We consider the system Applicative_05__Ex7_9. Alphabet: 0 : [] --> a cons : [b * c] --> c d : [a * a] --> c false : [] --> c filter : [b -> c * c] --> c gtr : [a * a] --> c if : [c * c * c] --> c len : [c] --> a nil : [] --> c s : [a] --> a sub : [a * a] --> a true : [] --> c Rules: if(true, x, y) => x if(false, x, y) => y sub(x, 0) => x sub(s(x), s(y)) => sub(x, y) gtr(0, x) => false gtr(s(x), 0) => true gtr(s(x), s(y)) => gtr(x, y) d(x, 0) => true d(s(x), s(y)) => if(gtr(x, y), false, d(s(x), sub(y, x))) len(nil) => 0 len(cons(x, y)) => s(len(y)) filter(f, nil) => nil filter(f, cons(x, y)) => if(f x, cons(x, filter(f, 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 the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs. After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] sub#(s(X), s(Y)) =#> sub#(X, Y) 1] gtr#(s(X), s(Y)) =#> gtr#(X, Y) 2] d#(s(X), s(Y)) =#> if#(gtr(X, Y), false, d(s(X), sub(Y, X))) 3] d#(s(X), s(Y)) =#> gtr#(X, Y) 4] d#(s(X), s(Y)) =#> d#(s(X), sub(Y, X)) 5] d#(s(X), s(Y)) =#> sub#(Y, X) 6] len#(cons(X, Y)) =#> len#(Y) 7] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y)) 8] filter#(F, cons(X, Y)) =#> F(X) 9] filter#(F, cons(X, Y)) =#> filter#(F, Y) 10] filter#(F, cons(X, Y)) =#> filter#(F, Y) Rules R_0: if(true, X, Y) => X if(false, X, Y) => Y sub(X, 0) => X sub(s(X), s(Y)) => sub(X, Y) gtr(0, X) => false gtr(s(X), 0) => true gtr(s(X), s(Y)) => gtr(X, Y) d(X, 0) => true d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) len(nil) => 0 len(cons(X, Y)) => s(len(Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, Y)), filter(F, Y)) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0 * 1 : 1 * 2 : * 3 : 1 * 4 : 2, 3, 4, 5 * 5 : 0 * 6 : 6 * 7 : * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 * 9 : 7, 8, 9, 10 * 10 : 7, 8, 9, 10 This graph has the following strongly connected components: P_1: sub#(s(X), s(Y)) =#> sub#(X, Y) P_2: gtr#(s(X), s(Y)) =#> gtr#(X, Y) P_3: d#(s(X), s(Y)) =#> d#(s(X), sub(Y, X)) P_4: len#(cons(X, Y)) =#> len#(Y) P_5: filter#(F, cons(X, Y)) =#> F(X) filter#(F, cons(X, Y)) =#> filter#(F, Y) filter#(F, cons(X, Y)) =#> filter#(F, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f), (P_3, R_0, m, f), (P_4, R_0, m, f) and (P_5, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: filter#(F, cons(X, Y)) >? F(X) filter#(F, cons(X, Y)) >? filter#(F, Y) filter#(F, cons(X, Y)) >? filter#(F, Y) if(true, X, Y) >= X if(false, X, Y) >= Y sub(X, 0) >= X sub(s(X), s(Y)) >= sub(X, Y) gtr(0, X) >= false gtr(s(X), 0) >= true gtr(s(X), s(Y)) >= gtr(X, Y) d(X, 0) >= true d(s(X), s(Y)) >= if(gtr(X, Y), false, d(s(X), sub(Y, X))) len(nil) >= 0 len(cons(X, Y)) >= s(len(Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[false]] = _|_ [[gtr(x_1, x_2)]] = gtr [[if(x_1, x_2, x_3)]] = if(x_2, x_3) [[nil]] = _|_ [[sub(x_1, x_2)]] = sub(x_1) [[true]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, cons, d, filter, filter#, gtr, if, len, s, sub}, and the following precedence: gtr > filter > cons > len = s > d > filter# > if > @_{o -> o} > sub Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: filter#(F, cons(X, Y)) >= @_{o -> o}(F, X) filter#(F, cons(X, Y)) > filter#(F, Y) filter#(F, cons(X, Y)) >= filter#(F, Y) if(X, Y) >= X if(X, Y) >= Y sub(X) >= X sub(s(X)) >= sub(X) gtr >= _|_ gtr >= _|_ gtr >= gtr d(X, _|_) >= _|_ d(s(X), s(Y)) >= if(_|_, d(s(X), sub(Y))) len(_|_) >= _|_ len(cons(X, Y)) >= s(len(Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) With these choices, we have: 1] filter#(F, cons(X, Y)) >= @_{o -> o}(F, X) because [2], by (Star) 2] filter#*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter# > @_{o -> o}, [3] and [5], by (Copy) 3] filter#*(F, cons(X, Y)) >= F because [4], by (Select) 4] F >= F by (Meta) 5] filter#*(F, cons(X, Y)) >= X because [6], by (Select) 6] cons(X, Y) >= X because [7], by (Star) 7] cons*(X, Y) >= X because [8], by (Select) 8] X >= X by (Meta) 9] filter#(F, cons(X, Y)) > filter#(F, Y) because [10], by definition 10] filter#*(F, cons(X, Y)) >= filter#(F, Y) because filter# in Mul, [11] and [12], by (Stat) 11] F >= F by (Meta) 12] cons(X, Y) > Y because [13], by definition 13] cons*(X, Y) >= Y because [14], by (Select) 14] Y >= Y by (Meta) 15] filter#(F, cons(X, Y)) >= filter#(F, Y) because [10], by (Star) 16] if(X, Y) >= X because [17], by (Star) 17] if*(X, Y) >= X because [18], by (Select) 18] X >= X by (Meta) 19] if(X, Y) >= Y because [20], by (Star) 20] if*(X, Y) >= Y because [21], by (Select) 21] Y >= Y by (Meta) 22] sub(X) >= X because [23], by (Star) 23] sub*(X) >= X because [24], by (Select) 24] X >= X by (Meta) 25] sub(s(X)) >= sub(X) because [26], by (Star) 26] sub*(s(X)) >= sub(X) because [27], by (Select) 27] s(X) >= sub(X) because [28], by (Star) 28] s*(X) >= sub(X) because s > sub and [29], by (Copy) 29] s*(X) >= X because [30], by (Select) 30] X >= X by (Meta) 31] gtr >= _|_ by (Bot) 32] gtr >= _|_ by (Bot) 33] gtr >= gtr because gtr in Mul, by (Fun) 34] d(X, _|_) >= _|_ by (Bot) 35] d(s(X), s(Y)) >= if(_|_, d(s(X), sub(Y))) because [36], by (Star) 36] d*(s(X), s(Y)) >= if(_|_, d(s(X), sub(Y))) because d > if, [37] and [38], by (Copy) 37] d*(s(X), s(Y)) >= _|_ by (Bot) 38] d*(s(X), s(Y)) >= d(s(X), sub(Y)) because d in Mul, [39] and [41], by (Stat) 39] s(X) >= s(X) because s in Mul and [40], by (Fun) 40] X >= X by (Meta) 41] s(Y) > sub(Y) because [42], by definition 42] s*(Y) >= sub(Y) because s > sub and [43], by (Copy) 43] s*(Y) >= Y because [44], by (Select) 44] Y >= Y by (Meta) 45] len(_|_) >= _|_ by (Bot) 46] len(cons(X, Y)) >= s(len(Y)) because [47], by (Star) 47] len*(cons(X, Y)) >= s(len(Y)) because len = s, len in Mul and [48], by (Stat) 48] cons(X, Y) > len(Y) because [49], by definition 49] cons*(X, Y) >= len(Y) because cons > len and [50], by (Copy) 50] cons*(X, Y) >= Y because [51], by (Select) 51] Y >= Y by (Meta) 52] filter(F, _|_) >= _|_ by (Bot) 53] filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) because [54], by (Star) 54] filter*(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) because filter > if, [55] and [57], by (Copy) 55] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y)) because filter > cons, [56] and [57], by (Copy) 56] filter*(F, cons(X, Y)) >= X because [6], by (Select) 57] filter*(F, cons(X, Y)) >= filter(F, Y) because