0.85/0.88 YES 0.91/0.96 We consider the system theBenchmark. 0.91/0.96 0.91/0.96 Alphabet: 0.91/0.96 0.91/0.96 0 : [] --> a 0.91/0.96 ack : [a * a] --> a 0.91/0.96 cons : [c * d] --> d 0.91/0.96 false : [] --> b 0.91/0.96 filter : [c -> b * d] --> d 0.91/0.96 filter2 : [b * c -> b * c * d] --> d 0.91/0.96 map : [c -> c * d] --> d 0.91/0.96 nil : [] --> d 0.91/0.96 succ : [a] --> a 0.91/0.96 true : [] --> b 0.91/0.96 0.91/0.96 Rules: 0.91/0.96 0.91/0.96 ack(0, x) => succ(x) 0.91/0.96 ack(succ(x), y) => ack(x, succ(0)) 0.91/0.96 ack(succ(x), succ(y)) => ack(x, ack(succ(x), y)) 0.91/0.96 map(f, nil) => nil 0.91/0.96 map(f, cons(x, y)) => cons(f x, map(f, y)) 0.91/0.96 filter(f, nil) => nil 0.91/0.96 filter(f, cons(x, y)) => filter2(f x, f, x, y) 0.91/0.96 filter2(true, f, x, y) => cons(x, filter(f, y)) 0.91/0.96 filter2(false, f, x, y) => filter(f, y) 0.91/0.96 0.91/0.96 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.91/0.96 0.91/0.96 We use rule removal, following [Kop12, Theorem 2.23]. 0.91/0.96 0.91/0.96 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.91/0.96 0.91/0.96 ack(0, X) >? succ(X) 0.91/0.96 ack(succ(X), Y) >? ack(X, succ(0)) 0.91/0.96 ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, nil) >? nil 0.91/0.96 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.91/0.96 filter(F, nil) >? nil 0.91/0.96 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 0.91/0.96 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 0.91/0.96 filter2(false, F, X, Y) >? filter(F, Y) 0.91/0.96 0.91/0.96 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.91/0.96 0.91/0.96 Argument functions: 0.91/0.96 0.91/0.96 [[0]] = _|_ 0.91/0.96 [[filter(x_1, x_2)]] = filter(x_2, x_1) 0.91/0.96 [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_1, x_3) 0.91/0.96 [[nil]] = _|_ 0.91/0.96 0.91/0.96 We choose Lex = {ack, filter, filter2} and Mul = {@_{o -> o}, cons, false, map, succ, true}, and the following precedence: ack > false > filter = filter2 > map > @_{o -> o} > cons > succ > true 0.91/0.96 0.91/0.96 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.91/0.96 0.91/0.96 ack(_|_, X) >= succ(X) 0.91/0.96 ack(succ(X), Y) >= ack(X, succ(_|_)) 0.91/0.96 ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, _|_) >= _|_ 0.91/0.96 map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) 0.91/0.96 filter(F, _|_) >= _|_ 0.91/0.96 filter(F, cons(X, Y)) > filter2(@_{o -> o}(F, X), F, X, Y) 0.91/0.96 filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 0.91/0.96 filter2(false, F, X, Y) >= filter(F, Y) 0.91/0.96 0.91/0.96 With these choices, we have: 0.91/0.96 0.91/0.96 1] ack(_|_, X) >= succ(X) because [2], by (Star) 0.91/0.96 2] ack*(_|_, X) >= succ(X) because ack > succ and [3], by (Copy) 0.91/0.96 3] ack*(_|_, X) >= X because [4], by (Select) 0.91/0.96 4] X >= X by (Meta) 0.91/0.96 0.91/0.96 5] ack(succ(X), Y) >= ack(X, succ(_|_)) because [6], by (Star) 0.91/0.96 6] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [7], [10] and [12], by (Stat) 0.91/0.96 7] succ(X) > X because [8], by definition 0.91/0.96 8] succ*(X) >= X because [9], by (Select) 0.91/0.96 9] X >= X by (Meta) 0.91/0.96 10] ack*(succ(X), Y) >= X because [11], by (Select) 0.91/0.96 11] succ(X) >= X because [8], by (Star) 0.91/0.96 12] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [13], by (Copy) 0.91/0.96 13] ack*(succ(X), Y) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 14] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [15], by (Star) 0.91/0.96 15] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [16], [19] and [21], by (Stat) 0.91/0.96 16] succ(X) > X because [17], by definition 0.91/0.96 17] succ*(X) >= X because [18], by (Select) 