0.00/0.17 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 ack : [N * N] --> N 0.00/0.18 s : [N] --> N 0.00/0.18 z : [] --> N 0.00/0.18 0.00/0.18 Rules: 0.00/0.18 0.00/0.18 ack(z, x) => s(x) 0.00/0.18 ack(s(x), z) => ack(x, s(z)) 0.00/0.18 ack(s(x), s(y)) => ack(x, ack(s(x), y)) 0.00/0.18 0.00/0.18 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 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 ack(z, X) >? s(X) 0.00/0.18 ack(s(X), z) >? ack(X, s(z)) 0.00/0.18 ack(s(X), s(Y)) >? ack(X, ack(s(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 Argument functions: 0.00/0.18 0.00/0.18 [[z]] = _|_ 0.00/0.18 0.00/0.18 We choose Lex = {ack} and Mul = {s}, and the following precedence: ack > s 0.00/0.18 0.00/0.18 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.00/0.18 0.00/0.18 ack(_|_, X) > s(X) 0.00/0.18 ack(s(X), _|_) >= ack(X, s(_|_)) 0.00/0.18 ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) 0.00/0.18 0.00/0.18 With these choices, we have: 0.00/0.18 0.00/0.18 1] ack(_|_, X) > s(X) because [2], by definition 0.00/0.18 2] ack*(_|_, X) >= s(X) because ack > s and [3], by (Copy) 0.00/0.18 3] ack*(_|_, X) >= X because [4], by (Select) 0.00/0.18 4] X >= X by (Meta) 0.00/0.18 0.00/0.18 5] ack(s(X), _|_) >= ack(X, s(_|_)) because [6], by (Star) 0.00/0.18 6] ack*(s(X), _|_) >= ack(X, s(_|_)) because [7], [10] and [12], by (Stat) 0.00/0.18 7] s(X) > X because [8], by definition 0.00/0.18 8] s*(X) >= X because [9], by (Select) 0.00/0.18 9] X >= X by (Meta) 0.00/0.18 10] ack*(s(X), _|_) >= X because [11], by (Select) 0.00/0.18 11] s(X) >= X because [8], by (Star) 0.00/0.18 12] ack*(s(X), _|_) >= s(_|_) because ack > s and [13], by (Copy) 0.00/0.18 13] ack*(s(X), _|_) >= _|_ by (Bot) 0.00/0.18 0.00/0.18 14] ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [15], by (Star) 0.00/0.18 15] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [16], [19] and [21], by (Stat) 0.00/0.18 16] s(X) > X because [17], by definition 0.00/0.18 17] s*(X) >= X because [18], by (Select) 0.00/0.18 18] X >= X by (Meta) 0.00/0.18 19] ack*(s(X), s(Y)) >= X because [20], by (Select) 0.00/0.18 20] s(X) >= X because [17], by (Star) 0.00/0.18 21] ack*(s(X), s(Y)) >= ack(s(X), Y) because [22], [24], [27] and [28], by (Stat) 0.00/0.18 22] s(X) >= s(X) because s in Mul and [23], by (Fun) 0.00/0.18 23] X >= X by (Meta) 0.00/0.18 24] s(Y) > Y because [25], by definition 0.00/0.18 25] s*(Y) >= Y because [26], by (Select) 0.00/0.18 26] Y >= Y by (Meta) 0.00/0.18 27] ack*(s(X), s(Y)) >= s(X) because ack > s and [19], by (Copy) 0.00/0.18 28] ack*(s(X), s(Y)) >= Y because [29], by (Select) 0.00/0.18 29] s(Y) >= Y because [25], by (Star) 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 ack(z, X) => s(X) 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 ack(s(X), z) >? ack(X, s(z)) 0.00/0.18 ack(s(X), s(Y)) >? ack(X, ack(s(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 Argument functions: 0.00/0.18 0.00/0.18 [[z]] = _|_ 0.00/0.18 0.00/0.18 We choose Lex = {ack} and Mul = {s}, and the following precedence: ack > s 0.00/0.18 0.00/0.18 