filter in Mul, [11] and [12], by (Stat) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, minimal, formative) by (P_6, R_0, minimal, formative), where P_6 consists of: filter#(F, cons(X, Y)) =#> F(X) filter#(F, cons(X, Y)) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_6, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_0, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: filter#(F, cons(X, Y)) >? F(X) filter#(F, cons(X, Y)) >? filter#(F, Y) if(true, X, Y) >= X if(false, X, Y) >= Y sub(X, 0) >= X sub(s(X), s(Y)) >= sub(X, Y) gtr(0, X) >= false gtr(s(X), 0) >= true gtr(s(X), s(Y)) >= gtr(X, Y) d(X, 0) >= true d(s(X), s(Y)) >= if(gtr(X, Y), false, d(s(X), sub(Y, X))) len(nil) >= 0 len(cons(X, Y)) >= s(len(Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[if(x_1, x_2, x_3)]] = if(x_2, x_3) [[nil]] = _|_ [[sub(x_1, x_2)]] = sub(x_1) [[true]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, cons, d, false, filter, filter#, gtr, if, len, s, sub}, and the following precedence: filter# > gtr > filter > cons > @_{o -> o} > d > if > len > false > s > sub Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: filter#(F, cons(X, Y)) > @_{o -> o}(F, X) filter#(F, cons(X, Y)) >= filter#(F, Y) if(X, Y) >= X if(X, Y) >= Y sub(X) >= X sub(s(X)) >= sub(X) gtr(_|_, X) >= false gtr(s(X), _|_) >= _|_ gtr(s(X), s(Y)) >= gtr(X, Y) d(X, _|_) >= _|_ d(s(X), s(Y)) >= if(false, d(s(X), sub(Y))) len(_|_) >= _|_ len(cons(X, Y)) >= s(len(Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) With these choices, we have: 1] filter#(F, cons(X, Y)) > @_{o -> o}(F, X) because [2], by definition 2] filter#*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter# > @_{o -> o}, [3] and [5], by (Copy) 3] filter#*(F, cons(X, Y)) >= F because [4], by (Select) 4] F >= F by (Meta) 5] filter#*(F, cons(X, Y)) >= X because [6], by (Select) 6] cons(X, Y) >= X because [7], by (Star) 7] cons*(X, Y) >= X because [8], by (Select) 8] X >= X by (Meta) 9] filter#(F, cons(X, Y)) >= filter#(F, Y) because filter# in Mul, [10] and [11], by (Fun) 10] F >= F by (Meta) 11] cons(X, Y) >= Y because [12], by (Star) 12] cons*(X, Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] if(X, Y) >= X because [15], by (Star) 15] if*(X, Y) >= X because [16], by (Select) 16] X >= X by (Meta) 17] if(X, Y) >= Y because [18], by (Star) 18] if*(X, Y) >= Y because [19], by (Select) 19] Y >= Y by (Meta) 20] sub(X) >= X because [21], by (Star) 21] sub*(X) >= X because [22], by (Select) 22] X >= X by (Meta) 23] sub(s(X)) >= sub(X) because [24], by (Star) 24] sub*(s(X)) >= sub(X) because [25], by (Select) 25] s(X) >= sub(X) because [26], by (Star) 26] s*(X) >= sub(X) because s > sub and [27], by (Copy) 27] s*(X) >= X because [28], by (Select) 28] X >= X by (Meta) 29] gtr(_|_, X) >= false because [30], by (Star) 30] gtr*(_|_, X) >= false because gtr > false, by (Copy) 31] gtr(s(X), _|_) >= _|_ by (Bot) 32] gtr(s(X), s(Y)) >= gtr(X, Y) because [33], by (Star) 33] gtr*(s(X), s(Y)) >= gtr(X, Y) because gtr in Mul, [34] and [37], by (Stat) 34] s(X) > X because [35], by definition 35] s*(X) >= X because [36], by (Select) 36] X >= X