1.90/0.95 YES 1.90/0.95 We consider the system theBenchmark. 1.90/0.95 1.90/0.95 Alphabet: 1.90/0.95 1.90/0.95 0 : [] --> c 1.90/0.95 cons : [a * b] --> b 1.90/0.95 div : [c * c] --> c 1.90/0.95 map : [a -> a * b] --> b 1.90/0.95 minus : [c * c] --> c 1.90/0.95 nil : [] --> b 1.90/0.95 s : [c] --> c 1.90/0.95 1.90/0.95 Rules: 1.90/0.95 1.90/0.95 map(f, nil) => nil 1.90/0.95 map(f, cons(x, y)) => cons(f x, map(f, y)) 1.90/0.95 minus(x, 0) => x 1.90/0.95 minus(s(x), s(y)) => minus(x, y) 1.90/0.95 div(0, s(x)) => 0 1.90/0.95 div(s(x), s(y)) => s(div(minus(x, y), s(y))) 1.90/0.95 1.90/0.95 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.90/0.95 1.90/0.95 We observe that the rules contain a first-order subset: 1.90/0.95 1.90/0.95 minus(X, 0) => X 1.90/0.95 minus(s(X), s(Y)) => minus(X, Y) 1.90/0.95 div(0, s(X)) => 0 1.90/0.95 div(s(X), s(Y)) => s(div(minus(X, Y), s(Y))) 1.90/0.95 1.90/0.95 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. 1.90/0.95 1.90/0.95 According to the external first-order termination prover, this system is indeed terminating: 1.90/0.95 1.90/0.95 || proof of resources/system.trs 1.90/0.95 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.90/0.95 || 1.90/0.95 || 1.90/0.95 || Termination w.r.t. Q of the given QTRS could be proven: 1.90/0.95 || 1.90/0.95 || (0) QTRS 1.90/0.95 || (1) Overlay + Local Confluence [EQUIVALENT] 1.90/0.95 || (2) QTRS 1.90/0.95 || (3) DependencyPairsProof [EQUIVALENT] 1.90/0.95 || (4) QDP 1.90/0.95 || (5) DependencyGraphProof [EQUIVALENT] 1.90/0.95 || (6) AND 1.90/0.95 || (7) QDP 1.90/0.95 || (8) UsableRulesProof [EQUIVALENT] 1.90/0.95 || (9) QDP 1.90/0.95 || (10) QReductionProof [EQUIVALENT] 1.90/0.95 || (11) QDP 1.90/0.95 || (12) QDPSizeChangeProof [EQUIVALENT] 1.90/0.95 || (13) YES 1.90/0.95 || (14) QDP 1.90/0.95 || (15) UsableRulesProof [EQUIVALENT] 1.90/0.95 || (16) QDP 1.90/0.95 || (17) QReductionProof [EQUIVALENT] 1.90/0.95 || (18) QDP 1.90/0.95 || (19) QDPOrderProof [EQUIVALENT] 1.90/0.95 || (20) QDP 1.90/0.95 || (21) PisEmptyProof [EQUIVALENT] 1.90/0.95 || (22) YES 1.90/0.95 || 1.90/0.95 || 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (0) 1.90/0.95 || Obligation: 1.90/0.95 || Q restricted rewrite system: 1.90/0.95 || The TRS R consists of the following rules: 1.90/0.95 || 1.90/0.95 || minus(%X, 0) -> %X 1.90/0.95 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.95 || div(0, s(%X)) -> 0 1.90/0.95 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) 1.90/0.95 || 1.90/0.95 || Q is empty. 1.90/0.95 || 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (1) Overlay + Local Confluence (EQUIVALENT) 1.90/0.95 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (2) 1.90/0.95 || Obligation: 1.90/0.95 || Q restricted rewrite system: 1.90/0.95 || The TRS R consists of the following rules: 1.90/0.95 || 1.90/0.95 || minus(%X, 0) -> %X 1.90/0.95 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.95 || div(0, s(%X)) -> 0 1.90/0.95 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) 1.90/0.95 || 1.90/0.95 || The set Q consists of the following terms: 1.90/0.95 || 1.90/0.95 || minus(x0, 0) 1.90/0.95 || minus(s(x0), s(x1)) 1.90/0.95 || div(0, s(x0)) 1.90/0.95 || div(s(x0), s(x1)) 1.90/0.95 || 1.90/0.95 || 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (3) DependencyPairsProof (EQUIVALENT) 1.90/0.95 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (4) 1.90/0.95 || Obligation: 1.90/0.95 || Q DP problem: 1.90/0.95 || The TRS P consists of the following rules: 1.90/0.95 || 1.90/0.95 || MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) 1.90/0.95 || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) 1.90/0.95 || DIV(s(%X), s(%Y)) -> MINUS(%X, %Y) 1.90/0.95 || 1.90/0.95 || The TRS R consists of the following rules: 1.90/0.95 || 1.90/0.95 || minus(%X, 0) -> %X 1.90/0.95 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.95 || div(0, s(%X)) -> 0 1.90/0.95 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) 1.90/0.95 || 1.90/0.95 || The set Q consists of the following terms: 1.90/0.95 || 1.90/0.95 || minus(x0, 0) 1.90/0.95 || minus(s(x0), s(x1)) 1.90/0.95 || div(0, s(x0)) 1.90/0.95 || div(s(x0), s(x1)) 1.90/0.95 || 1.90/0.95 