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 cons : [c * d] --> d 0.00/0.18 f : [a] --> a 0.00/0.18 false : [] --> b 0.00/0.18 filter : [c -> b * d] --> d 0.00/0.18 filter2 : [b * c -> b * c * d] --> d 0.00/0.18 g : [a] --> a 0.00/0.18 map : [c -> c * d] --> d 0.00/0.18 nil : [] --> d 0.00/0.18 true : [] --> b 0.00/0.18 0.00/0.18 Rules: 0.00/0.18 0.00/0.18 f(f(x)) => g(f(x)) 0.00/0.18 g(g(x)) => f(x) 0.00/0.18 map(h, nil) => nil 0.00/0.18 map(h, cons(x, y)) => cons(h x, map(h, y)) 0.00/0.18 filter(h, nil) => nil 0.00/0.18 filter(h, cons(x, y)) => filter2(h x, h, x, y) 0.00/0.18 filter2(true, h, x, y) => cons(x, filter(h, y)) 0.00/0.18 filter2(false, h, x, y) => filter(h, 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 f(f(X)) >? g(f(X)) 0.00/0.18 g(g(X)) >? f(X) 0.00/0.18 map(F, nil) >? nil 0.00/0.18 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.00/0.18 filter(F, nil) >? nil 0.00/0.18 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 0.00/0.18 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 0.00/0.18 filter2(false, F, X, Y) >? filter(F, 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 cons = \y0y1.1 + y0 + y1 0.00/0.18 f = \y0.2y0 0.00/0.18 false = 3 0.00/0.18 filter = \G0y1.3y1 + G0(0) + 2y1G0(y1) 0.00/0.18 filter2 = \y0G1y2y3.y2 + 2y0 + 3y3 + G1(0) + 2y3G1(y3) 0.00/0.18 g = \y0.2y0 0.00/0.18 map = \G0y1.2 + 3y1 + G0(y1) + y1G0(y1) 0.00/0.18 nil = 0 0.00/0.18 true = 3 0.00/0.18 0.00/0.18 Using this interpretation, the requirements translate to: 0.00/0.18 0.00/0.18 [[f(f(_x0))]] = 4x0 >= 4x0 = [[g(f(_x0))]] 0.00/0.18 [[g(g(_x0))]] = 4x0 >= 2x0 = [[f(_x0)]] 0.00/0.18 [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 0.00/0.18 [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + F0(x2) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 0.00/0.18 [[filter(_F0, nil)]] = F0(0) >= 0 = [[nil]] 0.00/0.18 [[filter(_F0, cons(_x1, _x2))]] = 3 + 3x1 + 3x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 3x1 + 3x2 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 0.00/0.18 [[filter2(true, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 1 + x1 + 3x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 0.00/0.18 [[filter2(false, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 3x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 map(F, nil) => nil 0.00/0.18 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 0.00/0.18 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 0.00/0.18 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 0.00/0.18 filter2(false, F, X, Y) => filter(F, 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 f(f(X)) >? g(f(X)) 0.00/0.18 g(g(X)) >? f(X) 0.00/0.18 filter(F, nil) >? nil 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 f = \y0.2 + 3y0 0.00/0.18 filter = \G0y1.3 + 3y1 + G0(0) 0.00/0.18 g = \y0.2 + 2y0 0.00/0.18 nil = 0 0.00/0.18 0.00/0.18 Using this interpretation, the requirements translate to: 0.00/0.18 0.00/0.18 [[f(f(_x0))]] = 8 + 9x0 > 6 + 6x0 = [[g(f(_x0))]] 0.00/0.18 [[g(g(_x0))]] = 6 + 4x0 > 2 + 3x0 = [[f(_x0)]] 0.00/0.18 [[filter(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 f(f(X)) => g(f(X)) 0.00/0.18 g(g(X)) => f(X) 0.00/0.18 filter(F, nil) => nil 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