0.00/0.16 YES 0.00/0.18 We consider the system theBenchmark. 0.00/0.18 0.00/0.18 Alphabet: 0.00/0.18 0.00/0.18 0 : [] --> nat 0.00/0.18 plus : [] --> nat -> nat -> nat 0.00/0.18 rec : [] --> nat -> nat -> (nat -> nat -> nat) -> nat 0.00/0.18 s : [] --> nat -> nat 0.00/0.18 succ : [] --> nat -> nat -> nat 0.00/0.18 0.00/0.18 Rules: 0.00/0.18 0.00/0.18 rec 0 x (/\y./\z.f y z) => x 0.00/0.18 rec (s x) y (/\z./\u.f z u) => f x (rec x y (/\v./\w.f v w)) 0.00/0.18 succ x y => s y 0.00/0.18 plus x y => rec x y (/\z./\u.succ z u) 0.00/0.18 0.00/0.18 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. 0.00/0.18 0.00/0.18 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: 0.00/0.18 0.00/0.18 Alphabet: 0.00/0.18 0.00/0.18 0 : [] --> nat 0.00/0.18 plus : [nat * nat] --> nat 0.00/0.18 rec : [nat * nat * nat -> nat -> nat] --> nat 0.00/0.18 s : [nat] --> nat 0.00/0.18 succ : [nat * nat] --> nat 0.00/0.18 ~AP1 : [nat -> nat -> nat * nat] --> nat -> nat 0.00/0.18 0.00/0.18 Rules: 0.00/0.18 0.00/0.18 rec(0, X, /\x./\y.~AP1(F, x) y) => X 0.00/0.18 rec(s(X), Y, /\x./\y.~AP1(F, x) y) => ~AP1(F, X) rec(X, Y, /\z./\u.~AP1(F, z) u) 0.00/0.18 succ(X, Y) => s(Y) 0.00/0.18 plus(X, Y) => rec(X, Y, /\x./\y.succ(x, y)) 0.00/0.18 rec(0, X, /\x./\y.plus(x, y)) => X 0.00/0.18 rec(0, X, /\x./\y.succ(x, y)) => X 0.00/0.18 rec(s(X), Y, /\x./\y.plus(x, y)) => plus(X, rec(X, Y, /\z./\u.plus(z, u))) 0.00/0.18 rec(s(X), Y, /\x./\y.succ(x, y)) => succ(X, rec(X, Y, /\z./\u.succ(z, u))) 0.00/0.18 ~AP1(F, X) => F X 0.00/0.18 0.00/0.18 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: 0.00/0.18 0.00/0.18 Alphabet: 0.00/0.18 0.00/0.18 0 : [] --> nat 0.00/0.18 plus : [nat * nat] --> nat 0.00/0.18 rec : [nat * nat * nat -> nat -> nat] --> nat 0.00/0.18 s : [nat] --> nat 0.00/0.18 succ : [nat * nat] --> nat 0.00/0.18 0.00/0.18 Rules: 0.00/0.18 0.00/0.18 rec(0, X, /\x./\y.Y(x, y)) => X 0.00/0.18 rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 0.00/0.18 succ(X, Y) => s(Y) 0.00/0.18 plus(X, Y) => rec(X, Y, /\x./\y.succ(x, y)) 0.00/0.18 0.00/0.18 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.18 0.00/0.18 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.18 0.00/0.18 rec(0, X, /\x./\y.Y(x, y)) >? X 0.00/0.18 rec(s(X), Y, /\x./\y.Z(x, y)) >? Z(X, rec(X, Y, /\z./\u.Z(z, u))) 0.00/0.18 succ(X, Y) >? s(Y) 0.00/0.18 plus(X, Y) >? rec(X, Y, /\x./\y.succ(x, y)) 0.00/0.18 0.00/0.18 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.18 0.00/0.18 We choose Lex = {} and Mul = {0, plus, rec, s, succ}, and the following precedence: 0 > plus > rec > succ > s 0.00/0.18 0.00/0.18 With these choices, we have: 0.00/0.18 0.00/0.18 1] rec(0, X, /\x./\y.Y(x, y)) >= X because [2], by (Star) 0.00/0.18 2] rec*(0, X, /\x./\y.Y(x, y)) >= X because [3], by (Select) 0.00/0.18 3] X >= X by (Meta) 0.00/0.18 0.00/0.18 4] rec(s(X), Y, /\x./\y.Z(x, y)) > Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [5], by definition 0.00/0.18 5] rec*(s(X), Y, /\x./\y.Z(x, y)) >= Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [6], by (Select) 0.00/0.18 6] Z(rec*(s(X), Y, /\x./\y.Z(x, y)), rec*(s(X), Y, /\z./\u.Z(z, u))) >= Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [7] and [11], by (Meta) 0.00/0.18 7] rec*(s(X), Y, /\x./\y.Z(x, y)) >= X because [8], by (Select) 0.00/0.18 8] s(X) >= X because [9], by (Star) 0.00/0.18 9] s*(X) >= X because [10], by (Select) 0.00/0.18 10] X >= X by (Meta) 0.00/0.18 11] rec*(s(X), Y, /\x./\y.Z(x, y)) >= rec(X, Y, /\x./\y.Z(x, y)) because rec in