We consider the system Applicative_first_order_05__#3.18. Alphabet: 0 : [] --> a cons : [c * d] --> d double : [a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d minus : [a * a] --> a nil : [] --> d plus : [a * a] --> a s : [a] --> a true : [] --> b Rules: minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) double(0) => 0 double(s(x)) => s(s(double(x))) plus(0, x) => x plus(s(x), y) => s(plus(x, y)) plus(s(x), y) => plus(x, s(y)) plus(s(x), y) => s(plus(minus(x, y), double(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 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] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] double#(s(X)) =#> double#(X) 2] plus#(s(X), Y) =#> plus#(X, Y) 3] plus#(s(X), Y) =#> plus#(X, s(Y)) 4] plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y)) 5] plus#(s(X), Y) =#> minus#(X, Y) 6] plus#(s(X), Y) =#> double#(Y) 7] map#(F, cons(X, Y)) =#> F(X) 8] map#(F, cons(X, Y)) =#> map#(F, Y) 9] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 10] filter#(F, cons(X, Y)) =#> F(X) 11] filter2#(true, F, X, Y) =#> filter#(F, Y) 12] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) double(0) => 0 double(s(X)) => s(s(double(X))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) plus(s(X), Y) => plus(X, s(Y)) plus(s(X), Y) => s(plus(minus(X, Y), double(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) 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 : 2, 3, 4, 5, 6 * 3 : 2, 3, 4, 5, 6 * 4 : 2, 3, 4, 5, 6 * 5 : 0 * 6 : 1 * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 8 : 7, 8 * 9 : 11, 12 * 10 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 11 : 9, 10 * 12 : 9, 10 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: double#(s(X)) =#> double#(X) P_3: plus#(s(X), Y) =#> plus#(X, Y) plus#(s(X), Y) =#> plus#(X, s(Y)) plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y)) P_4: 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), (P_2, R_0, m, f), (P_3, R_0, m, f) and (P_4, 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) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). The formative rules of (P_4, R_0) are R_1 ::= 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_4, R_0, minimal, formative) by (P_4, R_1, minimal, formative). 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_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, 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) 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: cons = \y0y1.2 + y1 + 2y0 false = 3 filter = \G0y1.y1 + 2y1G0(y1) filter2 = \y0G1y2y3.2 + y3 + 2y2 + 2y3G1(y3) filter2# = \y0G1y2y3.y3 + 2y3G1(y3) + 2G1(y3) filter# = \G0y1.y1 + 2y1G0(y1) + 2G0(y1) map = \G0y1.2y1 + y1G0(y1) map# = \G0y1.3 + 2G0(y1) + y1G0(y1) true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= 3 + 2F0(x2) + x2F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) > x2 + 2x2F0(x2) + 2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) + 2F0(x2) >= x2 + 2x2F0(x2) + 2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) + 2F0(x2) >= x2 + 2x2F0(x2) + 2F0(x2) = [[filter#(_F0, _x2)]] [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= 2 + 2x2 + x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) >= 2 + x2 + 2x1 + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 + 2x1 + 2x2F0(x2) >= 2 + x2 + 2x1 + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x2 + 2x1 + 2x2F0(x2) >= x2 + 2x2F0(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_4, R_1, minimal, formative) by (P_5, R_1, minimal, formative), where P_5 consists of: map#(F, cons(X, Y)) =#> map#(F, Y) 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_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_5, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_5, 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 : * 2 : This graph has the following strongly connected components: P_6: map#(F, cons(X, Y)) =#> map#(F, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_5, R_1, m, f) by (P_6, R_1, 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) and (P_6, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_6, 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_6, 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 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). The formative rules of (P_3, R_0) are R_2 ::= minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) double(0) => 0 double(s(X)) => s(s(double(X))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) plus(s(X), Y) => plus(X, s(Y)) plus(s(X), Y) => s(plus(minus(X, Y), double(Y))) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_3, R_2, minimal, formative). 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_2, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_2, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_3, R_2) are: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) double(0) => 0 double(s(X)) => s(s(double(X))) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: plus#(s(X), Y) >? plus#(X, Y) plus#(s(X), Y) >? plus#(X, s(Y)) plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 double = \y0.2 + 2y0 minus = \y0y1.y0 plus# = \y0y1.2y0 s = \y0.2 + y0 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 4 + 2x0 > 2x0 = [[plus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 4 + 2x0 > 2x0 = [[plus#(_x0, s(_x1))]] [[plus#(s(_x0), _x1)]] = 4 + 2x0 > 2x0 = [[plus#(minus(_x0, _x1), double(_x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 2 + x0 >= x0 = [[minus(_x0, _x1)]] [[double(0)]] = 2 >= 0 = [[0]] [[double(s(_x0))]] = 6 + 2x0 >= 6 + 2x0 = [[s(s(double(_x0)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_2) by ({}, R_2). 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(double#) = 1 Thus, we can orient the dependency pairs as follows: nu(double#(s(X))) = s(X) |> X = nu(double#(X)) 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(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(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.