/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0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) NonInfProof [EQUIVALENT, 0 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(true, x, y) -> f(gt(x, y), s(x), s(s(y))) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The TRS R 2 is f(true, x, y) -> f(gt(x, y), s(x), s(s(y))) The signature Sigma is {f_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(true, x, y) -> f(gt(x, y), s(x), s(s(y))) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: f(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) ---------------------------------------- (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(true, x, y) -> F(gt(x, y), s(x), s(s(y))) F(true, x, y) -> GT(x, y) GT(s(u), s(v)) -> GT(u, v) The TRS R consists of the following rules: f(true, x, y) -> f(gt(x, y), s(x), s(s(y))) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: f(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) 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: GT(s(u), s(v)) -> GT(u, v) The TRS R consists of the following rules: f(true, x, y) -> f(gt(x, y), s(x), s(s(y))) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: f(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) 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: GT(s(u), s(v)) -> GT(u, v) R is empty. The set Q consists of the following terms: f(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) 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(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: GT(s(u), s(v)) -> GT(u, v) 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: *GT(s(u), s(v)) -> GT(u, v) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: F(true, x, y) -> F(gt(x, y), s(x), s(s(y))) The TRS R consists of the following rules: f(true, x, y) -> f(gt(x, y), s(x), s(s(y))) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: f(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) 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(true, x, y) -> F(gt(x, y), s(x), s(s(y))) The TRS R consists of the following rules: gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: f(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) 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(true, x0, x1) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: F(true, x, y) -> F(gt(x, y), s(x), s(s(y))) The TRS R consists of the following rules: gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) NonInfProof (EQUIVALENT) The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: Note that final constraints are written in bold face. For Pair F(true, x, y) -> F(gt(x, y), s(x), s(s(y))) the following chains were created: *We consider the chain F(true, x0, x1) -> F(gt(x0, x1), s(x0), s(s(x1))), F(true, x2, x3) -> F(gt(x2, x3), s(x2), s(s(x3))) which results in the following constraint: (1) (F(gt(x0, x1), s(x0), s(s(x1)))=F(true, x2, x3) ==> F(true, x2, x3)_>=_F(gt(x2, x3), s(x2), s(s(x3)))) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (gt(x0, x1)=true ==> F(true, s(x0), s(s(x1)))_>=_F(gt(s(x0), s(s(x1))), s(s(x0)), s(s(s(s(x1)))))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x0, x1)=true which results in the following new constraints: (3) (true=true ==> F(true, s(s(x5)), s(s(0)))_>=_F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0)))))) (4) (gt(x7, x6)=true & (gt(x7, x6)=true ==> F(true, s(x7), s(s(x6)))_>=_F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6)))))) ==> F(true, s(s(x7)), s(s(s(x6))))_>=_F(gt(s(s(x7)), s(s(s(x6)))), s(s(s(x7))), s(s(s(s(s(x6))))))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (F(true, s(s(x5)), s(s(0)))_>=_F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0)))))) We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gt(x7, x6)=true ==> F(true, s(x7), s(s(x6)))_>=_F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6)))))) with sigma = [ ] which results in the following new constraint: (6) (F(true, s(x7), s(s(x6)))_>=_F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6))))) ==> F(true, s(s(x7)), s(s(s(x6))))_>=_F(gt(s(s(x7)), s(s(s(x6)))), s(s(s(x7))), s(s(s(s(s(x6))))))) To summarize, we get the following constraints P__>=_ for the following pairs. *F(true, x, y) -> F(gt(x, y), s(x), s(s(y))) *(F(true, s(s(x5)), s(s(0)))_>=_F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0)))))) *(F(true, s(x7), s(s(x6)))_>=_F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6))))) ==> F(true, s(s(x7)), s(s(s(x6))))_>=_F(gt(s(s(x7)), s(s(s(x6)))), s(s(s(x7))), s(s(s(s(s(x6))))))) The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. Using the following integer polynomial ordering the resulting constraints can be solved Polynomial interpretation [NONINF]: POL(0) = 0 POL(F(x_1, x_2, x_3)) = -1 - x_1 + x_2 - x_3 POL(c) = -1 POL(false) = 0 POL(gt(x_1, x_2)) = 0 POL(s(x_1)) = 1 + x_1 POL(true) = 0 The following pairs are in P_>: F(true, x, y) -> F(gt(x, y), s(x), s(s(y))) The following pairs are in P_bound: F(true, x, y) -> F(gt(x, y), s(x), s(s(y))) The following rules are usable: false -> gt(0, v) true -> gt(s(u), 0) gt(u, v) -> gt(s(u), s(v)) ---------------------------------------- (20) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: gt(0, x0) gt(s(x0), 0) gt(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (22) YES