2.81/1.64 YES 2.81/1.64 We consider the system theBenchmark. 2.81/1.64 2.81/1.64 Alphabet: 2.81/1.64 2.81/1.64 0 : [] --> b 2.81/1.64 cons : [c * d] --> d 2.81/1.64 false : [] --> a 2.81/1.64 filter : [c -> a * d] --> d 2.81/1.64 filter2 : [a * c -> a * c * d] --> d 2.81/1.64 if : [a * b * b] --> b 2.81/1.64 le : [b * b] --> a 2.81/1.64 map : [c -> c * d] --> d 2.81/1.64 minus : [b * b] --> b 2.81/1.64 nil : [] --> d 2.81/1.64 p : [b] --> b 2.81/1.64 s : [b] --> b 2.81/1.64 true : [] --> a 2.81/1.64 2.81/1.64 Rules: 2.81/1.64 2.81/1.64 p(0) => 0 2.81/1.64 p(s(x)) => x 2.81/1.64 le(0, x) => true 2.81/1.64 le(s(x), 0) => false 2.81/1.64 le(s(x), s(y)) => le(x, y) 2.81/1.64 minus(x, y) => if(le(x, y), x, y) 2.81/1.64 if(true, x, y) => 0 2.81/1.64 if(false, x, y) => s(minus(p(x), y)) 2.81/1.64 map(f, nil) => nil 2.81/1.64 map(f, cons(x, y)) => cons(f x, map(f, y)) 2.81/1.64 filter(f, nil) => nil 2.81/1.64 filter(f, cons(x, y)) => filter2(f x, f, x, y) 2.81/1.64 filter2(true, f, x, y) => cons(x, filter(f, y)) 2.81/1.64 filter2(false, f, x, y) => filter(f, y) 2.81/1.64 2.81/1.64 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 2.81/1.64 2.81/1.64 We observe that the rules contain a first-order subset: 2.81/1.64 2.81/1.65 p(0) => 0 2.81/1.65 p(s(X)) => X 2.81/1.65 le(0, X) => true 2.81/1.65 le(s(X), 0) => false 2.81/1.65 le(s(X), s(Y)) => le(X, Y) 2.81/1.65 minus(X, Y) => if(le(X, Y), X, Y) 2.81/1.65 if(true, X, Y) => 0 2.81/1.65 if(false, X, Y) => s(minus(p(X), Y)) 2.81/1.65 2.81/1.65 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. 2.81/1.65 2.81/1.65 According to the external first-order termination prover, this system is indeed terminating: 2.81/1.65 2.81/1.65 || proof of resources/system.trs 2.81/1.65 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || Termination w.r.t. Q of the given QTRS could be proven: 2.81/1.65 || 2.81/1.65 || (0) QTRS 2.81/1.65 || (1) Overlay + Local Confluence [EQUIVALENT] 2.81/1.65 || (2) QTRS 2.81/1.65 || (3) DependencyPairsProof [EQUIVALENT] 2.81/1.65 || (4) QDP 2.81/1.65 || (5) DependencyGraphProof [EQUIVALENT] 2.81/1.65 || (6) AND 2.81/1.65 || (7) QDP 2.81/1.65 || (8) UsableRulesProof [EQUIVALENT] 2.81/1.65 || (9) QDP 2.81/1.65 || (10) QReductionProof [EQUIVALENT] 2.81/1.65 || (11) QDP 2.81/1.65 || (12) QDPSizeChangeProof [EQUIVALENT] 2.81/1.65 || (13) YES 2.81/1.65 || (14) QDP 2.81/1.65 || (15) UsableRulesProof [EQUIVALENT] 2.81/1.65 || (16) QDP 2.81/1.65 || (17) QReductionProof [EQUIVALENT] 2.81/1.65 || (18) QDP 2.81/1.65 || (19) QDPOrderProof [EQUIVALENT] 2.81/1.65 || (20) QDP 2.81/1.65 || (21) DependencyGraphProof [EQUIVALENT] 2.81/1.65 || (22) TRUE 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (0) 2.81/1.65 || Obligation: 2.81/1.65 || Q restricted rewrite system: 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) 2.81/1.65 || if(true, %X, %Y) -> 0 2.81/1.65 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) 2.81/1.65 || 2.81/1.65 || Q is empty. 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (1) Overlay + Local Confluence (EQUIVALENT) 2.81/1.65 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (2) 2.81/1.65 || Obligation: 2.81/1.65 || Q restricted rewrite system: 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) 2.81/1.65 || if(true, %X, %Y) -> 0 2.81/1.65 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) 2.81/1.65 || 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (3) DependencyPairsProof (EQUIVALENT) 2.81/1.65 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (4) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 2.81/1.65 || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) 2.81/1.65 || MINUS(%X, %Y) -> LE(%X, %Y) 2.81/1.65 || IF(false, %X, %Y) -> MINUS(p(%X), %Y) 2.81/1.65 || IF(false, %X, %Y) -> P(%X) 2.81/1.65 || 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) 2.81/1.65 || if(true, %X, %Y) -> 0 2.81/1.65 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) 2.81/1.65 || 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (5) DependencyGraphProof (EQUIVALENT) 2.81/1.65 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (6) 2.81/1.65 || Complex Obligation (AND) 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (7) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 2.81/1.65 || 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) 2.81/1.65 || if(true, %X, %Y) -> 0 2.81/1.65 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) 2.81/1.65 || 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (8) UsableRulesProof (EQUIVALENT) 2.81/1.65 || 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. