/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) ATransformationProof [EQUIVALENT, 0 ms] (20) QDP (21) NonTerminationLoopProof [COMPLETE, 0 ms] (22) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(tt, x) -> f(isList(x), x) isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The TRS R 2 is f(tt, x) -> f(isList(x), x) The signature Sigma is {f_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(tt, x) -> f(isList(x), x) isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The set Q consists of the following terms: f(tt, x0) isList(Cons(x0, xs)) isList(nil) ---------------------------------------- (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: F(tt, x) -> F(isList(x), x) F(tt, x) -> ISLIST(x) ISLIST(Cons(x, xs)) -> ISLIST(xs) The TRS R consists of the following rules: f(tt, x) -> f(isList(x), x) isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The set Q consists of the following terms: f(tt, x0) isList(Cons(x0, xs)) isList(nil) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, x) -> F(isList(x), x) The TRS R consists of the following rules: f(tt, x) -> f(isList(x), x) isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The set Q consists of the following terms: f(tt, x0) isList(Cons(x0, xs)) isList(nil) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) 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. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, x) -> F(isList(x), x) The TRS R consists of the following rules: isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The set Q consists of the following terms: f(tt, x0) isList(Cons(x0, xs)) isList(nil) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) 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]. f(tt, x0) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, x) -> F(isList(x), x) The TRS R consists of the following rules: isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The set Q consists of the following terms: isList(Cons(x0, xs)) isList(nil) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(tt, x) -> F(isList(x), x) at position [0] we obtained the following new rules [LPAR04]: (F(tt, Cons(x0, xs)) -> F(isList(xs), Cons(x0, xs)),F(tt, Cons(x0, xs)) -> F(isList(xs), Cons(x0, xs))) (F(tt, nil) -> F(tt, nil),F(tt, nil) -> F(tt, nil)) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, Cons(x0, xs)) -> F(isList(xs), Cons(x0, xs)) F(tt, nil) -> F(tt, nil) The TRS R consists of the following rules: isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The set Q consists of the following terms: isList(Cons(x0, xs)) isList(nil) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, nil) -> F(tt, nil) The TRS R consists of the following rules: isList(Cons(x, xs)) -> isList(xs) isList(nil) -> tt The set Q consists of the following terms: isList(Cons(x0, xs)) isList(nil) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, nil) -> F(tt, nil) R is empty. The set Q consists of the following terms: isList(Cons(x0, xs)) isList(nil) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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]. isList(Cons(x0, xs)) isList(nil) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, nil) -> F(tt, nil) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: tt(nil) -> tt(nil) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = tt(nil) evaluates to t =tt(nil) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from tt(nil) to tt(nil). ---------------------------------------- (22) NO