We consider the system AotoYamada_05__011. Alphabet: 0 : [] --> b cons : [b * a] --> a curry : [b -> b -> b * b] --> b -> b 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) curry(f, x) y => f x y inc => map(curry(plus, s(0))) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). Symbol curry is an encoding for application that is only used in innocuous ways. We can simplify the program (without losing non-termination) by removing it. This gives: 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))) We observe that the rules contain a first-order subset: plus(0) X => X plus(s(X)) Y => s(plus(X) Y) Moreover, the system is orthogonal. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || proof of resources/system.trs || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 || || || Termination w.r.t. Q of the given QTRS could be proven: || || (0) QTRS || (1) QTRSRRRProof [EQUIVALENT] || (2) QTRS || (3) RisEmptyProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(0) = 2 || POL(plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 || POL(s(x_1)) = 1 + x_1 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (3) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. || ---------------------------------------- || || (4) || YES || 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] map(F) cons(X, Y) =#> F(X) 1] map(F) cons(X, Y) =#> map(F) Y 2] map(F) cons(X, Y) =#> map#(F) 3] inc X =#> map(plus(s(0))) X 4] inc# =#> map#(plus(s(0))) 5] 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, 2, 3, 4, 5 * 1 : 0, 1, 2 * 2 : * 3 : 0, 1, 2 * 4 : * 5 : This graph has the following strongly connected components: P_1: 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). 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). The formative rules of (P_1, 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_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, 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_1, R_1) by ({}, R_1). 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.