3.00/1.49 YES 3.05/1.50 We consider the system theBenchmark. 3.05/1.50 3.05/1.50 Alphabet: 3.05/1.50 3.05/1.50 0 : [] --> b 3.05/1.50 cons : [c * d] --> d 3.05/1.50 false : [] --> a 3.05/1.50 filter : [c -> a * d] --> d 3.05/1.50 filter2 : [a * c -> a * c * d] --> d 3.05/1.50 gcd : [b * b] --> b 3.05/1.50 if!fac6220gcd : [a * b * b] --> b 3.05/1.50 le : [b * b] --> a 3.05/1.50 map : [c -> c * d] --> d 3.05/1.50 minus : [b * b] --> b 3.05/1.50 nil : [] --> d 3.05/1.50 pred : [b] --> b 3.05/1.50 s : [b] --> b 3.05/1.50 true : [] --> a 3.05/1.50 3.05/1.50 Rules: 3.05/1.50 3.05/1.50 le(0, x) => true 3.05/1.50 le(s(x), 0) => false 3.05/1.50 le(s(x), s(y)) => le(x, y) 3.05/1.50 pred(s(x)) => x 3.05/1.50 minus(x, 0) => x 3.05/1.50 minus(x, s(y)) => pred(minus(x, y)) 3.05/1.50 gcd(0, x) => x 3.05/1.50 gcd(s(x), 0) => s(x) 3.05/1.50 gcd(s(x), s(y)) => if!fac6220gcd(le(y, x), s(x), s(y)) 3.05/1.50 if!fac6220gcd(true, s(x), s(y)) => gcd(minus(x, y), s(y)) 3.05/1.50 if!fac6220gcd(false, s(x), s(y)) => gcd(minus(y, x), s(x)) 3.05/1.50 map(f, nil) => nil 3.05/1.50 map(f, cons(x, y)) => cons(f x, map(f, y)) 3.05/1.50 filter(f, nil) => nil 3.05/1.50 filter(f, cons(x, y)) => filter2(f x, f, x, y) 3.05/1.50 filter2(true, f, x, y) => cons(x, filter(f, y)) 3.05/1.50 filter2(false, f, x, y) => filter(f, y) 3.05/1.50 3.05/1.50 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 3.05/1.50 3.05/1.50 We observe that the rules contain a first-order subset: 3.05/1.50 3.05/1.50 le(0, X) => true 3.05/1.50 le(s(X), 0) => false 3.05/1.50 le(s(X), s(Y)) => le(X, Y) 3.05/1.50 pred(s(X)) => X 3.05/1.50 minus(X, 0) => X 3.05/1.50 minus(X, s(Y)) => pred(minus(X, Y)) 3.05/1.50 gcd(0, X) => X 3.05/1.50 gcd(s(X), 0) => s(X) 3.05/1.50 gcd(s(X), s(Y)) => if!fac6220gcd(le(Y, X), s(X), s(Y)) 3.05/1.50 if!fac6220gcd(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 3.05/1.50 if!fac6220gcd(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 3.05/1.50 3.05/1.50 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. 3.05/1.50 3.05/1.50 According to the external first-order termination prover, this system is indeed terminating: 3.05/1.50 3.05/1.50 || proof of resources/system.trs 3.05/1.50 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 3.05/1.50 || 3.05/1.50 || 3.05/1.50 || Termination w.r.t. Q of the given QTRS could be proven: 3.05/1.50 || 3.05/1.50 || (0) QTRS 3.05/1.50 || (1) Overlay + Local Confluence [EQUIVALENT] 3.05/1.50 || (2) QTRS 3.05/1.50 || (3) DependencyPairsProof [EQUIVALENT] 3.05/1.50 || (4) QDP 3.05/1.50 || (5) DependencyGraphProof [EQUIVALENT] 3.05/1.50 || (6) AND 3.05/1.50 || (7) QDP 3.05/1.50 || (8) UsableRulesProof [EQUIVALENT] 3.05/1.50 || (9) QDP 3.05/1.50 || (10) QReductionProof [EQUIVALENT] 3.05/1.50 || (11) QDP 3.05/1.50 || (12) QDPSizeChangeProof [EQUIVALENT] 3.05/1.50 || (13) YES 3.05/1.50 || (14) QDP 3.05/1.50 || (15) UsableRulesProof [EQUIVALENT] 3.05/1.50 || (16) QDP 3.05/1.50 || (17) QReductionProof [EQUIVALENT] 3.05/1.50 || (18) QDP 3.05/1.50 || (19) QDPSizeChangeProof [EQUIVALENT] 3.05/1.50 || (20) YES 3.05/1.50 || (21) QDP 3.05/1.50 || (22) UsableRulesProof [EQUIVALENT] 3.05/1.50 || (23) QDP 3.05/1.50 || (24) QReductionProof [EQUIVALENT] 3.05/1.50 || (25) QDP 3.05/1.50 || (26) QDPOrderProof [EQUIVALENT] 3.05/1.50 || (27) QDP 3.05/1.50 || (28) DependencyGraphProof [EQUIVALENT] 3.05/1.50 || (29) TRUE 3.05/1.50 || 3.05/1.50 || 