We consider the system Applicative_05__mapDivMinusHard. Alphabet: 0 : [] --> c cons : [a * b] --> b div : [c * c] --> c map : [a -> a * b] --> b minus : [c * c] --> c nil : [] --> b p : [c] --> c s : [c] --> c Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) minus(x, 0) => x minus(s(x), s(y)) => minus(p(s(x)), p(s(y))) p(s(x)) => x div(0, s(x)) => 0 div(s(x), s(y)) => s(div(minus(x, y), s(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: minus(X, 0) => X minus(s(X), s(Y)) => minus(p(s(X)), p(s(Y))) p(s(X)) => X div(0, s(X)) => 0 div(s(X), s(Y)) => s(div(minus(X, Y), s(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) Overlay + Local Confluence [EQUIVALENT] || (2) QTRS || (3) DependencyPairsProof [EQUIVALENT] || (4) QDP || (5) DependencyGraphProof [EQUIVALENT] || (6) AND || (7) QDP || (8) UsableRulesProof [EQUIVALENT] || (9) QDP || (10) QReductionProof [EQUIVALENT] || (11) QDP || (12) MRRProof [EQUIVALENT] || (13) QDP || (14) DependencyGraphProof [EQUIVALENT] || (15) TRUE || (16) QDP || (17) UsableRulesProof [EQUIVALENT] || (18) QDP || (19) QReductionProof [EQUIVALENT] || (20) QDP || (21) QDPOrderProof [EQUIVALENT] || (22) QDP || (23) PisEmptyProof [EQUIVALENT] || (24) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || div(0, s(%X)) -> 0 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) || || Q is empty. || || ---------------------------------------- || || (1) Overlay + Local Confluence (EQUIVALENT) || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || div(0, s(%X)) -> 0 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || div(0, s(x0)) || div(s(x0), s(x1)) || || || ---------------------------------------- || || (3) DependencyPairsProof (EQUIVALENT) || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. || ---------------------------------------- || || (4) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(p(s(%X)), p(s(%Y))) || MINUS(s(%X), s(%Y)) -> P(s(%X)) || MINUS(s(%X), s(%Y)) -> P(s(%Y)) || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) || DIV(s(%X), s(%Y)) -> MINUS(%X, %Y) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || div(0, s(%X)) -> 0 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || div(0, s(x0)) || div(s(x0), s(x1)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (5) DependencyGraphProof (EQUIVALENT) || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes. || ---------------------------------------- || || (6) || Complex Obligation (AND) || || ---------------------------------------- || || (7) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(p(s(%X)), p(s(%Y))) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || div(0, s(%X)) -> 0 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || div(0, s(x0)) || div(s(x0), s(x1)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (8) UsableRulesProof (EQUIVALENT) || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. || ---------------------------------------- || || (9) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(p(s(%X)), p(s(%Y))) || || The TRS R consists of the following rules: || || p(s(%X)) -> %X || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || div(0, s(x0)) || div(s(x0), s(x1)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (10) QReductionProof (EQUIVALENT) || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. || || minus(x0, 0) || minus(s(x0), s(x1)) || div(0, s(x0)) || div(s(x0), s(x1)) || || || ---------------------------------------- || || (11) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(p(s(%X)), p(s(%Y))) || || The TRS R consists of the following rules: || || p(s(%X)) -> %X || || The set Q consists of the following terms: || || p(s(x0)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (12) MRRProof (EQUIVALENT) || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. || || || Strictly oriented rules of the TRS R: || || p(s(%X)) -> %X || || Used ordering: Polynomial interpretation [POLO]: || || POL(MINUS(x_1, x_2)) = 2*x_1 + 2*x_2 || POL(p(x_1)) = x_1 || POL(s(x_1)) = 2 + x_1 || || || ---------------------------------------- || || (13) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(p(s(%X)), p(s(%Y))) || || R is empty. || The set Q consists of the following terms: || || p(s(x0)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (14) DependencyGraphProof (EQUIVALENT) || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. || ---------------------------------------- || || (15) || TRUE || || ---------------------------------------- || || (16) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || div(0, s(%X)) -> 0 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || div(0, s(x0)) || div(s(x0), s(x1)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (17) UsableRulesProof (EQUIVALENT) || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. || ---------------------------------------- || || (18) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || div(0, s(x0)) || div(s(x0), s(x1)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (19) QReductionProof (EQUIVALENT) || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. || || div(0, s(x0)) || div(s(x0), s(x1)) || || || ---------------------------------------- || || (20) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (21) QDPOrderProof (EQUIVALENT) || We use the reduction pair processor [LPAR04,JAR06]. || || || The following pairs can be oriented strictly and are deleted. || || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) || The remaining pairs can at least be oriented weakly. || Used ordering: Polynomial interpretation [POLO]: || || POL(0) = 0 || POL(DIV(x_1, x_2)) = x_1 || POL(minus(x_1, x_2)) = x_1 || POL(p(x_1)) = x_1 || POL(s(x_1)) = 1 + x_1 || || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || || || ---------------------------------------- || || (22) || Obligation: || Q DP problem: || P is empty. || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(p(s(%X)), p(s(%Y))) || p(s(%X)) -> %X || || The set Q consists of the following terms: || || minus(x0, 0) || minus(s(x0), s(x1)) || p(s(x0)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (23) PisEmptyProof (EQUIVALENT) || The TRS P is empty. Hence, there is no (P,Q,R) chain. || ---------------------------------------- || || (24) || 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) Rules R_0: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) minus(X, 0) => X minus(s(X), s(Y)) => minus(p(s(X)), p(s(Y))) p(s(X)) => X div(0, s(X)) => 0 div(s(X), s(Y)) => s(div(minus(X, Y), s(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). The formative rules of (P_0, R_0) are R_1 ::= map(F, cons(X, Y)) => cons(F X, map(F, Y)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative). Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_0, 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) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.1 + y0 + 2y1 map = \G0y1.y1 + 2y1G0(y1) map# = \G0y1.3 + 2y1 + G0(y1) Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 5 + 2x1 + 4x2 + F0(1 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 5 + 2x1 + 4x2 + F0(1 + x1 + 2x2) > 3 + 2x2 + F0(x2) = [[map#(_F0, _x2)]] [[map(_F0, cons(_x1, _x2))]] = 1 + x1 + 2x2 + 2x1F0(1 + x1 + 2x2) + 2F0(1 + x1 + 2x2) + 4x2F0(1 + x1 + 2x2) >= 1 + 2x2 + 4x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, 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.