0.91/0.96 18] X >= X by (Meta) 0.91/0.96 19] ack*(succ(X), succ(Y)) >= X because [20], by (Select) 0.91/0.96 20] succ(X) >= X because [17], by (Star) 0.91/0.96 21] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [22], [24], [27] and [28], by (Stat) 0.91/0.96 22] succ(X) >= succ(X) because succ in Mul and [23], by (Fun) 0.91/0.96 23] X >= X by (Meta) 0.91/0.96 24] succ(Y) > Y because [25], by definition 0.91/0.96 25] succ*(Y) >= Y because [26], by (Select) 0.91/0.96 26] Y >= Y by (Meta) 0.91/0.96 27] ack*(succ(X), succ(Y)) >= succ(X) because [22], by (Select) 0.91/0.96 28] ack*(succ(X), succ(Y)) >= Y because [29], by (Select) 0.91/0.96 29] succ(Y) >= Y because [25], by (Star) 0.91/0.96 0.91/0.96 30] map(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 31] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [32], by (Star) 0.91/0.96 32] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [33] and [40], by (Copy) 0.91/0.96 33] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [34] and [36], by (Copy) 0.91/0.96 34] map*(F, cons(X, Y)) >= F because [35], by (Select) 0.91/0.96 35] F >= F by (Meta) 0.91/0.96 36] map*(F, cons(X, Y)) >= X because [37], by (Select) 0.91/0.96 37] cons(X, Y) >= X because [38], by (Star) 0.91/0.96 38] cons*(X, Y) >= X because [39], by (Select) 0.91/0.96 39] X >= X by (Meta) 0.91/0.96 40] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [41] and [42], by (Stat) 0.91/0.96 41] F >= F by (Meta) 0.91/0.96 42] cons(X, Y) > Y because [43], by definition 0.91/0.96 43] cons*(X, Y) >= Y because [44], by (Select) 0.91/0.96 44] Y >= Y by (Meta) 0.91/0.96 0.91/0.96 45] filter(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 46] filter(F, cons(X, Y)) > filter2(@_{o -> o}(F, X), F, X, Y) because [47], by definition 0.91/0.96 47] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [48], [51], [52], [54] and [58], by (Stat) 0.91/0.96 48] cons(X, Y) > Y because [49], by definition 0.91/0.96 49] cons*(X, Y) >= Y because [50], by (Select) 0.91/0.96 50] Y >= Y by (Meta) 0.91/0.96 51] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [52] and [54], by (Copy) 0.91/0.96 52] filter*(F, cons(X, Y)) >= F because [53], by (Select) 0.91/0.96 53] F >= F by (Meta) 0.91/0.96 54] filter*(F, cons(X, Y)) >= X because [55], by (Select) 0.91/0.96 55] cons(X, Y) >= X because [56], by (Star) 0.91/0.96 56] cons*(X, Y) >= X because [57], by (Select) 0.91/0.96 57] X >= X by (Meta) 0.91/0.96 58] filter*(F, cons(X, Y)) >= Y because [59], by (Select) 0.91/0.96 59] cons(X, Y) >= Y because [49], by (Star) 0.91/0.96 0.91/0.96 60] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [61], by (Star) 0.91/0.96 61] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [62] and [64], by (Copy) 0.91/0.96 62] filter2*(true, F, X, Y) >= X because [63], by (Select) 0.91/0.96 63] X >= X by (Meta) 0.91/0.96 64] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [65], [66], [67] and [68], by (Stat) 0.91/0.96 65] F >= F by (Meta) 0.91/0.96 66] Y >= Y by (Meta) 0.91/0.96 67] filter2*(true, F, X, Y) >= F because [65], by (Select) 0.91/0.96 68] filter2*(true, F, X, Y) >= Y because [66], by (Select) 0.91/0.96 0.91/0.96 69] filter2(false, F, X, Y) >= filter(F, Y) because [70], by (Star) 0.91/0.96 70] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [71], [72], [73] and [74], by (Stat) 0.91/0.96 71] F >= F by (Meta) 0.91/0.96 72] Y >= Y by (Meta) 0.91/0.96 73] filter2*(false, F, X, Y) >= F because [71], by (Select) 0.91/0.96 74] filter2*(false, F, X, Y) >= Y because [72], by (Select) 0.91/0.96 0.91/0.96 We can thus remove the following rules: 0.91/0.96 0.91/0.96 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 0.91/0.96 0.91/0.96 We use rule removal, following [Kop12, Theorem 2.23]. 