0.00/0.16 YES 0.00/0.17 We consider the system theBenchmark. 0.00/0.17 0.00/0.17 Alphabet: 0.00/0.17 0.00/0.17 app : [list * list] --> list 0.00/0.17 cons : [nat * list] --> list 0.00/0.17 hshuffle : [nat -> nat * list] --> list 0.00/0.17 nil : [] --> list 0.00/0.17 reverse : [list] --> list 0.00/0.17 0.00/0.17 Rules: 0.00/0.17 0.00/0.17 app(nil, x) => x 0.00/0.17 app(cons(x, y), z) => cons(x, app(y, z)) 0.00/0.17 reverse(nil) => nil 0.00/0.17 reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) 0.00/0.17 hshuffle(f, nil) => nil 0.00/0.17 hshuffle(f, cons(x, y)) => cons(f x, hshuffle(f, reverse(y))) 0.00/0.17 0.00/0.17 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.17 0.00/0.17 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.17 0.00/0.17 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.17 0.00/0.17 app(nil, X) >? X 0.00/0.17 app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 0.00/0.17 reverse(nil) >? nil 0.00/0.17 reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 0.00/0.17 hshuffle(F, nil) >? nil 0.00/0.17 hshuffle(F, cons(X, Y)) >? cons(F X, hshuffle(F, reverse(Y))) 0.00/0.17 0.00/0.17 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.17 0.00/0.17 The following interpretation satisfies the requirements: 0.00/0.17 0.00/0.17 app = \y0y1.y0 + y1 0.00/0.17 cons = \y0y1.2 + y0 + y1 0.00/0.17 hshuffle = \G0y1.2y1 + G0(0) + y1G0(y1) 0.00/0.17 nil = 0 0.00/0.17 reverse = \y0.1 + y0 0.00/0.17 0.00/0.17 Using this interpretation, the requirements translate to: 0.00/0.17 0.00/0.17 [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 0.00/0.17 [[app(cons(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 0.00/0.17 [[reverse(nil)]] = 1 > 0 = [[nil]] 0.00/0.17 [[reverse(cons(_x0, _x1))]] = 3 + x0 + x1 >= 3 + x0 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 0.00/0.17 [[hshuffle(_F0, nil)]] = F0(0) >= 0 = [[nil]] 0.00/0.17 [[hshuffle(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 4 + x1 + 2x2 + F0(0) + F0(x1) + F0(1 + x2) + x2F0(1 + x2) = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 0.00/0.17 0.00/0.17 We can thus remove the following rules: 0.00/0.17 0.00/0.17 reverse(nil) => nil 0.00/0.17 0.00/0.17 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.17 0.00/0.17 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.17 0.00/0.17 app(nil, X) >? X 0.00/0.17 app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 0.00/0.17 reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 0.00/0.17 hshuffle(F, nil) >? nil 0.00/0.17 hshuffle(F, cons(X, Y)) >? cons(F X, hshuffle(F, reverse(Y))) 0.00/0.17 0.00/0.17 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.17 0.00/0.17 The following interpretation satisfies the requirements: 0.00/0.17 0.00/0.17 app = \y0y1.y0 + y1 0.00/0.17 cons = \y0y1.1 + y0 + y1 0.00/0.17 hshuffle = \G0y1.2y1 + G0(y1) + 2y1G0(y1) 0.00/0.17 nil = 0 0.00/0.17 reverse = \y0.y0 0.00/0.17 0.00/0.17 Using this interpretation, the requirements translate to: 0.00/0.17 0.00/0.17 [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 0.00/0.17 [[app(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 0.00/0.17 [[reverse(cons(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 0.00/0.17 [[hshuffle(_F0, nil)]] = F0(0) >= 0 = [[nil]] 0.00/0.17 [[hshuffle(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) > 1 + x1 + 2x2 + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 0.00/0.17 0.00/0.17 We can thus remove the following rules: 0.00/0.17 0.00/0.17 hshuffle(F, cons(X, Y)) => cons(F X, hshuffle(F, reverse(Y))) 0.00/0.17 0.00/0.17 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.17 0.00/0.17 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.17 0.00/0.17 app(nil, X) >? X 0.00/0.17 app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 0.00/0.17 reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 0.00/0.17 hshuffle(F, nil) >? nil 0.00/0.17 0.00/0.17 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.17 0.00/0.17 The following interpretation satisfies the requirements: 0.00/0.17 0.00/0.17 app = \y0y1.y0 + y1 0.00/0.17 cons = \y0y1.2 + y1 + 3y0 0.00/0.17 hshuffle = \G0y1.3 + 3y1 + G0(0) 0.00/0.17 nil = 0 0.00/0.17 reverse = \y0.2 + 3y0 0.00/0.17 0.00/0.17 Using this interpretation, the requirements translate to: 0.00/0.17 0.00/0.17 [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 0.00/0.17 [[app(cons(_x0, _x1), _x2)]] = 2 + x1 + x2 + 3x0 >= 2 + x1 + x2 + 3x0 = [[cons(_x0, app(_x1, _x2))]] 0.00/0.17 [[reverse(cons(_x0, _x1))]] = 8 + 3x1 + 9x0 > 4 + 3x0 + 3x1 = [[app(reverse(_x1), cons(_x0, nil))]] 0.00/0.17 [[hshuffle(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 0.00/0.17 0.00/0.17 We can thus remove the following rules: 0.00/0.17 0.00/0.17 reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 0.00/0.17 hshuffle(F, nil) => nil 0.00/0.17 0.00/0.17 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.17 0.00/0.17 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.17 0.00/0.17 app(nil, X) >? X 0.00/0.17 app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 0.00/0.17 0.00/0.17 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.17 0.00/0.17 The following interpretation satisfies the requirements: 0.00/0.17 0.00/0.17 app = \y0y1.3 + y1 + 3y0 0.00/0.17 cons = \y0y1.3 + y0 + y1 0.00/0.17 nil = 3 0.00/0.17 0.00/0.17 Using this interpretation, the requirements translate to: 0.00/0.17 0.00/0.17 [[app(nil, _x0)]] = 12 + x0 > x0 = [[_x0]] 0.00/0.17 [[app(cons(_x0, _x1), _x2)]] = 12 + x2 + 3x0 + 3x1 > 6 + x0 + x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] 0.00/0.17 0.00/0.17 We can thus remove the following rules: 0.00/0.17 0.00/0.17 app(nil, X) => X 0.00/0.17 app(cons(X, Y), Z) => cons(X, app(Y, Z)) 0.00/0.17 0.00/0.17 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. 0.00/0.17 0.00/0.17 0.00/0.17 +++ Citations +++ 0.00/0.17 0.00/0.17 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.17 EOF