0.00/0.11 YES 0.00/0.12 We consider the system theBenchmark. 0.00/0.12 0.00/0.12 Alphabet: 0.00/0.12 0.00/0.12 append : [list * list] --> list 0.00/0.12 cons : [nat * list] --> list 0.00/0.12 map : [nat -> nat * list] --> list 0.00/0.12 mirror : [list] --> list 0.00/0.12 nil : [] --> list 0.00/0.12 reverse : [list] --> list 0.00/0.12 shuffle : [list] --> list 0.00/0.12 0.00/0.12 Rules: 0.00/0.12 0.00/0.12 append(nil, x) => x 0.00/0.12 append(cons(x, y), z) => cons(x, append(y, z)) 0.00/0.12 reverse(nil) => nil 0.00/0.12 shuffle(nil) => nil 0.00/0.12 shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 0.00/0.12 mirror(nil) => nil 0.00/0.12 mirror(cons(x, y)) => append(cons(x, mirror(y)), cons(x, nil)) 0.00/0.12 map(f, nil) => nil 0.00/0.12 map(f, cons(x, y)) => cons(f x, map(f, y)) 0.00/0.12 0.00/0.12 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.12 0.00/0.12 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.12 0.00/0.12 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.12 0.00/0.12 append(nil, X) >? X 0.00/0.12 append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 0.00/0.12 reverse(nil) >? nil 0.00/0.12 shuffle(nil) >? nil 0.00/0.12 shuffle(cons(X, Y)) >? cons(X, shuffle(reverse(Y))) 0.00/0.12 mirror(nil) >? nil 0.00/0.12 mirror(cons(X, Y)) >? append(cons(X, mirror(Y)), cons(X, nil)) 0.00/0.12 map(F, nil) >? nil 0.00/0.12 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.00/0.12 0.00/0.12 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.12 0.00/0.12 The following interpretation satisfies the requirements: 0.00/0.12 0.00/0.12 append = \y0y1.y0 + y1 0.00/0.12 cons = \y0y1.1 + y0 + y1 0.00/0.12 map = \G0y1.3y1 + 2G0(0) + y1G0(y1) 0.00/0.12 mirror = \y0.2y0 0.00/0.12 nil = 0 0.00/0.12 reverse = \y0.y0 0.00/0.12 shuffle = \y0.2y0 0.00/0.12 0.00/0.12 Using this interpretation, the requirements translate to: 0.00/0.12 0.00/0.12 [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 0.00/0.12 [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 0.00/0.12 [[reverse(nil)]] = 0 >= 0 = [[nil]] 0.00/0.12 [[shuffle(nil)]] = 0 >= 0 = [[nil]] 0.00/0.12 [[shuffle(cons(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + x0 + 2x1 = [[cons(_x0, shuffle(reverse(_x1)))]] 0.00/0.12 [[mirror(nil)]] = 0 >= 0 = [[nil]] 0.00/0.12 [[mirror(cons(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 0.00/0.12 [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 0.00/0.12 [[map(_F0, cons(_x1, _x2))]] = 3 + 3x1 + 3x2 + F0(1 + x1 + x2) + 2F0(0) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 1 + x1 + 3x2 + F0(x1) + 2F0(0) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 0.00/0.12 0.00/0.12 We can thus remove the following rules: 0.00/0.12 0.00/0.12 shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 0.00/0.12 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 0.00/0.12 0.00/0.12 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.12 0.00/0.12 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.12 0.00/0.12 append(nil, X) >? X 0.00/0.12 append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 0.00/0.12 reverse(nil) >? nil 0.00/0.12 shuffle(nil) >? nil 0.00/0.12 mirror(nil) >? nil 0.00/0.12 mirror(cons(X, Y)) >? append(cons(X, mirror(Y)), cons(X, nil)) 0.00/0.12 map(F, nil) >? nil 0.00/0.12 0.00/0.12 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.12 0.00/0.12 The following interpretation satisfies the requirements: 0.00/0.12 0.00/0.12 append = \y0y1.y0 + y1 0.00/0.12 cons = \y0y1.2 + y1 + 3y0 0.00/0.12 map = \G0y1.3 + 3y1 + G0(0) 0.00/0.12 mirror = \y0.2 + 3y0 0.00/0.12 nil = 0 0.00/0.12 reverse = \y0.3 + 3y0 0.00/0.12 shuffle = \y0.3 + 3y0 0.00/0.12 0.00/0.12 Using this interpretation, the requirements translate to: 0.00/0.12 0.00/0.12 [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 0.00/0.12 [[append(cons(_x0, _x1), _x2)]] = 2 + x1 + x2 + 3x0 >= 2 + x1 + x2 + 3x0 = [[cons(_x0, append(_x1, _x2))]] 0.00/0.12 [[reverse(nil)]] = 3 > 0 = [[nil]] 0.00/0.12 [[shuffle(nil)]] = 3 > 0 = [[nil]] 0.00/0.12 [[mirror(nil)]] = 2 > 0 = [[nil]] 0.00/0.12 [[mirror(cons(_x0, _x1))]] = 8 + 3x1 + 9x0 > 6 + 3x1 + 6x0 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] 0.00/0.12 [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 0.00/0.12 0.00/0.12 We can thus remove the following rules: 0.00/0.12 0.00/0.12 reverse(nil) => nil 0.00/0.12 shuffle(nil) => nil 0.00/0.12 mirror(nil) => nil 0.00/0.12 mirror(cons(X, Y)) => append(cons(X, mirror(Y)), cons(X, nil)) 0.00/0.12 map(F, nil) => nil 0.00/0.12 0.00/0.12 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.12 0.00/0.12 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.12 0.00/0.12 append(nil, X) >? X 0.00/0.12 append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 0.00/0.12 0.00/0.12 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.12 0.00/0.12 The following interpretation satisfies the requirements: 0.00/0.12 0.00/0.12 append = \y0y1.3 + y1 + 3y0 0.00/0.12 cons = \y0y1.3 + y0 + y1 0.00/0.12 nil = 3 0.00/0.12 0.00/0.12 Using this interpretation, the requirements translate to: 0.00/0.12 0.00/0.12 [[append(nil, _x0)]] = 12 + x0 > x0 = [[_x0]] 0.00/0.12 [[append(cons(_x0, _x1), _x2)]] = 12 + x2 + 3x0 + 3x1 > 6 + x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] 0.00/0.12 0.00/0.12 We can thus remove the following rules: 0.00/0.12 0.00/0.12 append(nil, X) => X 0.00/0.12 append(cons(X, Y), Z) => cons(X, append(Y, Z)) 0.00/0.12 0.00/0.12 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. 0.00/0.12 0.00/0.12 0.00/0.12 +++ Citations +++ 0.00/0.12 0.00/0.12 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.12 EOF