0.91/0.96 0.91/0.96 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.91/0.96 0.91/0.96 ack(0, X) >? succ(X) 0.91/0.96 ack(succ(X), Y) >? ack(X, succ(0)) 0.91/0.96 ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, nil) >? nil 0.91/0.96 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.91/0.96 filter(F, nil) >? nil 0.91/0.96 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 0.91/0.96 filter2(false, F, X, Y) >? filter(F, Y) 0.91/0.96 0.91/0.96 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.91/0.96 0.91/0.96 Argument functions: 0.91/0.96 0.91/0.96 [[nil]] = _|_ 0.91/0.96 0.91/0.96 We choose Lex = {ack} and Mul = {0, @_{o -> o}, cons, false, filter, filter2, map, succ, true}, and the following precedence: filter2 > ack > succ > 0 > false > map > @_{o -> o} > cons > filter > true 0.91/0.96 0.91/0.96 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.91/0.96 0.91/0.96 ack(0, X) >= succ(X) 0.91/0.96 ack(succ(X), Y) >= ack(X, succ(0)) 0.91/0.96 ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, _|_) >= _|_ 0.91/0.96 map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) 0.91/0.96 filter(F, _|_) >= _|_ 0.91/0.96 filter2(true, F, X, Y) > cons(X, filter(F, Y)) 0.91/0.96 filter2(false, F, X, Y) >= filter(F, Y) 0.91/0.96 0.91/0.96 With these choices, we have: 0.91/0.96 0.91/0.96 1] ack(0, X) >= succ(X) because [2], by (Star) 0.91/0.96 2] ack*(0, X) >= succ(X) because ack > succ and [3], by (Copy) 0.91/0.96 3] ack*(0, X) >= X because [4], by (Select) 0.91/0.96 4] X >= X by (Meta) 0.91/0.96 0.91/0.96 5] ack(succ(X), Y) >= ack(X, succ(0)) because [6], by (Star) 0.91/0.96 6] ack*(succ(X), Y) >= ack(X, succ(0)) because [7], [10] and [12], by (Stat) 0.91/0.96 7] succ(X) > X because [8], by definition 0.91/0.96 8] succ*(X) >= X because [9], by (Select) 0.91/0.96 9] X >= X by (Meta) 0.91/0.96 10] ack*(succ(X), Y) >= X because [11], by (Select) 0.91/0.96 11] succ(X) >= X because [8], by (Star) 0.91/0.96 12] ack*(succ(X), Y) >= succ(0) because ack > succ and [13], by (Copy) 0.91/0.96 13] ack*(succ(X), Y) >= 0 because ack > 0, by (Copy) 0.91/0.96 0.91/0.96 14] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [15], by (Star) 0.91/0.96 15] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [16], [19] and [21], by (Stat) 0.91/0.96 16] succ(X) > X because [17], by definition 0.91/0.96 17] succ*(X) >= X because [18], by (Select) 0.91/0.96 18] X >= X by (Meta) 0.91/0.96 19] ack*(succ(X), succ(Y)) >= X because [20], by (Select) 0.91/0.96 20] succ(X) >= X because [17], by (Star) 0.91/0.96 21] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [22], [24], [27] and [28], by (Stat) 0.91/0.96 22] succ(X) >= succ(X) because succ in Mul and [23], by (Fun) 0.91/0.96 23] X >= X by (Meta) 0.91/0.96 24] succ(Y) > Y because [25], by definition 0.91/0.96 25] succ*(Y) >= Y because [26], by (Select) 0.91/0.96 26] Y >= Y by (Meta) 0.91/0.96 27] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [19], by (Copy) 0.91/0.96 28] ack*(succ(X), succ(Y)) >= Y because [29], by (Select) 0.91/0.96 29] succ(Y) >= Y because [25], by (Star) 0.91/0.96 0.91/0.96 30] map(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 31] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) because [32], by definition 0.91/0.96 32] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [33] and [40], by (Copy) 0.91/0.96 33] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [34] and [36], by (Copy) 0.91/0.96 34] map*(F, cons(X, Y)) >= F because [35], by (Select) 0.91/0.96 35] F >= F by (Meta) 0.91/0.96 36] map*(F, cons(X, Y)) >= X because [37], by (Select) 0.91/0.96 37] cons(X, Y) >= X because [38], by (Star) 