We consider the system AotoYamada_05__028. Alphabet: cons : [a * c] --> c consif : [b * a * c] --> c false : [] --> b filter : [a -> b * c] --> c nil : [] --> c true : [] --> b Rules: consif(true, x, y) => cons(x, y) consif(false, x, y) => y filter(f, nil) => nil filter(f, cons(x, y)) => consif(f x, x, filter(f, y)) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: consif(true, X, Y) => cons(X, Y) consif(false, X, Y) => 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: || || consif(true, %X, %Y) -> cons(%X, %Y) || consif(false, %X, %Y) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(cons(x_1, x_2)) = 1 + x_1 + x_2 || POL(consif(x_1, x_2, x_3)) = 2 + x_1 + 2*x_2 + 2*x_3 || POL(false) = 1 || POL(true) = 2 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || consif(true, %X, %Y) -> cons(%X, %Y) || consif(false, %X, %Y) -> %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] filter#(F, cons(X, Y)) =#> consif#(F X, X, filter(F, Y)) 1] filter#(F, cons(X, Y)) =#> F(X) 2] filter#(F, cons(X, Y)) =#> filter#(F, Y) Rules R_0: consif(true, X, Y) => cons(X, Y) consif(false, X, Y) => Y filter(F, nil) => nil filter(F, cons(X, Y)) => consif(F X, X, 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 : * 1 : 0, 1, 2 * 2 : 0, 1, 2 This graph has the following strongly connected components: P_1: filter#(F, cons(X, Y)) =#> F(X) 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). 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 ::= consif(true, X, Y) => cons(X, Y) consif(false, X, Y) => Y filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) 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: filter#(F, cons(X, Y)) >? F(X) filter#(F, cons(X, Y)) >? filter#(F, Y) consif(true, X, Y) >= cons(X, Y) consif(false, X, Y) >= Y filter(F, cons(X, Y)) >= consif(F X, X, filter(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.y0 + y1 consif = \y0y1y2.y1 + y2 false = 0 filter = \G0y1.y1 filter# = \G0y1.3 + 2G0(y1) true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 3 + 2F0(x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filter#(_F0, cons(_x1, _x2))]] = 3 + 2F0(x1 + x2) >= 3 + 2F0(x2) = [[filter#(_F0, _x2)]] [[consif(true, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] [[consif(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] [[filter(_F0, cons(_x1, _x2))]] = x1 + x2 >= x1 + x2 = [[consif(_F0 _x1, _x1, filter(_F0, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of: filter#(F, cons(X, Y)) =#> filter#(F, Y) Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, 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 [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_1, minimal, f) by ({}, R_1, 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 +++ [FuhKop19] C. Fuhs, and C. Kop. A static higher-order dependency pair framework. In Proceedings of ESOP 2019, 2019. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.