We consider the system AotoYamada_05__016. Alphabet: 0 : [] --> a cons : [a * c] --> c false : [] --> b filter : [a -> b] --> c -> c filtersub : [b * a -> b * c] --> c neq : [a] --> a -> b nil : [] --> c nonzero : [] --> c -> c s : [a] --> a true : [] --> b Rules: neq(0) 0 => false neq(0) s(x) => true neq(s(x)) 0 => true neq(s(x)) s(y) => neq(x) y filter(f) nil => nil filter(f) cons(x, y) => filtersub(f x, f, cons(x, y)) filtersub(true, f, cons(x, y)) => cons(x, filter(f) y) filtersub(false, f, cons(x, y)) => filter(f) y nonzero => filter(neq(0)) 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: neq(0) 0 => false neq(0) s(X) => true neq(s(X)) 0 => true neq(s(X)) s(Y) => neq(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: || || neq(0, 0) -> false || neq(0, s(%X)) -> true || neq(s(%X), 0) -> true || neq(s(%X), s(%Y)) -> neq(%X, %Y) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(0) = 0 || POL(false) = 1 || POL(neq(x_1, x_2)) = 2 + 2*x_1 + x_2 || POL(s(x_1)) = 1 + 2*x_1 || POL(true) = 0 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || neq(0, 0) -> false || neq(0, s(%X)) -> true || neq(s(%X), 0) -> true || neq(s(%X), s(%Y)) -> neq(%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] filter(F) cons(X, Y) =#> filtersub#(F X, F, cons(X, Y)) 1] filter(F) cons(X, Y) =#> F(X) 2] filtersub#(true, F, cons(X, Y)) =#> filter(F) Y 3] filtersub#(true, F, cons(X, Y)) =#> filter#(F) 4] filtersub#(false, F, cons(X, Y)) =#> filter(F) Y 5] filtersub#(false, F, cons(X, Y)) =#> filter#(F) 6] nonzero X =#> filter(neq(0)) X 7] nonzero# =#> filter#(neq(0)) 8] nonzero# =#> neq#(0) Rules R_0: neq(0) 0 => false neq(0) s(X) => true neq(s(X)) 0 => true neq(s(X)) s(Y) => neq(X) Y filter(F) nil => nil filter(F) cons(X, Y) => filtersub(F X, F, cons(X, Y)) filtersub(true, F, cons(X, Y)) => cons(X, filter(F) Y) filtersub(false, F, cons(X, Y)) => filter(F) Y nonzero => filter(neq(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 : 2, 3, 4, 5 * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8 * 2 : 0, 1 * 3 : * 4 : 0, 1 * 5 : * 6 : 0, 1 * 7 : * 8 : This graph has the following strongly connected components: P_1: filter(F) cons(X, Y) =#> filtersub#(F X, F, cons(X, Y)) filter(F) cons(X, Y) =#> F(X) filtersub#(true, F, cons(X, Y)) =#> filter(F) Y filtersub#(false, F, cons(X, Y)) =#> filter(F) Y nonzero X =#> filter(neq(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 ::= neq(0) 0 => false neq(0) s(X) => true neq(s(X)) 0 => true neq(s(X)) s(Y) => neq(X) Y filter(F) cons(X, Y) => filtersub(F X, F, cons(X, Y)) filtersub(true, F, cons(X, Y)) => cons(X, filter(F) Y) filtersub(false, F, cons(X, Y)) => filter(F) Y nonzero X => filter(neq(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: filter(F, cons(X, Y)) >? filtersub#(F X, F, cons(X, Y)) filter(F, cons(X, Y)) >? F(X) filtersub#(true, F, cons(X, Y)) >? filter(F, Y) filtersub#(false, F, cons(X, Y)) >? filter(F, Y) nonzero(X) >? filter(neq(0), X) neq(0) 0 >= false neq(0) s(X) >= true neq(s(X)) 0 >= true neq(s(X)) s(Y) >= neq(X) Y filter(F, cons(X, Y)) >= filtersub(F X, F, cons(X, Y)) filtersub(true, F, cons(X, Y)) >= cons(X, filter(F, Y)) filtersub(false, F, cons(X, Y)) >= filter(F, Y) nonzero(X) >= filter(neq(0), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( nonzero(X) ) = #argfun-nonzero#(filter(neq(0), X), filter(neq(0), X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: #argfun-nonzero# = \y0y1.3 + max(y0, y1) 0 = 3 cons = \y0y1.3 + y0 + y1 false = 0 filter = \G0y1.1 + y1 + G0(y1) filtersub = \y0G1y2.1 + y2 + G1(y2) filtersub# = \y0G1y2.1 + y2 + G1(y2) neq = \y0y1.3 nonzero = \y0.0 s = \y0.3 + 2y0 true = 0 Using this interpretation, the requirements translate to: [[filter(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(3 + x1 + x2) >= 4 + x1 + x2 + F0(3 + x1 + x2) = [[filtersub#(_F0 _x1, _F0, cons(_x1, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(3 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filtersub#(true, _F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(3 + x1 + x2) > 1 + x2 + F0(x2) = [[filter(_F0, _x2)]] [[filtersub#(false, _F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(3 + x1 + x2) > 1 + x2 + F0(x2) = [[filter(_F0, _x2)]] [[#argfun-nonzero#(filter(neq(0), _x0), filter(neq(0), _x0))]] = 7 + x0 > 4 + x0 = [[filter(neq(0), _x0)]] [[neq(0) 0]] = 3 >= 0 = [[false]] [[neq(0) s(_x0)]] = max(3, 3 + 2x0) >= 0 = [[true]] [[neq(s(_x0)) 0]] = 3 >= 0 = [[true]] [[neq(s(_x0)) s(_x1)]] = max(3, 3 + 2x1) >= max(3, x1) = [[neq(_x0) _x1]] [[filter(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(3 + x1 + x2) >= 4 + x1 + x2 + F0(3 + x1 + x2) = [[filtersub(_F0 _x1, _F0, cons(_x1, _x2))]] [[filtersub(true, _F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(3 + x1 + x2) >= 4 + x1 + x2 + F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filtersub(false, _F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(3 + x1 + x2) >= 1 + x2 + F0(x2) = [[filter(_F0, _x2)]] [[#argfun-nonzero#(filter(neq(0), _x0), filter(neq(0), _x0))]] = 7 + x0 >= 4 + x0 = [[filter(neq(0), _x0)]] 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) =#> filtersub#(F X, F, cons(X, 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 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.