0.91/0.96 38] cons*(X, Y) >= X because [39], by (Select) 0.91/0.96 39] X >= X by (Meta) 0.91/0.96 40] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [41] and [42], by (Stat) 0.91/0.96 41] F >= F by (Meta) 0.91/0.96 42] cons(X, Y) > Y because [43], by definition 0.91/0.96 43] cons*(X, Y) >= Y because [44], by (Select) 0.91/0.96 44] Y >= Y by (Meta) 0.91/0.96 0.91/0.96 45] filter(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 46] filter2(true, F, X, Y) > cons(X, filter(F, Y)) because [47], by definition 0.91/0.96 47] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [48] and [50], by (Copy) 0.91/0.96 48] filter2*(true, F, X, Y) >= X because [49], by (Select) 0.91/0.96 49] X >= X by (Meta) 0.91/0.96 50] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 > filter, [51] and [53], by (Copy) 0.91/0.96 51] filter2*(true, F, X, Y) >= F because [52], by (Select) 0.91/0.96 52] F >= F by (Meta) 0.91/0.96 53] filter2*(true, F, X, Y) >= Y because [54], by (Select) 0.91/0.96 54] Y >= Y by (Meta) 0.91/0.96 0.91/0.96 55] filter2(false, F, X, Y) >= filter(F, Y) because [56], by (Star) 0.91/0.96 56] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 > filter, [57] and [59], by (Copy) 0.91/0.96 57] filter2*(false, F, X, Y) >= F because [58], by (Select) 0.91/0.96 58] F >= F by (Meta) 0.91/0.96 59] filter2*(false, F, X, Y) >= Y because [60], by (Select) 0.91/0.96 60] Y >= Y by (Meta) 0.91/0.96 0.91/0.96 We can thus remove the following rules: 0.91/0.96 0.91/0.96 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 0.91/0.96 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 0.91/0.96 0.91/0.96 We use rule removal, following [Kop12, Theorem 2.23]. 0.91/0.96 0.91/0.96 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.91/0.96 0.91/0.96 ack(0, X) >? succ(X) 0.91/0.96 ack(succ(X), Y) >? ack(X, succ(0)) 0.91/0.96 ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, nil) >? nil 0.91/0.96 filter(F, nil) >? nil 0.91/0.96 filter2(false, F, X, Y) >? filter(F, Y) 0.91/0.96 0.91/0.96 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.91/0.96 0.91/0.96 Argument functions: 0.91/0.96 0.91/0.96 [[0]] = _|_ 0.91/0.96 [[nil]] = _|_ 0.91/0.96 0.91/0.96 We choose Lex = {ack} and Mul = {false, filter, filter2, map, succ}, and the following precedence: ack > false > filter2 > filter > map > succ 0.91/0.96 0.91/0.96 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.91/0.96 0.91/0.96 ack(_|_, X) >= succ(X) 0.91/0.96 ack(succ(X), Y) > ack(X, succ(_|_)) 0.91/0.96 ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, _|_) >= _|_ 0.91/0.96 filter(F, _|_) >= _|_ 0.91/0.96 filter2(false, F, X, Y) >= filter(F, Y) 0.91/0.96 0.91/0.96 With these choices, we have: 0.91/0.96 0.91/0.96 1] ack(_|_, X) >= succ(X) because [2], by (Star) 0.91/0.96 2] ack*(_|_, X) >= succ(X) because ack > succ and [3], by (Copy) 0.91/0.96 3] ack*(_|_, X) >= X because [4], by (Select) 0.91/0.96 4] X >= X by (Meta) 0.91/0.96 0.91/0.96 5] ack(succ(X), Y) > ack(X, succ(_|_)) because [6], by definition 0.91/0.96 6] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [7], [10] and [12], by (Stat) 0.91/0.96 7] succ(X) > X because [8], by definition 0.91/0.96 8] succ*(X) >= X because [9], by (Select) 0.91/0.96 9] X >= X by (Meta) 0.91/0.96 10] ack*(succ(X), Y) >= X because [11], by (Select) 0.91/0.96 11] succ(X) >= X because [8], by (Star) 0.91/0.96 12] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [13], by (Copy) 0.91/0.96 13] ack*(succ(X), Y) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 14] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [15], by (Star) 0.91/0.96 15] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [16], [19] and [21], by (Stat) 0.91/0.96 