by (Meta) 37] s(Y) > Y because [38], by definition 38] s*(Y) >= Y because [39], by (Select) 39] Y >= Y by (Meta) 40] d(X, _|_) >= _|_ by (Bot) 41] d(s(X), s(Y)) >= if(false, d(s(X), sub(Y))) because [42], by (Star) 42] d*(s(X), s(Y)) >= if(false, d(s(X), sub(Y))) because d > if, [43] and [44], by (Copy) 43] d*(s(X), s(Y)) >= false because d > false, by (Copy) 44] d*(s(X), s(Y)) >= d(s(X), sub(Y)) because d in Mul, [45] and [47], by (Stat) 45] s(X) >= s(X) because s in Mul and [46], by (Fun) 46] X >= X by (Meta) 47] s(Y) > sub(Y) because [48], by definition 48] s*(Y) >= sub(Y) because s > sub and [49], by (Copy) 49] s*(Y) >= Y because [50], by (Select) 50] Y >= Y by (Meta) 51] len(_|_) >= _|_ by (Bot) 52] len(cons(X, Y)) >= s(len(Y)) because [53], by (Star) 53] len*(cons(X, Y)) >= s(len(Y)) because len > s and [54], by (Copy) 54] len*(cons(X, Y)) >= len(Y) because len in Mul and [55], by (Stat) 55] cons(X, Y) > Y because [56], by definition 56] cons*(X, Y) >= Y because [57], by (Select) 57] Y >= Y by (Meta) 58] filter(F, _|_) >= _|_ by (Bot) 59] filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) because [60], by (Star) 60] filter*(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) because filter > if, [61] and [63], by (Copy) 61] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y)) because filter > cons, [62] and [63], by (Copy) 62] filter*(F, cons(X, Y)) >= X because [6], by (Select) 63] filter*(F, cons(X, Y)) >= filter(F, Y) because filter in Mul, [10] and [64], by (Stat) 64] cons(X, Y) > Y because [65], by definition 65] cons*(X, Y) >= Y because [13], by (Select) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_0, minimal, formative) by (P_7, R_0, minimal, formative), where P_7 consists of: filter#(F, cons(X, Y)) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_7, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_7, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(filter#) = 2 Thus, we can orient the dependency pairs as follows: nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_7, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(len#) = 1 Thus, we can orient the dependency pairs as follows: nu(len#(cons(X, Y))) = cons(X, Y) |> Y = nu(len#(Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_3, R_0) are: sub(X, 0) => X sub(s(X), s(Y)) => sub(X, Y) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: d#(s(X), s(Y)) >? d#(s(X), sub(Y, X)) sub(X, 0) >= X sub(s(X), s(Y)) >= sub(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 d# = \y0y1.3y1 s = \y0.3 + 3y0 sub = \y0y1.y0 Using this interpretation, the requirements translate to: [[d#(s(_x0), s(_x1))]] = 9 + 9x1 > 3x1 = [[d#(s(_x0), sub(_x1, _x0))]] [[sub(_x0, 0)]] = x0 >= x0 = [[_x0]] [[sub(s(_x0), s(_x1))]] = 3 + 3x0 >= x0 = [[sub(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(gtr#) = 1 Thus, we can orient the dependency pairs as follows: nu(gtr#(s(X), s(Y))) = s(X) |> X = nu(gtr#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(sub#) = 1 Thus, we can orient the dependency pairs as follows: nu(sub#(s(X), s(Y))) = s(X) |> X = nu(sub#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.