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]] = _|_ [[@_{o -> o}(x_1, x_2)]] = @_{o -> o}(x_2, x_1) [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_3, x_1) We choose Lex = {@_{o -> o}, ack, cons, false, filter, filter2, succ, true} and Mul = {map, nil}, and the following precedence: ack > map > nil > succ > true > false > filter = filter2 > @_{o -> o} > 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, nil) > nil map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) filter(F, nil) >= nil 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 (Star) 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 (Star) 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 [13], by (Select) 13] succ(X) >= succ(_|_) because [14], by (Fun) 14] X >= _|_ by (Bot) 15] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [16], by (Star) 16] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [17], [20] and [22], by (Stat) 17] succ(X) > X because [18], by definition 18] succ*(X) >= X because [19], by (Select) 19] X >= X by (Meta) 20] ack*(succ(X), succ(Y)) >= X because [21], by (Select) 21] succ(X) >= X because [18], by (Star) 22] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [23], [25], [28] and [29], by (Stat) 23] succ(X) >= succ(X) because [24], by (Fun) 24] X >= X by (Meta) 25] succ(Y) > Y because [26], by definition 26] succ*(Y) >= Y because [27], by (Select) 27] Y >= Y by (Meta) 28] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [20], by (Copy) 29] ack*(succ(X), succ(Y)) >= Y because [30], by (Select) 30] succ(Y) >= Y because [26], by (Star) 31] map(F, nil) > nil because [32], by definition 32] map*(F, nil) >= nil because map > nil, by (Copy) 33] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) because [34], by definition 34] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [35] and [42], by (Copy) 35] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [36] and [38], by (Copy) 36] map*(F, cons(X, Y)) >= F because [37], by (Select) 37] F >= F by (Meta) 38] map*(F, cons(X, Y)) >= X because [39], by (Select) 39] cons(X, Y) >= X because [40], by (Star) 40] cons*(X, Y) >= X because [41], by (Select) 41] X >= X by (Meta) 42] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [43] and [44], by (Stat) 43] F >= F by (Meta) 44] cons(X, Y) > Y because [45], by definition 45] cons*(X, Y) >= Y because [46], by (Select) 46] Y >= Y by (Meta) 47] filter(F, nil) >= nil because [48], by (Star) 48] filter*(F, nil) >= nil because [49], by (Select) 49] nil >= nil by (Fun) 50] filter(F, cons(X, Y)) > filter2(@_{o -> o}(F, X), F, X, Y) because [51], by definition 51] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [52], [53], [56], [57], [58] and [62], by (Stat) 52] F >= F by (Meta) 53] cons(X, Y) > Y because [54], by definition 54] cons*(X, Y) >= Y because [55], by (Select) 55] Y >= Y by (Meta) 56] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [57] and [58], by (Copy) 57] filter*(F, cons(X, Y)) >= F because [52], by (Select) 58] filter*(F, cons(X, Y)) >= X because [59], by (Select) 59] cons(X, Y) >= X because [60], by (Star) 60] cons*(X, Y) >= X because [61], by (Select) 61] X >= X by (Meta) 62] filter*(F, cons(X, Y)) >= Y because [63], by (Select) 63] cons(X, Y) >= Y because [54], by (Star) 64] filter2(true, F, X, Y) > cons(X, filter(F, Y)) because [65], by definition 65] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [66] and [68], by (Copy) 66] filter2*(true, F, X, Y) >= X because [67], by (Select) 67] X >= X by (Meta) 68] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [69], [70], [71], [72], [69] and [70], by (Stat) 69] F >= F by (Meta) 70] Y >= Y by (Meta) 71] filter2*(true, F, X, Y) >= F because [69], by (Select) 72] filter2*(true, F, X, Y) >= Y because [70], by (Select) 73] filter2(false, F, X, Y) > filter(F, Y) because [74], by definition 74] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [75], [76], [77], [78], [75] and [76], by (Stat) 75] F >= F by (Meta) 76] Y >= Y by (Meta) 77] filter2*(false, F, X, Y) >= F because [75], by (Select) 78] filter2*(false, F, X, Y) >= Y because [76], by (Select) We can thus remove the following rules: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) 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 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)) filter(F, nil) >? nil We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {ack} and Mul = {filter, nil, succ}, and the following precedence: nil > ack > succ > filter 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)) filter(F, nil) > nil With these choices, we have: 1] ack(_|_, X) >= succ(X) because [2], by (Star) 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 [13], by (Select) 13] succ(X) >= succ(_|_) because succ in Mul and [14], by (Fun) 14] X >= _|_ by (Bot) 15] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [16], by (Star) 16] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [17], [20] and [22], by (Stat) 17] succ(X) > X because [18], by definition 18] succ*(X) >= X because [19], by (Select) 19] X >= X by (Meta) 20] ack*(succ(X), succ(Y)) >= X because [21], by (Select) 21] succ(X) >= X because [18], by (Star) 22] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [23], [25], [28] and [29], by (Stat) 23] succ(X) >= succ(X) because succ in Mul and [24], by (Fun) 24] X >= X by (Meta) 25] succ(Y) > Y because [26], by definition 26] succ*(Y) >= Y because [27], by (Select) 27] Y >= Y by (Meta) 28] ack*(succ(X), succ(Y)) >= succ(X) because [23], by (Select) 29] ack*(succ(X), succ(Y)) >= Y because [30], by (Select) 30] succ(Y) >= Y because [26], by (Star) 31] filter(F, nil) > nil because [32], by definition 32] filter*(F, nil) >= nil because [33], by (Select) 33] nil >= nil by (Fun) We can thus remove the following rules: ack(succ(X), Y) => ack(X, succ(0)) filter(F, nil) => nil 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), succ(Y)) >? ack(X, ack(succ(X), Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {0, ack, succ} and Mul = {}, and the following precedence: 0 > ack > succ With these choices, we have: 1] ack(0, X) > succ(X) because [2], by definition 2] ack*(0, X) >= succ(X) because ack > succ and [3], by (Copy) 3] ack*(0, X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [6], by (Star) 6] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) 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), succ(Y)) >= X because [11], by (Select) 11] succ(X) >= X because [8], by (Star) 12] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [13], [15], [18] and [19], by (Stat) 13] succ(X) >= succ(X) because [14], by (Fun) 14] X >= X by (Meta) 15] succ(Y) > Y because [16], by definition 16] succ*(Y) >= Y because [17], by (Select) 17] Y >= Y by (Meta) 18] ack*(succ(X), succ(Y)) >= succ(X) because [13], by (Select) 19] ack*(succ(X), succ(Y)) >= Y because [20], by (Select) 20] succ(Y) >= Y because [16], by (Star) We can thus remove the following rules: ack(0, X) => succ(X) 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)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {ack, succ} and Mul = {}, and the following precedence: ack = succ 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 [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) We can thus remove the following rules: ack(succ(X), succ(Y)) => ack(X, ack(succ(X), Y)) 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.