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (5) DependencyGraphProof (EQUIVALENT) 1.90/0.95 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (6) 1.90/0.95 || Complex Obligation (AND) 1.90/0.95 || 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (7) 1.90/0.95 || Obligation: 1.90/0.95 || Q DP problem: 1.90/0.95 || The TRS P consists of the following rules: 1.90/0.95 || 1.90/0.95 || MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) 1.90/0.95 || 1.90/0.95 || The TRS R consists of the following rules: 1.90/0.95 || 1.90/0.95 || minus(%X, 0) -> %X 1.90/0.95 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.95 || div(0, s(%X)) -> 0 1.90/0.95 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) 1.90/0.95 || 1.90/0.95 || The set Q consists of the following terms: 1.90/0.95 || 1.90/0.95 || minus(x0, 0) 1.90/0.95 || minus(s(x0), s(x1)) 1.90/0.95 || div(0, s(x0)) 1.90/0.95 || div(s(x0), s(x1)) 1.90/0.95 || 1.90/0.95 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (8) UsableRulesProof (EQUIVALENT) 1.90/0.95 || 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. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (9) 1.90/0.95 || Obligation: 1.90/0.95 || Q DP problem: 1.90/0.95 || The TRS P consists of the following rules: 1.90/0.95 || 1.90/0.95 || MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) 1.90/0.95 || 1.90/0.95 || R is empty. 1.90/0.95 || The set Q consists of the following terms: 1.90/0.95 || 1.90/0.95 || minus(x0, 0) 1.90/0.95 || minus(s(x0), s(x1)) 1.90/0.95 || div(0, s(x0)) 1.90/0.95 || div(s(x0), s(x1)) 1.90/0.95 || 1.90/0.95 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (10) QReductionProof (EQUIVALENT) 1.90/0.95 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 1.90/0.95 || 1.90/0.95 || minus(x0, 0) 1.90/0.95 || minus(s(x0), s(x1)) 1.90/0.95 || div(0, s(x0)) 1.90/0.95 || div(s(x0), s(x1)) 1.90/0.95 || 1.90/0.95 || 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (11) 1.90/0.95 || Obligation: 1.90/0.95 || Q DP problem: 1.90/0.95 || The TRS P consists of the following rules: 1.90/0.95 || 1.90/0.95 || MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) 1.90/0.95 || 1.90/0.95 || R is empty. 1.90/0.95 || Q is empty. 1.90/0.95 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (12) QDPSizeChangeProof (EQUIVALENT) 1.90/0.95 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 1.90/0.95 || 1.90/0.95 || From the DPs we obtained the following set of size-change graphs: 1.90/0.95 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) 1.90/0.95 || The graph contains the following edges 1 > 1, 2 > 2 1.90/0.95 || 1.90/0.95 || 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (13) 1.90/0.95 || YES 1.90/0.95 || 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (14) 1.90/0.95 || Obligation: 1.90/0.95 || Q DP problem: 1.90/0.95 || The TRS P consists of the following rules: 1.90/0.95 || 1.90/0.95 || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) 1.90/0.95 || 1.90/0.95 || The TRS R consists of the following rules: 1.90/0.95 || 1.90/0.95 || minus(%X, 0) -> %X 1.90/0.95 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.95 || div(0, s(%X)) -> 0 1.90/0.95 || div(s(%X), s(%Y)) -> s(div(minus(%X, %Y), s(%Y))) 1.90/0.95 || 1.90/0.95 || The set Q consists of the following terms: 1.90/0.95 || 1.90/0.95 || minus(x0, 0) 1.90/0.95 || minus(s(x0), s(x1)) 1.90/0.95 || div(0, s(x0)) 1.90/0.95 || div(s(x0), s(x1)) 1.90/0.95 || 1.90/0.95 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.95 || ---------------------------------------- 1.90/0.95 || 1.90/0.95 || (15) UsableRulesProof (EQUIVALENT) 1.90/0.96 || 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. 