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (0) 3.05/1.50 || Obligation: 3.05/1.50 || Q restricted rewrite system: 3.05/1.50 || The TRS R consists of the following rules: 3.05/1.50 || 3.05/1.50 || le(0, %X) -> true 3.05/1.50 || le(s(%X), 0) -> false 3.05/1.50 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.50 || pred(s(%X)) -> %X 3.05/1.50 || minus(%X, 0) -> %X 3.05/1.50 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.50 || gcd(0, %X) -> %X 3.05/1.50 || gcd(s(%X), 0) -> s(%X) 3.05/1.50 || gcd(s(%X), s(%Y)) -> if!fac6220gcd(le(%Y, %X), s(%X), s(%Y)) 3.05/1.50 || if!fac6220gcd(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y)) 3.05/1.50 || if!fac6220gcd(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X)) 3.05/1.50 || 3.05/1.50 || Q is empty. 3.05/1.50 || 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (1) Overlay + Local Confluence (EQUIVALENT) 3.05/1.50 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (2) 3.05/1.50 || Obligation: 3.05/1.50 || Q restricted rewrite system: 3.05/1.50 || The TRS R consists of the following rules: 3.05/1.50 || 3.05/1.50 || le(0, %X) -> true 3.05/1.50 || le(s(%X), 0) -> false 3.05/1.50 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.50 || pred(s(%X)) -> %X 3.05/1.50 || minus(%X, 0) -> %X 3.05/1.50 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.50 || gcd(0, %X) -> %X 3.05/1.50 || gcd(s(%X), 0) -> s(%X) 3.05/1.50 || gcd(s(%X), s(%Y)) -> if!fac6220gcd(le(%Y, %X), s(%X), s(%Y)) 3.05/1.50 || if!fac6220gcd(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y)) 3.05/1.50 || if!fac6220gcd(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X)) 3.05/1.50 || 3.05/1.50 || The set Q consists of the following terms: 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.50 || minus(x0, s(x1)) 3.05/1.50 || gcd(0, x0) 3.05/1.50 || gcd(s(x0), 0) 3.05/1.50 || gcd(s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.50 || 3.05/1.50 || 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (3) DependencyPairsProof (EQUIVALENT) 3.05/1.50 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (4) 3.05/1.50 || Obligation: 3.05/1.50 || Q DP problem: 3.05/1.50 || The TRS P consists of the following rules: 3.05/1.50 || 3.05/1.50 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 3.05/1.50 || MINUS(%X, s(%Y)) -> PRED(minus(%X, %Y)) 3.05/1.50 || MINUS(%X, s(%Y)) -> MINUS(%X, %Y) 3.05/1.50 || GCD(s(%X), s(%Y)) -> IF!FAC6220GCD(le(%Y, %X), s(%X), s(%Y)) 3.05/1.50 || GCD(s(%X), s(%Y)) -> LE(%Y, %X) 3.05/1.50 || IF!FAC6220GCD(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y)) 3.05/1.50 || IF!FAC6220GCD(true, s(%X), s(%Y)) -> MINUS(%X, %Y) 3.05/1.50 || IF!FAC6220GCD(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X)) 3.05/1.50 || IF!FAC6220GCD(false, s(%X), s(%Y)) -> MINUS(%Y, %X) 3.05/1.50 || 3.05/1.50 || The TRS R consists of the following rules: 3.05/1.50 || 3.05/1.50 || le(0, %X) -> true 3.05/1.50 || le(s(%X), 0) -> false 3.05/1.50 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.50 || pred(s(%X)) -> %X 3.05/1.50 || minus(%X, 0) -> %X 3.05/1.50 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.50 || gcd(0, %X) -> %X 3.05/1.50 || gcd(s(%X), 0) -> s(%X) 3.05/1.50 || gcd(s(%X), s(%Y)) -> if!fac6220gcd(le(%Y, %X), s(%X), s(%Y)) 3.05/1.50 || if!fac6220gcd(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y)) 3.05/1.50 || if!fac6220gcd(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X)) 3.05/1.50 || 3.05/1.50 || The set Q consists of the following terms: 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.50 || minus(x0, s(x1)) 3.05/1.50 || gcd(0, x0) 3.05/1.50 || gcd(s(x0), 0) 3.05/1.50 || gcd(s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.50 || 3.05/1.50 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (5) DependencyGraphProof (EQUIVALENT) 3.05/1.50 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (6) 3.05/1.50 || Complex Obligation (AND) 3.05/1.50 || 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (7) 3.05/1.50 || Obligation: 3.05/1.50 || Q DP problem: 3.05/1.50 || The TRS P consists of the following rules: 3.05/1.50 || 3.05/1.50 || MINUS(%X, s(%Y)) -> MINUS(%X, %Y) 3.05/1.50 || 3.05/1.50 || The TRS R consists of the following rules: 3.05/1.50 || 3.05/1.50 || le(0, %X) -> true 3.05/1.50 || le(s(%X), 0) -> false 3.05/1.50 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.50 || pred(s(%X)) -> %X 3.05/1.50 || minus(%X, 0) -> %X 3.05/1.50 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.50 || gcd(0, %X) -> %X 3.05/1.50 || gcd(s(%X), 0) -> s(%X) 3.05/1.50 || gcd(s(%X), s(%Y)) -> if!fac6220gcd(le(%Y, %X), s(%X), s(%Y)) 3.05/1.50 || if!fac6220gcd(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y)) 3.05/1.50 || if!fac6220gcd(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X)) 3.05/1.50 || 3.05/1.50 || The set Q consists of the following terms: 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.50 || minus(x0, s(x1)) 3.05/1.50 || gcd(0, x0) 3.05/1.50 || gcd(s(x0), 0) 3.05/1.50 || gcd(s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.50 || 3.05/1.50 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (8) UsableRulesProof (EQUIVALENT) 3.05/1.50 || 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. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (9) 3.05/1.50 || Obligation: 3.05/1.50 || Q DP problem: 3.05/1.50 || The TRS P consists of the following rules: 3.05/1.50 || 3.05/1.50 || MINUS(%X, s(%Y)) -> MINUS(%X, %Y) 3.05/1.50 || 3.05/1.50 || R is empty. 3.05/1.50 || The set Q consists of the following terms: 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.50 || minus(x0, s(x1)) 3.05/1.50 || gcd(0, x0) 3.05/1.50 || gcd(s(x0), 0) 3.05/1.50 || gcd(s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.50 || 3.05/1.50 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (10) QReductionProof (EQUIVALENT) 3.05/1.50 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.50 || minus(x0, s(x1)) 3.05/1.50 || gcd(0, x0) 3.05/1.50 || gcd(s(x0), 0) 3.05/1.50 || gcd(s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.50 || 3.05/1.50 || 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (11) 3.05/1.50 || Obligation: 3.05/1.50 || Q DP problem: 3.05/1.50 || The TRS P consists of the following rules: 3.05/1.50 || 3.05/1.50 || MINUS(%X, s(%Y)) -> MINUS(%X, %Y) 3.05/1.50 || 3.05/1.50 || R is empty. 3.05/1.50 || Q is empty. 3.05/1.50 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (12) QDPSizeChangeProof (EQUIVALENT) 3.05/1.50 || 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. 