Mul, [12], [14] and [15], by (Stat) 0.00/0.18 12] s(X) > X because [13], by definition 0.00/0.18 13] s*(X) >= X because [10], by (Select) 0.00/0.18 14] Y >= Y by (Meta) 0.00/0.18 15] /\x./\z.Z(x, z) >= /\x./\z.Z(x, z) because [16], by (Abs) 0.00/0.18 16] /\z.Z(y, z) >= /\z.Z(y, z) because [17], by (Abs) 0.00/0.18 17] Z(y, x) >= Z(y, x) because [18] and [19], by (Meta) 0.00/0.18 18] y >= y by (Var) 0.00/0.18 19] x >= x by (Var) 0.00/0.18 0.00/0.18 20] succ(X, Y) >= s(Y) because [21], by (Star) 0.00/0.18 21] succ*(X, Y) >= s(Y) because succ > s and [22], by (Copy) 0.00/0.18 22] succ*(X, Y) >= Y because [23], by (Select) 0.00/0.18 23] Y >= Y by (Meta) 0.00/0.18 0.00/0.18 24] plus(X, Y) >= rec(X, Y, /\x./\y.succ(x, y)) because [25], by (Star) 0.00/0.18 25] plus*(X, Y) >= rec(X, Y, /\x./\y.succ(x, y)) because plus > rec, [26], [28] and [30], by (Copy) 0.00/0.18 26] plus*(X, Y) >= X because [27], by (Select) 0.00/0.18 27] X >= X by (Meta) 0.00/0.18 28] plus*(X, Y) >= Y because [29], by (Select) 0.00/0.18 29] Y >= Y by (Meta) 0.00/0.18 30] plus*(X, Y) >= /\y./\z.succ(y, z) because [31], by (F-Abs) 0.00/0.18 31] plus*(X, Y, x) >= /\z.succ(x, z) because [32], by (F-Abs) 0.00/0.18 32] plus*(X, Y, x, y) >= succ(x, y) because plus > succ, [33] and [35], by (Copy) 0.00/0.18 33] plus*(X, Y, x, y) >= x because [34], by (Select) 0.00/0.18 34] x >= x by (Var) 0.00/0.18 35] plus*(X, Y, x, y) >= y because [36], by (Select) 0.00/0.18 36] y >= y by (Var) 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 0.00/0.18 0.00/0.18 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.18 0.00/0.18 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.18 0.00/0.18 rec(0, X, /\x./\y.Y(x, y)) >? X 0.00/0.18 succ(X, Y) >? s(Y) 0.00/0.18 plus(X, Y) >? rec(X, Y, /\x./\y.succ(x, y)) 0.00/0.18 0.00/0.18 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.18 0.00/0.18 The following interpretation satisfies the requirements: 0.00/0.18 0.00/0.18 0 = 3 0.00/0.18 plus = \y0y1.3 + 3y0 + 3y1 0.00/0.18 rec = \y0y1G2.y0 + y1 + G2(0,0) 0.00/0.18 s = \y0.y0 0.00/0.18 succ = \y0y1.y0 + y1 0.00/0.18 0.00/0.18 Using this interpretation, the requirements translate to: 0.00/0.18 0.00/0.18 [[rec(0, _x0, /\x./\y._x1(x, y))]] = 3 + x0 + F1(0,0) > x0 = [[_x0]] 0.00/0.18 [[succ(_x0, _x1)]] = x0 + x1 >= x1 = [[s(_x1)]] 0.00/0.18 [[plus(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[rec(_x0, _x1, /\x./\y.succ(x, y))]] 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 rec(0, X, /\x./\y.Y(x, y)) => X 0.00/0.18 plus(X, Y) => rec(X, Y, /\x./\y.succ(x, y)) 0.00/0.18 0.00/0.18 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.18 0.00/0.18 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.18 0.00/0.18 succ(X, Y) >? s(Y) 0.00/0.18 0.00/0.18 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.18 0.00/0.18 The following interpretation satisfies the requirements: 0.00/0.18 0.00/0.18 s = \y0.y0 0.00/0.18 succ = \y0y1.3 + y0 + 3y1 0.00/0.18 0.00/0.18 Using this interpretation, the requirements translate to: 0.00/0.18 0.00/0.18 [[succ(_x0, _x1)]] = 3 + x0 + 3x1 > x1 = [[s(_x1)]] 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 succ(X, Y) => s(Y) 0.00/0.18 0.00/0.18 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. 0.00/0.18 0.00/0.18 0.00/0.18 +++ Citations +++ 0.00/0.18 0.00/0.18 [Kop11] C. Kop. Simplifying Algebraic Functional Systems. In Proceedings of CAI 2011, volume 6742 of LNCS. 201--215, Springer, 2011. 0.00/0.18 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.18 EOF