1.90/0.96 || ---------------------------------------- 1.90/0.96 || 1.90/0.96 || (16) 1.90/0.96 || Obligation: 1.90/0.96 || Q DP problem: 1.90/0.96 || The TRS P consists of the following rules: 1.90/0.96 || 1.90/0.96 || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) 1.90/0.96 || 1.90/0.96 || The TRS R consists of the following rules: 1.90/0.96 || 1.90/0.96 || minus(%X, 0) -> %X 1.90/0.96 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.96 || 1.90/0.96 || The set Q consists of the following terms: 1.90/0.96 || 1.90/0.96 || minus(x0, 0) 1.90/0.96 || minus(s(x0), s(x1)) 1.90/0.96 || div(0, s(x0)) 1.90/0.96 || div(s(x0), s(x1)) 1.90/0.96 || 1.90/0.96 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.96 || ---------------------------------------- 1.90/0.96 || 1.90/0.96 || (17) QReductionProof (EQUIVALENT) 1.90/0.96 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 1.90/0.96 || 1.90/0.96 || div(0, s(x0)) 1.90/0.96 || div(s(x0), s(x1)) 1.90/0.96 || 1.90/0.96 || 1.90/0.96 || ---------------------------------------- 1.90/0.96 || 1.90/0.96 || (18) 1.90/0.96 || Obligation: 1.90/0.96 || Q DP problem: 1.90/0.96 || The TRS P consists of the following rules: 1.90/0.96 || 1.90/0.96 || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) 1.90/0.96 || 1.90/0.96 || The TRS R consists of the following rules: 1.90/0.96 || 1.90/0.96 || minus(%X, 0) -> %X 1.90/0.96 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.96 || 1.90/0.96 || The set Q consists of the following terms: 1.90/0.96 || 1.90/0.96 || minus(x0, 0) 1.90/0.96 || minus(s(x0), s(x1)) 1.90/0.96 || 1.90/0.96 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.96 || ---------------------------------------- 1.90/0.96 || 1.90/0.96 || (19) QDPOrderProof (EQUIVALENT) 1.90/0.96 || We use the reduction pair processor [LPAR04,JAR06]. 1.90/0.96 || 1.90/0.96 || 1.90/0.96 || The following pairs can be oriented strictly and are deleted. 1.90/0.96 || 1.90/0.96 || DIV(s(%X), s(%Y)) -> DIV(minus(%X, %Y), s(%Y)) 1.90/0.96 || The remaining pairs can at least be oriented weakly. 1.90/0.96 || Used ordering: Polynomial interpretation [POLO]: 1.90/0.96 || 1.90/0.96 || POL(0) = 0 1.90/0.96 || POL(DIV(x_1, x_2)) = x_1 1.90/0.96 || POL(minus(x_1, x_2)) = x_1 1.90/0.96 || POL(s(x_1)) = 1 + x_1 1.90/0.96 || 1.90/0.96 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 1.90/0.96 || 1.90/0.96 || minus(%X, 0) -> %X 1.90/0.96 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.96 || 1.90/0.96 || 1.90/0.96 || ---------------------------------------- 1.90/0.96 || 1.90/0.96 || (20) 1.90/0.96 || Obligation: 1.90/0.96 || Q DP problem: 1.90/0.96 || P is empty. 1.90/0.96 || The TRS R consists of the following rules: 1.90/0.96 || 1.90/0.96 || minus(%X, 0) -> %X 1.90/0.96 || minus(s(%X), s(%Y)) -> minus(%X, %Y) 1.90/0.96 || 1.90/0.96 || The set Q consists of the following terms: 1.90/0.96 || 1.90/0.96 || minus(x0, 0) 1.90/0.96 || minus(s(x0), s(x1)) 1.90/0.96 || 1.90/0.96 || We have to consider all minimal (P,Q,R)-chains. 1.90/0.96 || ---------------------------------------- 1.90/0.96 || 1.90/0.96 || (21) PisEmptyProof (EQUIVALENT) 1.90/0.96 || The TRS P is empty. Hence, there is no (P,Q,R) chain. 1.90/0.96 || ---------------------------------------- 1.90/0.96 || 1.90/0.96 || (22) 1.90/0.96 || YES 1.90/0.96 || 1.90/0.96 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 [Kop13]). 1.90/0.96 1.90/0.96 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 1.90/0.96 1.90/0.96 Dependency Pairs P_0: 1.90/0.96 1.90/0.96 0] map#(F, cons(X, Y)) =#> map#(F, Y) 1.90/0.96 1.90/0.96 Rules R_0: 1.90/0.96 1.90/0.96 map(F, nil) => nil 1.90/0.96 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 1.90/0.96 minus(X, 0) => X 1.90/0.96 minus(s(X), s(Y)) => minus(X, Y) 1.90/0.96 div(0, s(X)) => 0 1.90/0.96 div(s(X), s(Y)) => s(div(minus(X, Y), s(Y))) 1.90/0.96 1.90/0.96 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 1.90/0.96 1.90/0.96 We consider the dependency pair problem (P_0, R_0, static, formative). 1.90/0.96 1.90/0.96 We apply the subterm criterion with the following projection function: 1.90/0.96 1.90/0.96 nu(map#) = 2 1.90/0.96 1.90/0.96 Thus, we can orient the dependency pairs as follows: 1.90/0.96 1.90/0.96 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 1.90/0.96 1.90/0.96 By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_0, R_0, static, f) by ({}, R_0, static, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 1.90/0.96 1.90/0.96 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 1.90/0.96 1.90/0.96 1.90/0.96 +++ Citations +++ 1.90/0.96 1.90/0.96 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 1.90/0.96 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 1.90/0.96 [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. 1.90/0.96 EOF