16] succ(X) > X because [17], by definition 0.91/0.96 17] succ*(X) >= X because [18], by (Select) 0.91/0.96 18] X >= X by (Meta) 0.91/0.96 19] ack*(succ(X), succ(Y)) >= X because [20], by (Select) 0.91/0.96 20] succ(X) >= X because [17], by (Star) 0.91/0.96 21] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [22], [24], [27] and [28], by (Stat) 0.91/0.96 22] succ(X) >= succ(X) because succ in Mul and [23], by (Fun) 0.91/0.96 23] X >= X by (Meta) 0.91/0.96 24] succ(Y) > Y because [25], by definition 0.91/0.96 25] succ*(Y) >= Y because [26], by (Select) 0.91/0.96 26] Y >= Y by (Meta) 0.91/0.96 27] ack*(succ(X), succ(Y)) >= succ(X) because [22], by (Select) 0.91/0.96 28] ack*(succ(X), succ(Y)) >= Y because [29], by (Select) 0.91/0.96 29] succ(Y) >= Y because [25], by (Star) 0.91/0.96 0.91/0.96 30] map(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 31] filter(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 32] filter2(false, F, X, Y) >= filter(F, Y) because [33], by (Star) 0.91/0.96 33] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 > filter, [34] and [36], by (Copy) 0.91/0.96 34] filter2*(false, F, X, Y) >= F because [35], by (Select) 0.91/0.96 35] F >= F by (Meta) 0.91/0.96 36] filter2*(false, F, X, Y) >= Y because [37], by (Select) 0.91/0.96 37] Y >= Y by (Meta) 0.91/0.96 0.91/0.96 We can thus remove the following rules: 0.91/0.96 0.91/0.96 ack(succ(X), Y) => ack(X, succ(0)) 0.91/0.96 0.91/0.96 We use rule removal, following [Kop12, Theorem 2.23]. 0.91/0.96 0.91/0.96 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.91/0.96 0.91/0.96 ack(0, X) >? succ(X) 0.91/0.96 ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, nil) >? nil 0.91/0.96 filter(F, nil) >? nil 0.91/0.96 filter2(false, F, X, Y) >? filter(F, Y) 0.91/0.96 0.91/0.96 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.91/0.96 0.91/0.96 Argument functions: 0.91/0.96 0.91/0.96 [[nil]] = _|_ 0.91/0.96 0.91/0.96 We choose Lex = {ack} and Mul = {0, false, filter, filter2, map, succ}, and the following precedence: 0 > false > filter2 > filter > map > ack > succ 0.91/0.96 0.91/0.96 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.91/0.96 0.91/0.96 ack(0, X) > succ(X) 0.91/0.96 ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, _|_) >= _|_ 0.91/0.96 filter(F, _|_) >= _|_ 0.91/0.96 filter2(false, F, X, Y) > filter(F, Y) 0.91/0.96 0.91/0.96 With these choices, we have: 0.91/0.96 0.91/0.96 1] ack(0, X) > succ(X) because [2], by definition 0.91/0.96 2] ack*(0, X) >= succ(X) because ack > succ and [3], by (Copy) 0.91/0.96 3] ack*(0, X) >= X because [4], by (Select) 0.91/0.96 4] X >= X by (Meta) 0.91/0.96 0.91/0.96 5] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [6], by (Star) 0.91/0.96 6] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [7], [10] and [12], by (Stat) 0.91/0.96 7] succ(X) > X because [8], by definition 0.91/0.96 8] succ*(X) >= X because [9], by (Select) 0.91/0.96 9] X >= X by (Meta) 0.91/0.96 10] ack*(succ(X), succ(Y)) >= X because [11], by (Select) 0.91/0.96 11] succ(X) >= X because [8], by (Star) 0.91/0.96 12] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [13], [15], [18] and [19], by (Stat) 0.91/0.96 13] succ(X) >= succ(X) because succ in Mul and [14], by (Fun) 0.91/0.96 14] X >= X by (Meta) 0.91/0.96 15] succ(Y) > Y because [16], by definition 0.91/0.96 16] succ*(Y) >= Y because [17], by (Select) 0.91/0.96 17] Y >= Y by (Meta) 0.91/0.96 18] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [10], by (Copy) 0.91/0.96 19] ack*(succ(X), succ(Y)) >= Y because [20], by (Select) 0.91/0.96 20] succ(Y) >= Y because [16], by (Star) 0.91/0.96 0.91/0.96 21] map(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 