We consider the system Applicative_first_order_05__29. Alphabet: 0 : [] --> a ack : [a * a] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d succ : [a] --> a true : [] --> b Rules: ack(0, x) => succ(x) ack(succ(x), y) => ack(x, succ(0)) ack(succ(x), succ(y)) => ack(x, ack(succ(x), y)) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(f, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): ack(0, X) >? succ(X) ack(succ(X), Y) >? ack(X, succ(0)) ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) filter(F, nil) >? nil filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) filter2(true, F, X, Y) >? cons(X, filter(F, Y)) filter2(false, F, X, Y) >? filter(F, Y) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[filter(x_1, x_2)]] = filter(x_2, x_1) [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_1, x_3) [[nil]] = _|_ We choose Lex = {ack, filter, filter2} and Mul = {@_{o -> o}, cons, false, map, succ, true}, and the following precedence: ack > succ > filter = filter2 > map > @_{o -> o} > false > true > cons Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack(_|_, X) > succ(X) ack(succ(X), Y) > ack(X, succ(_|_)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) filter2(true, F, X, Y) > cons(X, filter(F, Y)) filter2(false, F, X, Y) > filter(F, Y) With these choices, we have: 1] ack(_|_, X) > succ(X) because [2], by definition 2] ack*(_|_, X) >= succ(X) because ack > succ and [3], by (Copy) 3] ack*(_|_, X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] ack(succ(X), Y) > ack(X, succ(_|_)) because [6], by definition 6] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [7], [10] and [12], by (Stat) 7] succ(X) > X because [8], by definition 8] succ*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] ack*(succ(X), Y) >= X because [11], by (Select) 11] succ(X) >= X because [8], by (Star) 12] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [13], by (Copy) 13] ack*(succ(X), Y) >= _|_ by (Bot) 14] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [15], by (Star) 15] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [16], [19] and [21], by (Stat) 16] succ(X) > X because [17], by definition 17] succ*(X) >= X because [18], by (Select) 18] X >= X by (Meta) 19] ack*(succ(X), succ(Y)) >= X because [20], by (Select) 20] succ(X) >= X because [17], by (Star) 21] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [22], [24], [27] and [28], by (Stat) 22] succ(X) >= succ(X) because succ in Mul and [23], by (Fun) 23] X >= X by (Meta) 24] succ(Y) > Y because [25], by definition 25] succ*(Y) >= Y because [26], by (Select) 26] Y >= Y by (Meta) 27] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [19], by (Copy) 28] ack*(succ(X), succ(Y)) >= Y because [29], by (Select) 29] succ(Y) >= Y because [25], by (Star) 30] map(F, _|_) >= _|_ by (Bot) 31] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [32], by (Star) 32] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [33] and [40], by (Copy) 33] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [34] and [36], by (Copy) 34] map*(F, cons(X, Y)) >= F because [35], by (Select) 35] F >= F by (Meta) 36] map*(F, cons(X, Y)) >= X because [37], by (Select) 37] cons(X, Y) >= X because [38], by (Star) 38] cons*(X, Y) >= X because [39], by (Select) 39] X >= X by (Meta) 40] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [41] and [42], by (Stat) 41] F >= F by (Meta) 42] cons(X, Y) > Y because [43], by definition 43] cons*(X, Y) >= Y because [44], by (Select) 44] Y >= Y by (Meta) 45] filter(F, _|_) >= _|_ by (Bot) 46] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [47], by (Star) 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) 48] cons(X, Y) > Y because [49], by definition 49] cons*(X, Y) >= Y because [50], by (Select) 50] Y >= Y by (Meta) 51] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [52] and [54], by (Copy) 52] filter*(F, cons(X, Y)) >= F because [53], by (Select) 53] F >= F by (Meta) 54] filter*(F, cons(X, Y)) >= X because [55], by (Select) 55] cons(X, Y) >= X because [56], by (Star) 56] cons*(X, Y) >= X because [57], by (Select) 57] X >= X by (Meta) 58] filter*(F, cons(X, Y)) >= Y because [59], by (Select) 59] cons(X, Y) >= Y because [49], by (Star) 60] filter2(true, F, X, Y) > cons(X, filter(F, Y)) because [61], by definition 61] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [62] and [64], by (Copy) 62] filter2*(true, F, X, Y) >= X because [63], by (Select) 63] X >= X by (Meta) 64] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [65], [66], [67] and [68], by (Stat) 65] F >= F by (Meta) 66] Y >= Y by (Meta) 67] filter2*(true, F, X, Y) >= F because [65], by (Select) 68] filter2*(true, F, X, Y) >= Y because [66], by (Select) 69] filter2(false, F, X, Y) > filter(F, Y) because [70], by definition 70] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [71], [72], [73] and [74], by (Stat) 71] F >= F by (Meta) 72] Y >= Y by (Meta) 73] filter2*(false, F, X, Y) >= F because [71], by (Select) 74] filter2*(false, F, X, Y) >= Y because [72], by (Select) We can thus remove the following rules: ack(0, X) => succ(X) ack(succ(X), Y) => ack(X, succ(0)) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) filter(F, nil) >? nil filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {ack} and Mul = {@_{o -> o}, cons, filter, filter2, map, nil, succ}, and the following precedence: ack > map > filter > succ > filter2 > nil > @_{o -> o} > cons With these choices, we have: 1] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [2], by (Star) 2] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [3], [6] and [8], by (Stat) 3] succ(X) > X because [4], by definition 4] succ*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] ack*(succ(X), succ(Y)) >= X because [7], by (Select) 7] succ(X) >= X because [4], by (Star) 8] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [9], [11], [14] and [15], by (Stat) 9] succ(X) >= succ(X) because succ in Mul and [10], by (Fun) 10] X >= X by (Meta) 11] succ(Y) > Y because [12], by definition 12] succ*(Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [6], by (Copy) 15] ack*(succ(X), succ(Y)) >= Y because [16], by (Select) 16] succ(Y) >= Y because [12], by (Star) 17] map(F, nil) > nil because [18], by definition 18] map*(F, nil) >= nil because map > nil, by (Copy) 19] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [20], by (Star) 20] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [21] and [28], by (Copy) 21] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [22] and [24], by (Copy) 22] map*(F, cons(X, Y)) >= F because [23], by (Select) 23] F >= F by (Meta) 24] map*(F, cons(X, Y)) >= X because [25], by (Select) 25] cons(X, Y) >= X because [26], by (Star) 26] cons*(X, Y) >= X because [27], by (Select) 27] X >= X by (Meta) 28] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [29] and [30], by (Stat) 29] F >= F by (Meta) 30] cons(X, Y) > Y because [31], by definition 31] cons*(X, Y) >= Y because [32], by (Select) 32] Y >= Y by (Meta) 33] filter(F, nil) >= nil because [34], by (Star) 34] filter*(F, nil) >= nil because filter > nil, by (Copy) 35] filter(F, cons(X, Y)) > filter2(@_{o -> o}(F, X), F, X, Y) because [36], by definition 36] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter > filter2, [37], [38], [40] and [44], by (Copy) 37] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [38] and [40], by (Copy) 38] filter*(F, cons(X, Y)) >= F because [39], by (Select) 39] F >= F by (Meta) 40] filter*(F, cons(X, Y)) >= X because [41], by (Select) 41] cons(X, Y) >= X because [42], by (Star) 42] cons*(X, Y) >= X because [43], by (Select) 43] X >= X by (Meta) 44] filter*(F, cons(X, Y)) >= Y because [45], by (Select) 45] cons(X, Y) >= Y because [46], by (Star) 46] cons*(X, Y) >= Y because [47], by (Select) 47] Y >= Y by (Meta) We can thus remove the following rules: map(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) map(F, cons(X, Y)) >? cons(F X, map(F, Y)) filter(F, nil) >? nil We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[nil]] = _|_ We choose Lex = {ack} and Mul = {@_{o -> o}, cons, filter, map, succ}, and the following precedence: map > @_{o -> o} > ack > succ > filter > cons Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ With these choices, we have: 1] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [2], by (Star) 2] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [3], [6] and [8], by (Stat) 3] succ(X) > X because [4], by definition 4] succ*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] ack*(succ(X), succ(Y)) >= X because [7], by (Select) 7] succ(X) >= X because [4], by (Star) 8] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [9], [11], [14] and [15], by (Stat) 9] succ(X) >= succ(X) because succ in Mul and [10], by (Fun) 10] X >= X by (Meta) 11] succ(Y) > Y because [12], by definition 12] succ*(Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] ack*(succ(X), succ(Y)) >= succ(X) because [9], by (Select) 15] ack*(succ(X), succ(Y)) >= Y because [16], by (Select) 16] succ(Y) >= Y because [12], by (Star) 17] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) because [18], by definition 18] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [19] and [26], by (Copy) 19] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [20] and [22], by (Copy) 20] map*(F, cons(X, Y)) >= F because [21], by (Select) 21] F >= F by (Meta) 22] map*(F, cons(X, Y)) >= X because [23], by (Select) 23] cons(X, Y) >= X because [24], by (Star) 24] cons*(X, Y) >= X because [25], by (Select) 25] X >= X by (Meta) 26] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [27] and [28], by (Stat) 27] F >= F by (Meta) 28] cons(X, Y) > Y because [29], by definition 29] cons*(X, Y) >= Y because [30], by (Select) 30] Y >= Y by (Meta) 31] filter(F, _|_) >= _|_ by (Bot) We can thus remove the following rules: map(F, cons(X, Y)) => cons(F X, map(F, Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), Y)) filter(F, nil) >? nil We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[nil]] = _|_ We choose Lex = {ack} and Mul = {filter, succ}, and the following precedence: filter > ack > succ Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack(succ(X), succ(Y)) > ack(X, ack(succ(X), Y)) filter(F, _|_) >= _|_ With these choices, we have: 1] ack(succ(X), succ(Y)) > ack(X, ack(succ(X), Y)) because [2], by definition 2] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [3], [6] and [8], by (Stat) 3] succ(X) > X because [4], by definition 4] succ*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] ack*(succ(X), succ(Y)) >= X because [7], by (Select) 7] succ(X) >= X because [4], by (Star) 8] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [9], [11], [14] and [15], by (Stat) 9] succ(X) >= succ(X) because succ in Mul and [10], by (Fun) 10] X >= X by (Meta) 11] succ(Y) > Y because [12], by definition 12] succ*(Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [6], by (Copy) 15] ack*(succ(X), succ(Y)) >= Y because [16], by (Select) 16] succ(Y) >= Y because [12], by (Star) 17] filter(F, _|_) >= _|_ by (Bot) We can thus remove the following rules: ack(succ(X), succ(Y)) => ack(X, ack(succ(X), Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): filter(F, nil) >? nil We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[nil]] = _|_ We choose Lex = {} and Mul = {filter}, and the following precedence: filter Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: filter(F, _|_) > _|_ With these choices, we have: 1] filter(F, _|_) > _|_ because [2], by definition 2] filter*(F, _|_) >= _|_ by (Bot) We can thus remove the following rules: filter(F, nil) => nil 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. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.