0.13/1.11 YES 0.13/1.11 We consider the system theBenchmark. 0.13/1.11 0.13/1.11 Alphabet: 0.13/1.11 0.13/1.11 0 : [] --> nat 0.13/1.11 rec : [nat * a * nat -> a -> a] --> a 0.13/1.11 s : [nat] --> nat 0.13/1.11 0.13/1.11 Rules: 0.13/1.11 0.13/1.11 rec(0, x, f) => x 0.13/1.11 rec(s(x), y, f) => f x rec(x, y, f) 0.13/1.11 0.13/1.11 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.13/1.11 0.13/1.11 We use rule removal, following [Kop12, Theorem 2.23]. 0.13/1.11 0.13/1.11 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.13/1.11 0.13/1.11 rec(0, X, F) >? X 0.13/1.11 rec(s(X), Y, F) >? F X rec(X, Y, F) 0.13/1.11 0.13/1.11 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.13/1.11 0.13/1.11 We choose Lex = {} and Mul = {0, @_{o -> o -> o}, @_{o -> o}, rec, s}, and the following precedence: 0 > @_{o -> o -> o} = rec > @_{o -> o} > s 0.13/1.11 0.13/1.11 With these choices, we have: 0.13/1.11 0.13/1.11 1] rec(0, X, F) > X because [2], by definition 0.13/1.11 2] rec*(0, X, F) >= X because [3], by (Select) 0.13/1.11 3] X >= X by (Meta) 0.13/1.11 0.13/1.11 4] rec(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because [5], by (Star) 0.13/1.11 5] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because rec > @_{o -> o}, [6] and [11], by (Copy) 0.13/1.11 6] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X) because rec = @_{o -> o -> o}, rec in Mul, [7] and [10], by (Stat) 0.13/1.11 7] s(X) > X because [8], by definition 0.13/1.11 8] s*(X) >= X because [9], by (Select) 0.13/1.11 9] X >= X by (Meta) 0.13/1.11 10] F >= F by (Meta) 0.13/1.11 11] rec*(s(X), Y, F) >= rec(X, Y, F) because rec in Mul, [7], [12] and [10], by (Stat) 0.13/1.11 12] Y >= Y by (Meta) 0.13/1.11 0.13/1.11 We can thus remove the following rules: 0.13/1.11 0.13/1.11 rec(0, X, F) => X 0.13/1.11 0.13/1.11 We use rule removal, following [Kop12, Theorem 2.23]. 0.13/1.11 0.13/1.11 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.13/1.11 0.13/1.11 rec(s(X), Y, F) >? F X rec(X, Y, F) 0.13/1.11 0.13/1.11 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.13/1.11 0.13/1.11 We choose Lex = {} and Mul = {@_{o -> o -> o}, @_{o -> o}, rec, s}, and the following precedence: rec > @_{o -> o -> o} > @_{o -> o} > s 0.13/1.11 0.13/1.11 With these choices, we have: 0.13/1.11 0.13/1.11 1] rec(s(X), Y, F) > @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because [2], by definition 0.13/1.11 2] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because rec > @_{o -> o}, [3] and [10], by (Copy) 0.13/1.11 3] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X) because rec > @_{o -> o -> o}, [4] and [6], by (Copy) 0.13/1.11 4] rec*(s(X), Y, F) >= F because [5], by (Select) 0.13/1.11 5] F >= F by (Meta) 0.13/1.11 6] rec*(s(X), Y, F) >= X because [7], by (Select) 0.13/1.11 7] s(X) >= X because [8], by (Star) 0.13/1.11 8] s*(X) >= X because [9], by (Select) 0.13/1.11 9] X >= X by (Meta) 0.13/1.11 10] rec*(s(X), Y, F) >= rec(X, Y, F) because rec in Mul, [11], [13] and [14], by (Stat) 0.13/1.11 11] s(X) > X because [12], by definition 0.13/1.11 12] s*(X) >= X because [9], by (Select) 0.13/1.11 13] Y >= Y by (Meta) 0.13/1.11 14] F >= F by (Meta) 0.13/1.11 0.13/1.11 We can thus remove the following rules: 0.13/1.11 0.13/1.11 rec(s(X), Y, F) => F X rec(X, Y, F) 0.13/1.11 0.13/1.11 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. 0.13/1.11 0.13/1.11 0.13/1.11 +++ Citations +++ 0.13/1.11 0.13/1.11 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.13/1.11 EOF