3.05/1.50 || 3.05/1.50 || From the DPs we obtained the following set of size-change graphs: 3.05/1.50 || *MINUS(%X, s(%Y)) -> MINUS(%X, %Y) 3.05/1.50 || The graph contains the following edges 1 >= 1, 2 > 2 3.05/1.50 || 3.05/1.50 || 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (13) 3.05/1.50 || YES 3.05/1.50 || 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (14) 3.05/1.50 || Obligation: 3.05/1.50 || Q DP problem: 3.05/1.50 || The TRS P consists of the following rules: 3.05/1.50 || 3.05/1.50 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 3.05/1.50 || 3.05/1.50 || The TRS R consists of the following rules: 3.05/1.50 || 3.05/1.50 || le(0, %X) -> true 3.05/1.50 || le(s(%X), 0) -> false 3.05/1.50 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.50 || pred(s(%X)) -> %X 3.05/1.50 || minus(%X, 0) -> %X 3.05/1.50 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.50 || gcd(0, %X) -> %X 3.05/1.50 || gcd(s(%X), 0) -> s(%X) 3.05/1.50 || gcd(s(%X), s(%Y)) -> if!fac6220gcd(le(%Y, %X), s(%X), s(%Y)) 3.05/1.50 || if!fac6220gcd(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y)) 3.05/1.50 || if!fac6220gcd(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X)) 3.05/1.50 || 3.05/1.50 || The set Q consists of the following terms: 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.50 || minus(x0, s(x1)) 3.05/1.50 || gcd(0, x0) 3.05/1.50 || gcd(s(x0), 0) 3.05/1.50 || gcd(s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.50 || 3.05/1.50 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (15) UsableRulesProof (EQUIVALENT) 3.05/1.50 || 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. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (16) 3.05/1.50 || Obligation: 3.05/1.50 || Q DP problem: 3.05/1.50 || The TRS P consists of the following rules: 3.05/1.50 || 3.05/1.50 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 3.05/1.50 || 3.05/1.50 || R is empty. 3.05/1.50 || The set Q consists of the following terms: 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.50 || minus(x0, s(x1)) 3.05/1.50 || gcd(0, x0) 3.05/1.50 || gcd(s(x0), 0) 3.05/1.50 || gcd(s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.50 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.50 || 3.05/1.50 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.50 || ---------------------------------------- 3.05/1.50 || 3.05/1.50 || (17) QReductionProof (EQUIVALENT) 3.05/1.50 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 3.05/1.50 || 3.05/1.50 || le(0, x0) 3.05/1.50 || le(s(x0), 0) 3.05/1.50 || le(s(x0), s(x1)) 3.05/1.50 || pred(s(x0)) 3.05/1.50 || minus(x0, 0) 3.05/1.51 || minus(x0, s(x1)) 3.05/1.51 || gcd(0, x0) 3.05/1.51 || gcd(s(x0), 0) 3.05/1.51 || gcd(s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.51 || 3.05/1.51 || 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (18) 3.05/1.51 || Obligation: 3.05/1.51 || Q DP problem: 3.05/1.51 || The TRS P consists of the following rules: 3.05/1.51 || 3.05/1.51 || LE(s(%X), s(%Y)) -> LE(%X, %Y) 3.05/1.51 || 3.05/1.51 || R is empty. 3.05/1.51 || Q is empty. 3.05/1.51 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (19) QDPSizeChangeProof (EQUIVALENT) 3.05/1.51 || 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. 