We consider the system AotoYamada_05__004. Alphabet: 0 : [] --> b cons : [a * b] --> b nil : [] --> b plus : [b * b] --> b s : [b] --> b sumwith : [a -> b * b] --> b Rules: plus(0, x) => x plus(s(x), y) => s(plus(x, y)) sumwith(f, nil) => nil sumwith(f, cons(x, y)) => plus(f x, sumwith(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: 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: || Knuth-Bendix order [KBO] with precedence:plus_2 > s_1 > 0 || || and weight map: || || 0=1 || s_1=1 || plus_2=0 || || The variable weight is 1With 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] sumwith#(F, cons(X, Y)) =#> plus#(F X, sumwith(F, Y)) 1] sumwith#(F, cons(X, Y)) =#> F(X) 2] sumwith#(F, cons(X, Y)) =#> sumwith#(F, Y) Rules R_0: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) sumwith(F, nil) => nil sumwith(F, cons(X, Y)) => plus(F X, sumwith(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: sumwith#(F, cons(X, Y)) =#> F(X) sumwith#(F, cons(X, Y)) =#> sumwith#(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 ::= plus(0, X) => X sumwith(F, cons(X, Y)) => plus(F X, sumwith(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: sumwith#(F, cons(X, Y)) >? F(X) sumwith#(F, cons(X, Y)) >? sumwith#(F, Y) plus(0, X) >= X sumwith(F, cons(X, Y)) >= plus(F X, sumwith(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.3 + y0 + y1 plus = \y0y1.y1 sumwith = \G0y1.0 sumwith# = \G0y1.3 + 2y1G0(y1) Using this interpretation, the requirements translate to: [[sumwith#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 6F0(3 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[sumwith#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 6F0(3 + x1 + x2) >= 3 + 2x2F0(x2) = [[sumwith#(_F0, _x2)]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[sumwith(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[plus(_F0 _x1, sumwith(_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: sumwith#(F, cons(X, Y)) =#> sumwith#(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(sumwith#) = 2 Thus, we can orient the dependency pairs as follows: nu(sumwith#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(sumwith#(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.