We consider the system Applicative_first_order_05__perfect. Alphabet: 0 : [] --> b cons : [d * e] --> e f : [b * b * b * b] --> c false : [] --> c filter : [d -> c * e] --> e filter2 : [c * d -> c * d * e] --> e if : [a * c * c] --> c le : [b * b] --> a map : [d -> d * e] --> e minus : [b * b] --> b nil : [] --> e perfectp : [b] --> c s : [b] --> b true : [] --> c Rules: perfectp(0) => false perfectp(s(x)) => f(x, s(0), s(x), s(x)) f(0, x, 0, y) => true f(0, x, s(y), z) => false f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) map(g, nil) => nil map(g, cons(x, y)) => cons(g x, map(g, y)) filter(g, nil) => nil filter(g, cons(x, y)) => filter2(g x, g, x, y) filter2(true, g, x, y) => cons(x, filter(g, y)) filter2(false, g, x, y) => filter(g, 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] perfectp#(s(X)) =#> f#(X, s(0), s(X), s(X)) 1] f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) 2] f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) 3] f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) 4] map#(F, cons(X, Y)) =#> F(X) 5] map#(F, cons(X, Y)) =#> map#(F, Y) 6] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 7] filter#(F, cons(X, Y)) =#> F(X) 8] filter2#(true, F, X, Y) =#> filter#(F, Y) 9] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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) 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 : 2, 3 * 1 : 1, 2, 3 * 2 : * 3 : 1, 2, 3 * 4 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 5 : 4, 5 * 6 : 8, 9 * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 8 : 6, 7 * 9 : 6, 7 This graph has the following strongly connected components: P_1: f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) P_2: map#(F, cons(X, Y)) =#> F(X) map#(F, cons(X, Y)) =#> map#(F, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, 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) and (P_2, R_0, m, f). 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). The formative rules of (P_2, R_0) are R_1 ::= perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 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) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, 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: map#(F, cons(X, Y)) >? F(X) map#(F, cons(X, Y)) >? map#(F, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) perfectp(0) >= false perfectp(s(X)) >= f(X, s(0), s(X), s(X)) f(0, X, 0, Y) >= true f(0, X, s(Y), Z) >= false f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 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 orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.2 + 2y0 + 2y1 f = \y0y1y2y3.0 false = 0 filter = \G0y1.y1 filter2 = \y0G1y2y3.2 + 2y2 + 2y3 filter2# = \y0G1y2y3.2y3G1(y3) filter# = \G0y1.2y1G0(y1) map = \G0y1.y1 + 2y1G0(y1) map# = \G0y1.3 + 2G0(y1) minus = \y0y1.0 perfectp = \y0.3 s = \y0.0 true = 0 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 3 + 2F0(2 + 2x1 + 2x2) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 3 + 2F0(2 + 2x1 + 2x2) >= 3 + 2F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 4x1F0(2 + 2x1 + 2x2) + 4x2F0(2 + 2x1 + 2x2) + 4F0(2 + 2x1 + 2x2) >= 2x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 4x1F0(2 + 2x1 + 2x2) + 4x2F0(2 + 2x1 + 2x2) + 4F0(2 + 2x1 + 2x2) >= F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2F0(x2) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2F0(x2) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] [[perfectp(0)]] = 3 >= 0 = [[false]] [[perfectp(s(_x0))]] = 3 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + 4x1F0(2 + 2x1 + 2x2) + 4x2F0(2 + 2x1 + 2x2) + 4F0(2 + 2x1 + 2x2) >= 2 + 2x2 + 4x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 >= 2 + 2x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 >= 2 + 2x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 >= x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_1, minimal, formative) by (P_3, R_1, minimal, formative), where P_3 consists of: map#(F, cons(X, Y)) =#> map#(F, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_3, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_1, 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 : 3, 4 * 2 : 0, 1, 2, 3, 4 * 3 : 1, 2 * 4 : 1, 2 This graph has the following strongly connected components: P_4: map#(F, cons(X, Y)) =#> map#(F, Y) P_5: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_3, R_1, m, f) by (P_4, R_1, m, f) and (P_5, R_1, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_4, R_1, minimal, formative) and (P_5, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_1, 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)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) perfectp(0) >= false perfectp(s(X)) >= f(X, s(0), s(X), s(X)) f(0, X, 0, Y) >= true f(0, X, s(Y), Z) >= false f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 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 orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.2 + y0 + y1 f = \y0y1y2y3.2 false = 0 filter = \G0y1.2y1 filter2 = \y0G1y2y3.2 + y2 + 2y3 filter2# = \y0G1y2y3.2y3 + y3G1(y3) filter# = \G0y1.2y1 + y1G0(y1) map = \G0y1.y1 + 2y1G0(y1) minus = \y0y1.0 perfectp = \y0.3 s = \y0.0 true = 0 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > 2x2 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2 + x2F0(x2) >= 2x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2 + x2F0(x2) >= 2x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[perfectp(0)]] = 3 >= 0 = [[false]] [[perfectp(s(_x0))]] = 3 >= 2 = [[f(_x0, s(0), s(_x0), s(_x0))]] [[f(0, _x0, 0, _x1)]] = 2 >= 0 = [[true]] [[f(0, _x0, s(_x1), _x2)]] = 2 >= 0 = [[false]] [[f(s(_x0), 0, _x1, _x2)]] = 2 >= 2 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 4F0(2 + x1 + x2) >= 2 + x2 + 2x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 >= 2 + x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + 2x2 >= 2 + x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + 2x2 >= 2x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_1, minimal, formative) by (P_6, R_1, minimal, formative), where P_6 consists of: filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_4, R_1, minimal, formative) and (P_6, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_1, 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 : * 1 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_4, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_1, minimal, formative). We apply the subterm criterion with the following projection function: nu(map#) = 2 Thus, we can orient the dependency pairs as follows: nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_1, minimal, f) by ({}, R_1, 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(f#) = 1 Thus, we can orient the dependency pairs as follows: nu(f#(s(X), 0, Y, Z)) = s(X) |> X = nu(f#(X, Z, minus(Y, s(X)), Z)) nu(f#(s(X), s(Y), Z, U)) = s(X) |> X = nu(f#(X, U, Z, U)) 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.