We consider the system AotoYamada_05__007. Alphabet: 0 : [] --> b cons : [b * a] --> a inc : [] --> a -> a map : [b -> b] --> a -> a nil : [] --> a plus : [b] --> b -> b s : [b] --> b Rules: plus(0) x => x plus(s(x)) y => s(plus(x) y) map(f) nil => nil map(f) cons(x, y) => cons(f x, map(f) y) inc => map(plus(s(0))) 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] plus(s(X)) Y =#> plus(X) Y 1] plus(s(X)) Y =#> plus#(X) 2] map(F) cons(X, Y) =#> F(X) 3] map(F) cons(X, Y) =#> map(F) Y 4] map(F) cons(X, Y) =#> map#(F) 5] inc X =#> map(plus(s(0))) X 6] inc# =#> map#(plus(s(0))) 7] inc# =#> plus#(s(0)) Rules R_0: plus(0) X => X plus(s(X)) Y => s(plus(X) Y) map(F) nil => nil map(F) cons(X, Y) => cons(F X, map(F) Y) inc => map(plus(s(0))) 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 : 0, 1, 2, 3, 4, 5, 6, 7 * 3 : 2, 3, 4 * 4 : * 5 : 2, 3, 4 * 6 : * 7 : This graph has the following strongly connected components: P_1: plus(s(X)) Y =#> plus(X) Y P_2: map(F) cons(X, Y) =#> F(X) map(F) cons(X, Y) =#> map(F) Y inc X =#> map(plus(s(0))) X 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 ::= map(F) cons(X, Y) => cons(F X, map(F) Y) inc X => map(plus(s(0))) X 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) inc(X) >? map(plus(s(0)), X) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) inc(X) >= map(plus(s(0)), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( inc(X) ) = #argfun-inc#(map(plus(s(0)), X), map(plus(s(0)), X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: #argfun-inc# = \y0y1.3 + max(y0, y1) 0 = 0 cons = \y0y1.3 + y0 + y1 inc = \y0.0 map = \G0y1.2y1 + G0(y1) + 2G0(0) + y1G0(y1) plus = \y0y1.0 s = \y0.0 Using this interpretation, the requirements translate to: [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + 2F0(0) + 4F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + 2F0(0) + 4F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) > 2x2 + F0(x2) + 2F0(0) + x2F0(x2) = [[map(_F0, _x2)]] [[#argfun-inc#(map(plus(s(0)), _x0), map(plus(s(0)), _x0))]] = 3 + 2x0 > 2x0 = [[map(plus(s(0)), _x0)]] [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + 2F0(0) + 4F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) >= 3 + 2x2 + F0(x2) + 2F0(0) + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[#argfun-inc#(map(plus(s(0)), _x0), map(plus(s(0)), _x0))]] = 3 + 2x0 >= 2x0 = [[map(plus(s(0)), _x0)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_1) by ({}, R_1). 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(plus) = 1 Thus, we can orient the dependency pairs as follows: nu(plus(s(X)) Y) = s(X) |> X = nu(plus(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.