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (9) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 2.81/1.65 || 2.81/1.65 || R is empty. 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (10) QReductionProof (EQUIVALENT) 2.81/1.65 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (11) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 2.81/1.65 || 2.81/1.65 || R is empty. 2.81/1.65 || Q is empty. 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (12) QDPSizeChangeProof (EQUIVALENT) 2.81/1.65 || 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. 2.81/1.65 || 2.81/1.65 || From the DPs we obtained the following set of size-change graphs: 2.81/1.65 || *LE(s(%X), s(%Y)) -> LE(%X, %Y) 2.81/1.65 || The graph contains the following edges 1 > 1, 2 > 2 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (13) 2.81/1.65 || YES 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (14) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) 2.81/1.65 || IF(false, %X, %Y) -> MINUS(p(%X), %Y) 2.81/1.65 || 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) 2.81/1.65 || if(true, %X, %Y) -> 0 2.81/1.65 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) 2.81/1.65 || 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (15) UsableRulesProof (EQUIVALENT) 2.81/1.65 || 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. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (16) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) 2.81/1.65 || IF(false, %X, %Y) -> MINUS(p(%X), %Y) 2.81/1.65 || 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (17) QReductionProof (EQUIVALENT) 2.81/1.65 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 2.81/1.65 || 2.81/1.65 || minus(x0, x1) 2.81/1.65 || if(true, x0, x1) 2.81/1.65 || if(false, x0, x1) 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (18) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) 2.81/1.65 || IF(false, %X, %Y) -> MINUS(p(%X), %Y) 2.81/1.65 || 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (19) QDPOrderProof (EQUIVALENT) 2.81/1.65 || We use the reduction pair processor [LPAR04,JAR06]. 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || The following pairs can be oriented strictly and are deleted. 2.81/1.65 || 2.81/1.65 || IF(false, %X, %Y) -> MINUS(p(%X), %Y) 2.81/1.65 || The remaining pairs can at least be oriented weakly. 2.81/1.65 || Used ordering: Polynomial interpretation [POLO,RATPOLO]: 2.81/1.65 || 2.81/1.65 || POL(0) = [1/4] 2.81/1.65 || POL(IF(x_1, x_2, x_3)) = [1/2] + [1/4]x_1 + [1/4]x_2 + [1/4]x_3 2.81/1.65 || POL(MINUS(x_1, x_2)) = [1/2] + [1/2]x_1 + [1/4]x_2 2.81/1.65 || POL(false) = [1] 2.81/1.65 || POL(le(x_1, x_2)) = [1/2]x_1 2.81/1.65 || POL(p(x_1)) = [1/4] + [1/2]x_1 2.81/1.65 || POL(s(x_1)) = [4] + [4]x_1 2.81/1.65 || POL(true) = 0 2.81/1.65 || The value of delta used in the strict ordering is 1/8. 2.81/1.65 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 2.81/1.65 || 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || 2.81/1.65 || 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (20) 2.81/1.65 || Obligation: 2.81/1.65 || Q DP problem: 2.81/1.65 || The TRS P consists of the following rules: 2.81/1.65 || 2.81/1.65 || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) 2.81/1.65 || 2.81/1.65 || The TRS R consists of the following rules: 2.81/1.65 || 2.81/1.65 || p(0) -> 0 2.81/1.65 || p(s(%X)) -> %X 2.81/1.65 || le(0, %X) -> true 2.81/1.65 || le(s(%X), 0) -> false 2.81/1.65 || le(s(%X), s(%Y)) -> le(%X, %Y) 2.81/1.65 || 2.81/1.65 || The set Q consists of the following terms: 2.81/1.65 || 2.81/1.65 || p(0) 2.81/1.65 || p(s(x0)) 2.81/1.65 || le(0, x0) 2.81/1.65 || le(s(x0), 0) 2.81/1.65 || le(s(x0), s(x1)) 2.81/1.65 || 2.81/1.65 || We have to consider all minimal (P,Q,R)-chains. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (21) DependencyGraphProof (EQUIVALENT) 2.81/1.65 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 2.81/1.65 || ---------------------------------------- 2.81/1.65 || 2.81/1.65 || (22) 2.81/1.65 || TRUE 2.81/1.65 || 2.81/1.65 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]). 2.81/1.65 2.81/1.65 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 2.81/1.65 2.81/1.65 Dependency Pairs P_0: 2.81/1.65 2.81/1.65 0] map#(F, cons(X, Y)) =#> map#(F, Y) 2.81/1.65 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.81/1.65 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 2.81/1.65 3] filter2#(false, F, X, Y) =#> filter#(F, Y) 2.81/1.65 2.81/1.65 Rules R_0: 2.81/1.65 2.81/1.65 p(0) => 0 2.81/1.65 p(s(X)) => X 2.81/1.65 le(0, X) => true 2.81/1.65 le(s(X), 0) => false 2.81/1.65 le(s(X), s(Y)) => le(X, Y) 2.81/1.65 minus(X, Y) => if(le(X, Y), X, Y) 2.81/1.65 if(true, X, Y) => 0 2.81/1.65 if(false, X, Y) => s(minus(p(X), Y)) 2.81/1.65 map(F, nil) => nil 2.81/1.65 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 2.81/1.65 filter(F, nil) => nil 2.81/1.65 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 2.81/1.65 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 2.81/1.65 filter2(false, F, X, Y) => filter(F, Y) 2.81/1.65 2.81/1.65 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 2.81/1.65 2.81/1.65 We consider the dependency pair problem (P_0, R_0, static, formative). 2.81/1.65 2.81/1.65 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.81/1.65 2.81/1.65 * 0 : 0 2.81/1.65 * 1 : 2, 3 2.81/1.65 * 2 : 1 2.81/1.65 * 3 : 1 2.81/1.65 2.81/1.65 This graph has the following strongly connected components: 2.81/1.65 2.81/1.65 P_1: 2.81/1.65 2.81/1.65 map#(F, cons(X, Y)) =#> map#(F, Y) 2.81/1.65 2.81/1.65 P_2: 2.81/1.65 2.81/1.65 filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.81/1.65 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.81/1.65 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.81/1.65 2.81/1.65 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). 2.81/1.65 2.81/1.65 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 2.81/1.65 2.81/1.65 We consider the dependency pair problem (P_2, R_0, static, formative). 2.81/1.65 2.81/1.65 We apply the subterm criterion with the following projection function: 2.81/1.65 2.81/1.65 nu(filter2#) = 4 2.81/1.65 nu(filter#) = 2 2.81/1.65 2.81/1.65 Thus, we can orient the dependency pairs as follows: 2.81/1.65 2.81/1.65 nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 2.81/1.65 nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.81/1.65 nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.81/1.65 2.81/1.65 By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by (P_3, R_0, static, f), where P_3 contains: 2.81/1.65 2.81/1.65 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.81/1.65 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.81/1.65 2.81/1.65 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite. 2.81/1.65 2.81/1.65 We consider the dependency pair problem (P_3, R_0, static, formative). 2.81/1.65 2.81/1.65 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.81/1.65 2.81/1.65 * 0 : 2.81/1.65 * 1 : 2.81/1.65 2.81/1.65 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 2.81/1.65 2.81/1.65 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 2.81/1.65 2.81/1.65 We consider the dependency pair problem (P_1, R_0, static, formative). 2.81/1.65 2.81/1.65 We apply the subterm criterion with the following projection function: 2.81/1.65 2.81/1.65 nu(map#) = 2 2.81/1.65 2.81/1.65 Thus, we can orient the dependency pairs as follows: 2.81/1.65 2.81/1.65 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 2.81/1.65 2.81/1.65 By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 2.81/1.65 2.81/1.65 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 2.81/1.65 2.81/1.65 2.81/1.65 +++ Citations +++ 2.81/1.65 2.81/1.65 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 2.81/1.65 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 2.81/1.65 [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. 2.81/1.65 EOF