3.05/1.51 || 3.05/1.51 || From the DPs we obtained the following set of size-change graphs: 3.05/1.51 || *LE(s(%X), s(%Y)) -> LE(%X, %Y) 3.05/1.51 || The graph contains the following edges 1 > 1, 2 > 2 3.05/1.51 || 3.05/1.51 || 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (20) 3.05/1.51 || YES 3.05/1.51 || 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (21) 3.05/1.51 || Obligation: 3.05/1.51 || Q DP problem: 3.05/1.51 || The TRS P consists of the following rules: 3.05/1.51 || 3.05/1.51 || GCD(s(%X), s(%Y)) -> IF!FAC6220GCD(le(%Y, %X), s(%X), s(%Y)) 3.05/1.51 || IF!FAC6220GCD(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y)) 3.05/1.51 || IF!FAC6220GCD(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X)) 3.05/1.51 || 3.05/1.51 || The TRS R consists of the following rules: 3.05/1.51 || 3.05/1.51 || le(0, %X) -> true 3.05/1.51 || le(s(%X), 0) -> false 3.05/1.51 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.51 || pred(s(%X)) -> %X 3.05/1.51 || minus(%X, 0) -> %X 3.05/1.51 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.51 || gcd(0, %X) -> %X 3.05/1.51 || gcd(s(%X), 0) -> s(%X) 3.05/1.51 || gcd(s(%X), s(%Y)) -> if!fac6220gcd(le(%Y, %X), s(%X), s(%Y)) 3.05/1.51 || if!fac6220gcd(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y)) 3.05/1.51 || if!fac6220gcd(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X)) 3.05/1.51 || 3.05/1.51 || The set Q consists of the following terms: 3.05/1.51 || 3.05/1.51 || le(0, x0) 3.05/1.51 || le(s(x0), 0) 3.05/1.51 || le(s(x0), s(x1)) 3.05/1.51 || pred(s(x0)) 3.05/1.51 || minus(x0, 0) 3.05/1.51 || minus(x0, s(x1)) 3.05/1.51 || gcd(0, x0) 3.05/1.51 || gcd(s(x0), 0) 3.05/1.51 || gcd(s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.51 || 3.05/1.51 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (22) UsableRulesProof (EQUIVALENT) 3.05/1.51 || 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. 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (23) 3.05/1.51 || Obligation: 3.05/1.51 || Q DP problem: 3.05/1.51 || The TRS P consists of the following rules: 3.05/1.51 || 3.05/1.51 || GCD(s(%X), s(%Y)) -> IF!FAC6220GCD(le(%Y, %X), s(%X), s(%Y)) 3.05/1.51 || IF!FAC6220GCD(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y)) 3.05/1.51 || IF!FAC6220GCD(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X)) 3.05/1.51 || 3.05/1.51 || The TRS R consists of the following rules: 3.05/1.51 || 3.05/1.51 || minus(%X, 0) -> %X 3.05/1.51 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.51 || pred(s(%X)) -> %X 3.05/1.51 || le(0, %X) -> true 3.05/1.51 || le(s(%X), 0) -> false 3.05/1.51 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.51 || 3.05/1.51 || The set Q consists of the following terms: 3.05/1.51 || 3.05/1.51 || le(0, x0) 3.05/1.51 || le(s(x0), 0) 3.05/1.51 || le(s(x0), s(x1)) 3.05/1.51 || pred(s(x0)) 3.05/1.51 || minus(x0, 0) 3.05/1.51 || minus(x0, s(x1)) 3.05/1.51 || gcd(0, x0) 3.05/1.51 || gcd(s(x0), 0) 3.05/1.51 || gcd(s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.51 || 3.05/1.51 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (24) QReductionProof (EQUIVALENT) 3.05/1.51 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 3.05/1.51 || 3.05/1.51 || gcd(0, x0) 3.05/1.51 || gcd(s(x0), 0) 3.05/1.51 || gcd(s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(true, s(x0), s(x1)) 3.05/1.51 || if!fac6220gcd(false, s(x0), s(x1)) 3.05/1.51 || 3.05/1.51 || 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (25) 3.05/1.51 || Obligation: 3.05/1.51 || Q DP problem: 3.05/1.51 || The TRS P consists of the following rules: 3.05/1.51 || 3.05/1.51 || GCD(s(%X), s(%Y)) -> IF!FAC6220GCD(le(%Y, %X), s(%X), s(%Y)) 3.05/1.51 || IF!FAC6220GCD(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y)) 3.05/1.51 || IF!FAC6220GCD(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X)) 3.05/1.51 || 3.05/1.51 || The TRS R consists of the following