding for application that is only used in innocuous ways. We can simplify the program (without losing non-termination) by removing it. Additionally, we can remove some (now-)redundant rules. This gives: 0.00/0.24 0.00/0.24 Alphabet: 0.00/0.24 0.00/0.24 0 : [] --> nat 0.00/0.24 add : [nat * nat] --> nat 0.00/0.24 rec : [nat -> nat -> nat * nat * nat] --> nat 0.00/0.24 s : [nat] --> nat 0.00/0.24 0.00/0.24 Rules: 0.00/0.24 0.00/0.24 rec(/\x./\y.X(x, y), Y, 0) => Y 0.00/0.24 rec(/\x./\y.X(x, y), Y, s(Z)) => X(Z, rec(/\z./\u.X(z, u), Y, Z)) 0.00/0.24 add(X, Y) => rec(/\x./\y.s(y), X, Y) 0.00/0.24 add(X, 0) => X 0.00/0.24 add(X, s(Y)) => s(add(X, Y)) 0.00/0.24 0.00/0.24 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.24 0.00/0.24 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.24 0.00/0.24 rec(/\x./\y.X(x, y), Y, 0) >? Y 0.00/0.24 rec(/\x./\y.X(x, y), Y, s(Z)) >? X(Z, rec(/\z./\u.X(z, u), Y, Z)) 0.00/0.24 add(X, Y) >? rec(/\x./\y.s(y), X, Y) 0.00/0.24 add(X, 0) >? X 0.00/0.24 add(X, s(Y)) >? s(add(X, Y)) 0.00/0.24 0.00/0.24 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.24 0.00/0.24 We choose Lex = {} and Mul = {0, add, rec, s}, and the following precedence: 0 > add > rec > s 0.00/0.24 0.00/0.24 With these choices, we have: 0.00/0.24 0.00/0.24 1] rec(/\x./\y.X(x, y), Y, 0) >= Y because [2], by (Star) 0.00/0.24 2] rec*(/\x./\y.X(x, y), Y, 0) >= Y because [3], by (Select) 0.00/0.24 3] Y >= Y by (Meta) 0.00/0.24 0.00/0.24 4] rec(/\x./\y.X(x, y), Y, s(Z)) >= X(Z, rec(/\x./\y.X(x, y), Y, Z)) because [5], by (Star) 0.00/0.24 5] rec*(/\x./\y.X(x, y), Y, s(Z)) >= X(Z, rec(/\x./\y.X(x, y), Y, Z)) because [6], by (Select) 0.00/0.24 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) 0.00/0.24 7] rec*(/\x./\y.X(x, y), Y, s(Z)) >= Z because [8], by (Select) 0.00/0.24 8] s(Z) >= Z because [9], by (Star) 0.00/0.24 9] s*(Z) >= Z because [10], by (Select) 0.00/0.24 10] Z >= Z by (Meta) 0.00/0.24 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) 0.00/0.24 12] /\x./\z.X(x, z) >= /\x./\z.X(x, z) because [13], by (Abs) 0.00/0.24 13] /\z.X(y, z) >= /\z.X(y, z) because [14], by (Abs) 0.00/0.24 14] X(y, x) >= X(y, x) because [15] and [16], by (Meta) 0.00/0.24 15] y >= y by (Var) 0.00/0.24 16] x >= x by (Var) 0.00/0.24 17] Y >= Y by (Meta) 0.00/0.24 18] s(Z) > Z because [19], by definition 0.00/0.24 19] s*(Z) >= Z because [10], by (Select) 0.00/0.24 0.00/0.24 20] add(X, Y) >= rec(/\x./\y.s(y), X, Y) because [21], by (Star) 0.00/0.24 21] add*(X, Y) >= rec(/\x./\y.s(y), X, Y) because add > rec, [22], [27] and [29], by (Copy) 0.00/0.24 22] add*(X, Y) >= /\y./\z.s(z) because [23], by (F-Abs) 0.00/0.24 23] add*(X, Y, x) >= /\z.s(z) because [24], by (F-Abs) 0.00/0.24 24] add*(X, Y, x, y) >= s(y) because add > s and [25], by (Copy) 0.00/0.24 25] add*(X, Y, x, y) >= y because [26], by (Select) 0.00/0.24 26] y >= y by (Var) 0.00/0.24 27] add*(X, Y) >= X because [28], by (Select) 0.00/0.24 28] X >= X by (Meta) 0.00/0.24 29] add*(X, Y) >= Y because [30], by (Select) 0.00/0.24 30] Y >= Y by (Meta) 0.00/0.24 0.00/0.24 31] add(X, 0) > X because [32], by definition 0.00/0.24 32] add*(X, 0) >= X because [33], by (Select) 0.00/0.24 33] X >= X by (Meta) 0.00/0.24 0.00/0.24 34] add(X, s(Y)) > s(add(X, Y)) because [35], by definition 0.00/0.24 35] add*(X, s(Y)) >= s(add(X, Y)) because