0.00/0.33 YES 0.00/0.35 We consider the system theBenchmark. 0.00/0.35 0.00/0.35 Alphabet: 0.00/0.35 0.00/0.35 !facplus : [b * b] --> b 0.00/0.35 !factimes : [a * b] --> b 0.00/0.35 cons : [d * e] --> e 0.00/0.35 false : [] --> c 0.00/0.35 filter : [d -> c * e] --> e 0.00/0.35 filter2 : [c * d -> c * d * e] --> e 0.00/0.35 map : [d -> d * e] --> e 0.00/0.35 nil : [] --> e 0.00/0.35 true : [] --> c 0.00/0.35 0.00/0.35 Rules: 0.00/0.35 0.00/0.35 !factimes(x, !facplus(y, z)) => !facplus(!factimes(x, y), !factimes(x, z)) 0.00/0.35 map(f, nil) => nil 0.00/0.35 map(f, cons(x, y)) => cons(f x, map(f, y)) 0.00/0.35 filter(f, nil) => nil 0.00/0.35 filter(f, cons(x, y)) => filter2(f x, f, x, y) 0.00/0.35 filter2(true, f, x, y) => cons(x, filter(f, y)) 0.00/0.35 filter2(false, f, x, y) => filter(f, y) 0.00/0.35 0.00/0.35 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.35 0.00/0.35 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.35 0.00/0.35 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.35 0.00/0.35 !factimes(X, !facplus(Y, Z)) >? !facplus(!factimes(X, Y), !factimes(X, Z)) 0.00/0.35 map(F, nil) >? nil 0.00/0.35 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.00/0.35 filter(F, nil) >? nil 0.00/0.35 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 0.00/0.35 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 0.00/0.35 filter2(false, F, X, Y) >? filter(F, Y) 0.00/0.35 0.00/0.35 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.35 0.00/0.35 Argument functions: 0.00/0.35 0.00/0.35 [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) 0.00/0.35 [[nil]] = _|_ 0.00/0.35 0.00/0.35 We choose Lex = {@_{o -> o}, filter, filter2} and Mul = {!facplus, !factimes, cons, false, map, true}, and the following precedence: map > false > @_{o -> o} = filter = filter2 > !factimes > !facplus > cons > true 0.00/0.35 0.00/0.35 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.00/0.35 0.00/0.35 !factimes(X, !facplus(Y, Z)) > !facplus(!factimes(X, Y), !factimes(X, Z)) 0.00/0.35 map(F, _|_) >= _|_ 0.00/0.35 map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 0.00/0.35 filter(F, _|_) >= _|_ 0.00/0.35 filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) 0.00/0.35 filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 0.00/0.35 filter2(false, F, X, Y) > filter(F, Y) 0.00/0.35 0.00/0.35 With these choices, we have: 0.00/0.35 0.00/0.35 1] !factimes(X, !facplus(Y, Z)) > !facplus(!factimes(X, Y), !factimes(X, Z)) because [2], by definition 0.00/0.35 2] !factimes*(X, !facplus(Y, Z)) >= !facplus(!factimes(X, Y), !factimes(X, Z)) because !factimes > !facplus, [3] and [8], by (Copy) 0.00/0.35 3] !factimes*(X, !facplus(Y, Z)) >= !factimes(X, Y) because !factimes in Mul, [4] and [5], by (Stat) 0.00/0.35 4] X >= X by (Meta) 0.00/0.35 5] !facplus(Y, Z) > Y because [6], by definition 0.00/0.35 6] !facplus*(Y, Z) >= Y because [7], by (Select) 0.00/0.35 7] Y >= Y by (Meta) 0.00/0.35 8] !factimes*(X, !facplus(Y, Z)) >= !factimes(X, Z) because !factimes in Mul, [4] and [9], by (Stat) 0.00/0.35 9] !facplus(Y, Z) > Z because [10], by definition 0.00/0.35 10] !facplus*(Y, Z) >= Z because [11], by (Select) 0.00/0.35 11] Z >= Z by (Meta) 0.00/0.35 0.00/0.35 12] map(F, _|_) >= _|_ by (Bot) 0.00/0.35 0.00/0.35 13] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [14], by (Star) 0.00/0.35 14] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [15] and [22], by (Copy) 0.00/0.35 15] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [16] and [18], by (Copy) 0.00/0.35 16] map*(F, cons(X, Y)) >= F because [17], by (Select) 0.00/0.35 17] F >= F by (Meta) 0.00/0.35 18] map*(F, cons(X, Y)) >= X because [19], by (Select) 0.00/0.35 19] cons(X, Y) >= X because [20], by (Star) 0.00/0.35 20] cons*(X, Y) >= X because [21], by (Select) 0.00/0.35 21] X >= X by (Meta) 0.00/0.35 22] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [23] and [24], by (Stat) 0.00/0.35 23] F >= F by (Meta) 0.00/0.35 24] cons(X, Y) > Y because [25], by definition 0.00/0.35 25] cons*(X, Y) >= Y because [26], by (Select) 0.00/0.35 26] Y >= Y by (Meta) 0.00/0.35 0.00/0.35 27] filter(F, _|_) >= _|_ by (Bot) 0.00/0.35 0.00/0.35 28] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [29], by (Star) 0.00/0.35 29] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [30], [31], [34], [38], [39] and [41], by (Stat) 0.00/0.35 30] F >= F by (Meta) 0.00/0.35 31] cons(X, Y) > Y because [32], by definition 0.00/0.35 32] cons*(X, Y) >= Y because [33], by (Select) 0.00/0.35 33] Y >= Y by (Meta) 0.00/0.35 34] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter = @_{o -> o}, [30], [35], [38] and [39], by (Stat) 0.00/0.35 35] cons(X, Y) > X because [36], by definition 0.00/0.35 36] cons*(X, Y) >= X because [37], by (Select) 0.00/0.35 37] X >= X by (Meta) 0.00/0.35 38] filter*(F, cons(X, Y)) >= F because [30], by (Select) 0.00/0.35 39] filter*(F, cons(X, Y)) >= X because [40], by (Select) 0.00/0.35 40] cons(X, Y) >= X because [36], by (Star) 0.00/0.35 41] filter*(F, cons(X, Y)) >= Y because [42], by (Select) 0.00/0.35 42] cons(X, Y) >= Y because [32], by (Star) 0.00/0.35 0.00/0.35 43] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [44], by (Star) 0.00/0.35 44] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [45] and [47], by (Copy) 0.00/0.35 45] filter2*(true, F, X, Y) >= X because [46], by (Select) 0.00/0.35 46] X >= X by (Meta) 0.00/0.35 47] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [48], [49], [50] and [51], by (Stat) 0.00/0.35 48] F >= F by (Meta) 0.00/0.35 49] Y >= Y by (Meta) 0.00/0.35 50] filter2*(true, F, X, Y) >= F because [48], by (Select) 0.00/0.35 51] filter2*(true, F, X, Y) >= Y because [49], by (Select) 0.00/0.35 0.00/0.35 52] filter2(false, F, X, Y) > filter(F, Y) because [53], by definition 0.00/0.35 53] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [54], [55], [56] and [57], by (Stat) 0.00/0.35 54] F >= F by (Meta) 0.00/0.35 55] Y >= Y by (Meta) 0.00/0.35 56] filter2*(false, F, X, Y) >= F because [54], by (Select) 0.00/0.35 57] filter2*(false, F, X, Y) >= Y because [55], by (Select) 0.00/0.35 0.00/0.35 We can thus remove the following rules: 0.00/0.35 0.00/0.35 !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) 0.00/0.35 filter2(false, F, X, Y) => filter(F, Y) 0.00/0.35 0.00/0.35 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.35 0.00/0.35 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.35 0.00/0.35 map(F, nil) >? nil 0.00/0.35 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.00/0.35 filter(F, nil) >? nil 0.00/0.35 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 0.00/0.35 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 0.00/0.35 0.00/0.35 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.35 0.00/0.35 The following interpretation satisfies the requirements: 0.00/0.35 0.00/0.35 cons = \y0y1.3 + y0 + y1 0.00/0.35 filter = \G0y1.3y1 + G0(y1) + 3y1G0(y1) 0.00/0.35 filter2 = \y0G1y2y3.1 + y0 + 2y2 + 3y3 + G1(y3) + 2G1(0) + 3y3G1(y3) 0.00/0.35 map = \G0y1.3y1 + G0(0) + 2y1G0(y1) + 2G0(y1) 0.00/0.35 nil = 2 0.00/0.35 true = 3 0.00/0.35 0.00/0.35 Using this interpretation, the requirements translate to: 0.00/0.35 0.00/0.35 [[map(_F0, nil)]] = 6 + F0(0) + 6F0(2) > 2 = [[nil]] 0.00/0.35 [[map(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + F0(0) + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 8F0(3 + x1 + x2) > 3 + x1 + 3x2 + F0(0) + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 0.00/0.35 [[filter(_F0, nil)]] = 6 + 7F0(2) > 2 = [[nil]] 0.00/0.35 [[filter(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 10F0(3 + x1 + x2) > 1 + 3x1 + 3x2 + F0(x1) + F0(x2) + 2F0(0) + 3x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 0.00/0.35 [[filter2(true, _F0, _x1, _x2)]] = 4 + 2x1 + 3x2 + F0(x2) + 2F0(0) + 3x2F0(x2) > 3 + x1 + 3x2 + F0(x2) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 0.00/0.35 0.00/0.35 We can thus remove the following rules: 0.00/0.35 0.00/0.35 map(F, nil) => nil 0.00/0.35 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 0.00/0.35 filter(F, nil) => nil 0.00/0.35 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 0.00/0.35 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 0.00/0.35 0.00/0.35 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. 0.00/0.35 0.00/0.35 0.00/0.35 +++ Citations +++ 0.00/0.35 0.00/0.35 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.35 EOF