0.00/0.08 YES 0.00/0.08 We consider the system theBenchmark. 0.00/0.08 0.00/0.08 Alphabet: 0.00/0.08 0.00/0.08 and : [c * c] --> c 0.00/0.08 cons : [a * b] --> b 0.00/0.08 false : [] --> c 0.00/0.08 forall : [a -> c * b] --> c 0.00/0.08 forsome : [a -> c * b] --> c 0.00/0.08 nil : [] --> b 0.00/0.08 or : [c * c] --> c 0.00/0.08 true : [] --> c 0.00/0.08 0.00/0.08 Rules: 0.00/0.08 0.00/0.08 and(true, true) => true 0.00/0.08 and(true, false) => false 0.00/0.08 and(false, true) => false 0.00/0.08 and(false, false) => false 0.00/0.08 or(true, true) => true 0.00/0.08 or(true, false) => true 0.00/0.08 or(false, true) => true 0.00/0.08 or(false, false) => false 0.00/0.08 forall(f, nil) => true 0.00/0.08 forall(f, cons(x, y)) => and(f x, forall(f, y)) 0.00/0.08 forsome(f, nil) => false 0.00/0.08 forsome(f, cons(x, y)) => or(f x, forsome(f, y)) 0.00/0.08 0.00/0.08 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.08 0.00/0.08 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.08 0.00/0.08 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.08 0.00/0.08 and(true, true) >? true 0.00/0.08 and(true, false) >? false 0.00/0.08 and(false, true) >? false 0.00/0.08 and(false, false) >? false 0.00/0.08 or(true, true) >? true 0.00/0.08 or(true, false) >? true 0.00/0.08 or(false, true) >? true 0.00/0.08 or(false, false) >? false 0.00/0.08 forall(F, nil) >? true 0.00/0.08 forall(F, cons(X, Y)) >? and(F X, forall(F, Y)) 0.00/0.08 forsome(F, nil) >? false 0.00/0.08 forsome(F, cons(X, Y)) >? or(F X, forsome(F, Y)) 0.00/0.08 0.00/0.08 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.08 0.00/0.08 The following interpretation satisfies the requirements: 0.00/0.08 0.00/0.08 and = \y0y1.2 + y0 + y1 0.00/0.08 cons = \y0y1.3 + 3y0 + 3y1 0.00/0.08 false = 0 0.00/0.08 forall = \G0y1.3y1 + G0(0) + G0(y1) + 3y1G0(y1) 0.00/0.08 forsome = \G0y1.3y1 + 3y1G0(y1) + 3G0(y1) 0.00/0.08 nil = 3 0.00/0.08 or = \y0y1.2 + y0 + 2y1 0.00/0.08 true = 0 0.00/0.08 0.00/0.08 Using this interpretation, the requirements translate to: 0.00/0.08 0.00/0.08 [[and(true, true)]] = 2 > 0 = [[true]] 0.00/0.08 [[and(true, false)]] = 2 > 0 = [[false]] 0.00/0.08 [[and(false, true)]] = 2 > 0 = [[false]] 0.00/0.08 [[and(false, false)]] = 2 > 0 = [[false]] 0.00/0.08 [[or(true, true)]] = 2 > 0 = [[true]] 0.00/0.08 [[or(true, false)]] = 2 > 0 = [[true]] 0.00/0.08 [[or(false, true)]] = 2 > 0 = [[true]] 0.00/0.08 [[or(false, false)]] = 2 > 0 = [[false]] 0.00/0.08 [[forall(_F0, nil)]] = 9 + F0(0) + 10F0(3) > 0 = [[true]] 0.00/0.08 [[forall(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + F0(0) + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 10F0(3 + 3x1 + 3x2) > 2 + x1 + 3x2 + F0(0) + F0(x1) + F0(x2) + 3x2F0(x2) = [[and(_F0 _x1, forall(_F0, _x2))]] 0.00/0.08 [[forsome(_F0, nil)]] = 9 + 12F0(3) > 0 = [[false]] 0.00/0.08 [[forsome(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 12F0(3 + 3x1 + 3x2) > 2 + x1 + 6x2 + F0(x1) + 6x2F0(x2) + 6F0(x2) = [[or(_F0 _x1, forsome(_F0, _x2))]] 0.00/0.08 0.00/0.08 We can thus remove the following rules: 0.00/0.08 0.00/0.08 and(true, true) => true 0.00/0.08 and(true, false) => false 0.00/0.08 and(false, true) => false 0.00/0.08 and(false, false) => false 0.00/0.08 or(true, true) => true 0.00/0.08 or(true, false) => true 0.00/0.08 or(false, true) => true 0.00/0.08 or(false, false) => false 0.00/0.08 forall(F, nil) => true 0.00/0.08 forall(F, cons(X, Y)) => and(F X, forall(F, Y)) 0.00/0.08 forsome(F, nil) => false 0.00/0.08 forsome(F, cons(X, Y)) => or(F X, forsome(F, Y)) 0.00/0.08 0.00/0.08 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. 0.00/0.08 0.00/0.08 0.00/0.08 +++ Citations +++ 0.00/0.08 0.00/0.08 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.08 EOF