rules: 3.05/1.51 || 3.05/1.51 || minus(%X, 0) -> %X 3.05/1.51 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.51 || pred(s(%X)) -> %X 3.05/1.51 || le(0, %X) -> true 3.05/1.51 || le(s(%X), 0) -> false 3.05/1.51 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.51 || 3.05/1.51 || The set Q consists of the following terms: 3.05/1.51 || 3.05/1.51 || le(0, x0) 3.05/1.51 || le(s(x0), 0) 3.05/1.51 || le(s(x0), s(x1)) 3.05/1.51 || pred(s(x0)) 3.05/1.51 || minus(x0, 0) 3.05/1.51 || minus(x0, s(x1)) 3.05/1.51 || 3.05/1.51 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (26) QDPOrderProof (EQUIVALENT) 3.05/1.51 || We use the reduction pair processor [LPAR04,JAR06]. 3.05/1.51 || 3.05/1.51 || 3.05/1.51 || The following pairs can be oriented strictly and are deleted. 3.05/1.51 || 3.05/1.51 || IF!FAC6220GCD(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y)) 3.05/1.51 || IF!FAC6220GCD(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X)) 3.05/1.51 || The remaining pairs can at least be oriented weakly. 3.05/1.51 || Used ordering: Polynomial interpretation [POLO]: 3.05/1.51 || 3.05/1.51 || POL(0) = 0 3.05/1.51 || POL(GCD(x_1, x_2)) = 1 + x_1 + x_2 3.05/1.51 || POL(IF!FAC6220GCD(x_1, x_2, x_3)) = 1 + x_2 + x_3 3.05/1.51 || POL(false) = 0 3.05/1.51 || POL(le(x_1, x_2)) = 0 3.05/1.51 || POL(minus(x_1, x_2)) = x_1 3.05/1.51 || POL(pred(x_1)) = x_1 3.05/1.51 || POL(s(x_1)) = 1 + x_1 3.05/1.51 || POL(true) = 0 3.05/1.51 || 3.05/1.51 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 3.05/1.51 || 3.05/1.51 || minus(%X, 0) -> %X 3.05/1.51 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.51 || pred(s(%X)) -> %X 3.05/1.51 || 3.05/1.51 || 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (27) 3.05/1.51 || Obligation: 3.05/1.51 || Q DP problem: 3.05/1.51 || The TRS P consists of the following rules: 3.05/1.51 || 3.05/1.51 || GCD(s(%X), s(%Y)) -> IF!FAC6220GCD(le(%Y, %X), s(%X), s(%Y)) 3.05/1.51 || 3.05/1.51 || The TRS R consists of the following rules: 3.05/1.51 || 3.05/1.51 || minus(%X, 0) -> %X 3.05/1.51 || minus(%X, s(%Y)) -> pred(minus(%X, %Y)) 3.05/1.51 || pred(s(%X)) -> %X 3.05/1.51 || le(0, %X) -> true 3.05/1.51 || le(s(%X), 0) -> false 3.05/1.51 || le(s(%X), s(%Y)) -> le(%X, %Y) 3.05/1.51 || 3.05/1.51 || The set Q consists of the following terms: 3.05/1.51 || 3.05/1.51 || le(0, x0) 3.05/1.51 || le(s(x0), 0) 3.05/1.51 || le(s(x0), s(x1)) 3.05/1.51 || pred(s(x0)) 3.05/1.51 || minus(x0, 0) 3.05/1.51 || minus(x0, s(x1)) 3.05/1.51 || 3.05/1.51 || We have to consider all minimal (P,Q,R)-chains. 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (28) DependencyGraphProof (EQUIVALENT) 3.05/1.51 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 3.05/1.51 || ---------------------------------------- 3.05/1.51 || 3.05/1.51 || (29) 3.05/1.51 || TRUE 3.05/1.51 || 3.05/1.51 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]). 3.05/1.51 3.05/1.51 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 3.05/1.51 3.05/1.51 Dependency Pairs P_0: 3.05/1.51 3.05/1.51 0] map#(F, cons(X, Y)) =#> map#(F, Y) 3.05/1.51 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 3.05/1.51 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 3.05/1.51 3] filter2#(false, F, X, Y) =#> filter#(F, Y) 3.05/1.51 3.05/1.51 Rules R_0: 3.05/1.51 3.05/1.51 le(0, X) => true 3.05/1.51 le(s(X), 0) => false 3.05/1.51 le(s(X), s(Y)) => le(X, Y) 3.05/1.51 pred(s(X)) => X 3.05/1.51 