/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT,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) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) TransformationProof [EQUIVALENT, 0 ms] (20) QDP (21) TransformationProof [EQUIVALENT, 0 ms] (22) QDP (23) TransformationProof [EQUIVALENT, 0 ms] (24) QDP (25) TransformationProof [EQUIVALENT, 0 ms] (26) QDP (27) TransformationProof [EQUIVALENT, 0 ms] (28) QDP (29) TransformationProof [EQUIVALENT, 0 ms] (30) QDP (31) TransformationProof [EQUIVALENT, 0 ms] (32) QDP (33) TransformationProof [EQUIVALENT, 0 ms] (34) QDP (35) MNOCProof [EQUIVALENT, 0 ms] (36) QDP (37) NonTerminationLoopProof [COMPLETE, 5 ms] (38) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is id(s(x)) -> s(id(x)) id(0) -> 0 The TRS R 2 is f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The signature Sigma is {f_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) ---------------------------------------- (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(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> ID(s(s(s(s(s(s(s(s(x))))))))) ID(s(x)) -> ID(x) The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) 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 1 less node. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: ID(s(x)) -> ID(x) The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) 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: ID(s(x)) -> ID(x) R is empty. The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) 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]. f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: ID(s(x)) -> ID(x) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *ID(s(x)) -> ID(x) The graph contains the following edges 1 > 1 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) 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(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) 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]. f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) at position [0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y) at position [0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y) at position [0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y) at position [0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y) at position [0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y)) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y) at position [0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y)) ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y) at position [0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y)) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y) at position [0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)) ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) MNOCProof (EQUIVALENT) We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (37) 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 = F(s(s(s(s(s(s(s(s(x)))))))), y, y) evaluates to t =F(s(s(s(s(s(s(s(s(id(x))))))))), y, y) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [x / id(x)] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from F(s(s(s(s(s(s(s(s(x)))))))), y, y) to F(s(s(s(s(s(s(s(s(id(x))))))))), y, y). ---------------------------------------- (38) NO