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.00/0.18 0.00/0.18 ack(s(X), _|_) > ack(X, s(_|_)) 0.00/0.18 ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) 0.00/0.18 0.00/0.18 With these choices, we have: 0.00/0.18 0.00/0.18 1] ack(s(X), _|_) > ack(X, s(_|_)) because [2], by definition 0.00/0.18 2] ack*(s(X), _|_) >= ack(X, s(_|_)) because [3], [6] and [8], by (Stat) 0.00/0.18 3] s(X) > X because [4], by definition 0.00/0.18 4] s*(X) >= X because [5], by (Select) 0.00/0.18 5] X >= X by (Meta) 0.00/0.18 6] ack*(s(X), _|_) >= X because [7], by (Select) 0.00/0.18 7] s(X) >= X because [4], by (Star) 0.00/0.18 8] ack*(s(X), _|_) >= s(_|_) because ack > s and [9], by (Copy) 0.00/0.18 9] ack*(s(X), _|_) >= _|_ by (Bot) 0.00/0.18 0.00/0.18 10] ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [11], by (Star) 0.00/0.18 11] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [12], [15] and [17], by (Stat) 0.00/0.18 12] s(X) > X because [13], by definition 0.00/0.18 13] s*(X) >= X because [14], by (Select) 0.00/0.18 14] X >= X by (Meta) 0.00/0.18 15] ack*(s(X), s(Y)) >= X because [16], by (Select) 0.00/0.18 16] s(X) >= X because [13], by (Star) 0.00/0.18 17] ack*(s(X), s(Y)) >= ack(s(X), Y) because [18], [20], [23] and [24], by (Stat) 0.00/0.18 18] s(X) >= s(X) because s in Mul and [19], by (Fun) 0.00/0.18 19] X >= X by (Meta) 0.00/0.18 20] s(Y) > Y because [21], by definition 0.00/0.18 21] s*(Y) >= Y because [22], by (Select) 0.00/0.18 22] Y >= Y by (Meta) 0.00/0.18 23] ack*(s(X), s(Y)) >= s(X) because ack > s and [15], by (Copy) 0.00/0.18 24] ack*(s(X), s(Y)) >= Y because [25], by (Select) 0.00/0.18 25] s(Y) >= Y because [21], by (Star) 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 ack(s(X), z) => ack(X, s(z)) 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 ack(s(X), s(Y)) >? ack(X, ack(s(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 = {ack} and Mul = {s}, and the following precedence: ack > s 0.00/0.18 0.00/0.18 With these choices, we have: 0.00/0.18 0.00/0.18 1] ack(s(X), s(Y)) > ack(X, ack(s(X), Y)) because [2], by definition 0.00/0.18 2] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [3], [6] and [8], by (Stat) 0.00/0.18 3] s(X) > X because [4], by definition 0.00/0.18 4] s*(X) >= X because [5], by (Select) 0.00/0.18 5] X >= X by (Meta) 0.00/0.18 6] ack*(s(X), s(Y)) >= X because [7], by (Select) 0.00/0.18 7] s(X) >= X because [4], by (Star) 0.00/0.18 8] ack*(s(X), s(Y)) >= ack(s(X), Y) because [9], [11], [14] and [15], by (Stat) 0.00/0.18 9] s(X) >= s(X) because s in Mul and [10], by (Fun) 0.00/0.18 10] X >= X by (Meta) 0.00/0.18 11] s(Y) > Y because [12], by definition 0.00/0.18 12] s*(Y) >= Y because [13], by (Select) 0.00/0.18 13] Y >= Y by (Meta) 0.00/0.18 14] ack*(s(X), s(Y)) >= s(X) because ack > s and [6], by (Copy) 0.00/0.18 15] ack*(s(X), s(Y)) >= Y because [16], by (Select) 0.00/0.18 16] s(Y) >= Y because [12], by (Star) 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 ack(s(X), s(Y)) => ack(X, ack(s(X), 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 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.18 EOF