22] filter(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 23] filter2(false, F, X, Y) > filter(F, Y) because [24], by definition 0.91/0.96 24] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 > filter, [25] and [27], by (Copy) 0.91/0.96 25] filter2*(false, F, X, Y) >= F because [26], by (Select) 0.91/0.96 26] F >= F by (Meta) 0.91/0.96 27] filter2*(false, F, X, Y) >= Y because [28], by (Select) 0.91/0.96 28] Y >= Y by (Meta) 0.91/0.96 0.91/0.96 We can thus remove the following rules: 0.91/0.96 0.91/0.96 ack(0, X) => succ(X) 0.91/0.96 filter2(false, F, X, Y) => filter(F, Y) 0.91/0.96 0.91/0.96 We use rule removal, following [Kop12, Theorem 2.23]. 0.91/0.96 0.91/0.96 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.91/0.96 0.91/0.96 ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, nil) >? nil 0.91/0.96 filter(F, nil) >? nil 0.91/0.96 0.91/0.96 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.91/0.96 0.91/0.96 Argument functions: 0.91/0.96 0.91/0.96 [[nil]] = _|_ 0.91/0.96 0.91/0.96 We choose Lex = {ack} and Mul = {filter, map, succ}, and the following precedence: ack > filter > succ > map 0.91/0.96 0.91/0.96 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.91/0.96 0.91/0.96 ack(succ(X), succ(Y)) > ack(X, ack(succ(X), Y)) 0.91/0.96 map(F, _|_) >= _|_ 0.91/0.96 filter(F, _|_) >= _|_ 0.91/0.96 0.91/0.96 With these choices, we have: 0.91/0.96 0.91/0.96 1] ack(succ(X), succ(Y)) > ack(X, ack(succ(X), Y)) because [2], by definition 0.91/0.96 2] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [3], [6] and [8], by (Stat) 0.91/0.96 3] succ(X) > X because [4], by definition 0.91/0.96 4] succ*(X) >= X because [5], by (Select) 0.91/0.96 5] X >= X by (Meta) 0.91/0.96 6] ack*(succ(X), succ(Y)) >= X because [7], by (Select) 0.91/0.96 7] succ(X) >= X because [4], by (Star) 0.91/0.96 8] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [9], [11], [14] and [15], by (Stat) 0.91/0.96 9] succ(X) >= succ(X) because succ in Mul and [10], by (Fun) 0.91/0.96 10] X >= X by (Meta) 0.91/0.96 11] succ(Y) > Y because [12], by definition 0.91/0.96 12] succ*(Y) >= Y because [13], by (Select) 0.91/0.96 13] Y >= Y by (Meta) 0.91/0.96 14] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [6], by (Copy) 0.91/0.96 15] ack*(succ(X), succ(Y)) >= Y because [16], by (Select) 0.91/0.96 16] succ(Y) >= Y because [12], by (Star) 0.91/0.96 0.91/0.96 17] map(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 18] filter(F, _|_) >= _|_ by (Bot) 0.91/0.96 0.91/0.96 We can thus remove the following rules: 0.91/0.96 0.91/0.96 ack(succ(X), succ(Y)) => ack(X, ack(succ(X), Y)) 0.91/0.96 0.91/0.96 We use rule removal, following [Kop12, Theorem 2.23]. 0.91/0.96 0.91/0.96 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.91/0.96 0.91/0.96 map(F, nil) >? nil 0.91/0.96 filter(F, nil) >? nil 0.91/0.96 0.91/0.96 We orient these requirements with a polynomial interpretation in the natural numbers. 0.91/0.96 0.91/0.96 The following interpretation satisfies the requirements: 0.91/0.96 0.91/0.96 filter = \G0y1.3 + 3y1 + G0(0) 0.91/0.96 map = \G0y1.3 + 3y1 + G0(0) 0.91/0.96 nil = 0 0.91/0.96 0.91/0.96 Using this interpretation, the requirements translate to: 0.91/0.96 0.91/0.96 [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 0.91/0.96 [[filter(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 0.91/0.96 0.91/0.96 We can thus remove the following rules: 0.91/0.96 0.91/0.96 map(F, nil) => nil 0.91/0.96 filter(F, nil) => nil 0.91/0.96 0.91/0.96 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. 0.91/0.97 0.91/0.97 0.91/0.97 +++ Citations +++ 0.91/0.97 0.91/0.97 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.91/0.97 EOF