minus(X, 0) => X 3.05/1.51 minus(X, s(Y)) => pred(minus(X, Y)) 3.05/1.51 gcd(0, X) => X 3.05/1.51 gcd(s(X), 0) => s(X) 3.05/1.51 gcd(s(X), s(Y)) => if!fac6220gcd(le(Y, X), s(X), s(Y)) 3.05/1.51 if!fac6220gcd(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 3.05/1.51 if!fac6220gcd(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 3.05/1.51 map(F, nil) => nil 3.05/1.51 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 3.05/1.51 filter(F, nil) => nil 3.05/1.51 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 3.05/1.51 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 3.05/1.51 filter2(false, F, X, Y) => filter(F, Y) 3.05/1.51 3.05/1.51 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 3.05/1.51 3.05/1.51 We consider the dependency pair problem (P_0, R_0, static, formative). 3.05/1.51 3.05/1.51 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 3.05/1.51 3.05/1.51 * 0 : 0 3.05/1.51 * 1 : 2, 3 3.05/1.51 * 2 : 1 3.05/1.51 * 3 : 1 3.05/1.51 3.05/1.51 This graph has the following strongly connected components: 3.05/1.51 3.05/1.51 P_1: 3.05/1.51 3.05/1.51 map#(F, cons(X, Y)) =#> map#(F, Y) 3.05/1.51 3.05/1.51 P_2: 3.05/1.51 3.05/1.51 filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 3.05/1.51 filter2#(true, F, X, Y) =#> filter#(F, Y) 3.05/1.51 filter2#(false, F, X, Y) =#> filter#(F, Y) 3.05/1.51 3.05/1.51 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). 3.05/1.51 3.05/1.51 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 3.05/1.51 3.05/1.51 We consider the dependency pair problem (P_2, R_0, static, formative). 3.05/1.51 3.05/1.51 We apply the subterm criterion with the following projection function: 3.05/1.51 3.05/1.51 nu(filter2#) = 4 3.05/1.51 nu(filter#) = 2 3.05/1.51 3.05/1.51 Thus, we can orient the dependency pairs as follows: 3.05/1.51 3.05/1.51 nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 3.05/1.51 nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 3.05/1.51 nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 3.05/1.51 3.05/1.51 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: 3.05/1.51 3.05/1.51 filter2#(true, F, X, Y) =#> filter#(F, Y) 3.05/1.51 filter2#(false, F, X, Y) =#> filter#(F, Y) 3.05/1.51 3.05/1.51 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite. 3.05/1.51 3.05/1.51 We consider the dependency pair problem (P_3, R_0, static, formative). 3.05/1.51 3.05/1.51 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 3.05/1.51 3.05/1.51 * 0 : 3.05/1.51 * 1 : 3.05/1.51 3.05/1.51 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 3.05/1.51 3.05/1.51 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 3.05/1.51 3.05/1.51 We consider the dependency pair problem (P_1, R_0, static, formative). 3.05/1.51 3.05/1.51 We apply the subterm criterion with the following projection function: 3.05/1.51 3.05/1.51 nu(map#) = 2 3.05/1.51 3.05/1.51 Thus, we can orient the dependency pairs as follows: 3.05/1.51 3.05/1.51 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 3.05/1.51 3.05/1.51 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. 3.05/1.51 3.05/1.51 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 3.05/1.51 3.05/1.51 3.05/1.51 +++ Citations +++ 3.05/1.51 3.05/1.51 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 3.05/1.51 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 3.05/1.51 [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. 3.05/1.51 EOF