We consider the system AotoYamada_05__009. Alphabet: and : [c * c] --> c cons : [a * b] --> b false : [] --> c forall : [a -> c * b] --> c forsome : [a -> c * b] --> c nil : [] --> b or : [c * c] --> c true : [] --> c Rules: and(true, true) => true and(true, false) => false and(false, true) => false and(false, false) => false or(true, true) => true or(true, false) => true or(false, true) => true or(false, false) => false forall(f, nil) => true forall(f, cons(x, y)) => and(f x, forall(f, y)) forsome(f, nil) => false forsome(f, cons(x, y)) => or(f x, forsome(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: and(true, true) => true and(true, false) => false and(false, true) => false and(false, false) => false or(true, true) => true or(true, false) => true or(false, true) => true or(false, false) => false 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: || || and(true, true) -> true || and(true, false) -> false || and(false, true) -> false || and(false, false) -> false || or(true, true) -> true || or(true, false) -> true || or(false, true) -> true || or(false, false) -> false || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(and(x_1, x_2)) = 2*x_1 + 2*x_2 || POL(false) = 2 || POL(or(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 || POL(true) = 2 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || and(true, true) -> true || and(true, false) -> false || and(false, true) -> false || and(false, false) -> false || or(true, true) -> true || or(true, false) -> true || or(false, true) -> true || or(false, false) -> false || || || || || ---------------------------------------- || || (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 static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] forall#(F, cons(X, Y)) =#> and#(F X, forall(F, Y)) 1] forall#(F, cons(X, Y)) =#> forall#(F, Y) 2] forsome#(F, cons(X, Y)) =#> or#(F X, forsome(F, Y)) 3] forsome#(F, cons(X, Y)) =#> forsome#(F, Y) Rules R_0: and(true, true) => true and(true, false) => false and(false, true) => false and(false, false) => false or(true, true) => true or(true, false) => true or(false, true) => true or(false, false) => false forall(F, nil) => true forall(F, cons(X, Y)) => and(F X, forall(F, Y)) forsome(F, nil) => false forsome(F, cons(X, Y)) => or(F X, forsome(F, Y)) Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite. We consider the dependency pair problem (P_0, R_0, computable, 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 : * 3 : 2, 3 This graph has the following strongly connected components: P_1: forall#(F, cons(X, Y)) =#> forall#(F, Y) P_2: forsome#(F, cons(X, Y)) =#> forsome#(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) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_2, R_0, computable, formative) is finite. We consider the dependency pair problem (P_2, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(forsome#) = 2 Thus, we can orient the dependency pairs as follows: nu(forsome#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(forsome#(F, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by ({}, R_0, computable, 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, computable, formative) is finite. We consider the dependency pair problem (P_1, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(forall#) = 2 Thus, we can orient the dependency pairs as follows: nu(forall#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(forall#(F, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, 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. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.