add > s and [36], by (Copy) 0.00/0.24 36] add*(X, s(Y)) >= add(X, Y) because add in Mul, [37] and [38], by (Stat) 0.00/0.24 37] X >= X by (Meta) 0.00/0.24 38] s(Y) > Y because [39], by definition 0.00/0.24 39] s*(Y) >= Y because [40], by (Select) 0.00/0.24 40] Y >= Y by (Meta) 0.00/0.24 0.00/0.24 We can thus remove the following rules: 0.00/0.24 0.00/0.24 add(X, 0) => X 0.00/0.24 add(X, s(Y)) => s(add(X, Y)) 0.00/0.24 0.00/0.24 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.24 0.00/0.24 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.24 0.00/0.24 rec(/\x./\y.X(x, y), Y, 0) >? Y 0.00/0.24 rec(/\x./\y.X(x, y), Y, s(Z)) >? X(Z, rec(/\z./\u.X(z, u), Y, Z)) 0.00/0.24 add(X, Y) >? rec(/\x./\y.s(y), X, Y) 0.00/0.24 0.00/0.24 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.24 0.00/0.24 We choose Lex = {} and Mul = {0, add, rec, s}, and the following precedence: add > rec > s > 0 0.00/0.24 0.00/0.24 With these choices, we have: 0.00/0.24 0.00/0.24 1] rec(/\x./\y.X(x, y), Y, 0) > Y because [2], by definition 0.00/0.24 2] rec*(/\x./\y.X(x, y), Y, 0) >= Y because [3], by (Select) 0.00/0.24 3] Y >= Y by (Meta) 0.00/0.24 0.00/0.24 4] rec(/\x./\y.X(x, y), Y, s(Z)) > X(Z, rec(/\x./\y.X(x, y), Y, Z)) because [5], by definition 0.00/0.24 5] rec*(/\x./\y.X(x, y), Y, s(Z)) >= X(Z, rec(/\x./\y.X(x, y), Y, Z)) because [6], by (Select) 0.00/0.24 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) 0.00/0.24 7] rec*(/\x./\y.X(x, y), Y, s(Z)) >= Z because [8], by (Select) 0.00/0.24 8] s(Z) >= Z because [9], by (Star) 0.00/0.24 9] s*(Z) >= Z because [10], by (Select) 0.00/0.24 10] Z >= Z by (Meta) 0.00/0.24 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) 0.00/0.24 12] /\x./\z.X(x, z) >= /\x./\z.X(x, z) because [13], by (Abs) 0.00/0.24 13] /\z.X(y, z) >= /\z.X(y, z) because [14], by (Abs) 0.00/0.24 14] X(y, x) >= X(y, x) because [15] and [16], by (Meta) 0.00/0.24 15] y >= y by (Var) 0.00/0.24 16] x >= x by (Var) 0.00/0.24 17] Y >= Y by (Meta) 0.00/0.24 18] s(Z) > Z because [19], by definition 0.00/0.24 19] s*(Z) >= Z because [10], by (Select) 0.00/0.24 0.00/0.24 20] add(X, Y) > rec(/\x./\y.s(y), X, Y) because [21], by definition 0.00/0.24 21] add*(X, Y) >= rec(/\x./\y.s(y), X, Y) because add > rec, [22], [27] and [29], by (Copy) 0.00/0.24 22] add*(X, Y) >= /\y./\z.s(z) because [23], by (F-Abs) 0.00/0.24 23] add*(X, Y, x) >= /\z.s(z) because [24], by (F-Abs) 0.00/0.24 24] add*(X, Y, x, y) >= s(y) because add > s and [25], by (Copy) 0.00/0.24 25] add*(X, Y, x, y) >= y because [26], by (Select) 0.00/0.24 26] y >= y by (Var) 0.00/0.24 27] add*(X, Y) >= X because [28], by (Select) 0.00/0.24 28] X >= X by (Meta) 0.00/0.24 29] add*(X, Y) >= Y because [30], by (Select) 0.00/0.24 30] Y >= Y by (Meta) 0.00/0.24 0.00/0.24 We can thus remove the following rules: 0.00/0.24 0.00/0.24 rec(/\x./\y.X(x, y), Y, 0) => Y 0.00/0.25 rec(/\x./\y.X(x, y), Y, s(Z)) => X(Z, rec(/\z./\u.X(z, u), Y, Z)) 0.00/0.25 add(X, Y) => rec(/\x./\y.s(y), X, Y) 0.00/0.25 0.00/0.25 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. 0.00/0.25 0.00/0.25 0.00/0.25 +++ Citations +++ 0.00/0.25 0.00/0.25 [Kop11] C. Kop. Simplifying Algebraic Functional Systems. In Proceedings of CAI 2011, volume 6742 of LNCS. 201--215, Springer, 2011. 0.00/0.25 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.25 EOF