/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 23 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) MRRProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) f(minus(x)) -> minus(minus(minus(f(x)))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(+(x, y)) -> MINUS(minus(minus(x))) MINUS(+(x, y)) -> MINUS(minus(x)) MINUS(+(x, y)) -> MINUS(x) MINUS(+(x, y)) -> MINUS(minus(minus(y))) MINUS(+(x, y)) -> MINUS(minus(y)) MINUS(+(x, y)) -> MINUS(y) MINUS(*(x, y)) -> MINUS(minus(minus(x))) MINUS(*(x, y)) -> MINUS(minus(x)) MINUS(*(x, y)) -> MINUS(x) MINUS(*(x, y)) -> MINUS(minus(minus(y))) MINUS(*(x, y)) -> MINUS(minus(y)) MINUS(*(x, y)) -> MINUS(y) F(minus(x)) -> MINUS(minus(minus(f(x)))) F(minus(x)) -> MINUS(minus(f(x))) F(minus(x)) -> MINUS(f(x)) F(minus(x)) -> F(x) The TRS R consists of the following rules: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) f(minus(x)) -> minus(minus(minus(f(x)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(+(x, y)) -> MINUS(minus(x)) MINUS(+(x, y)) -> MINUS(minus(minus(x))) MINUS(+(x, y)) -> MINUS(x) MINUS(+(x, y)) -> MINUS(minus(minus(y))) MINUS(+(x, y)) -> MINUS(minus(y)) MINUS(+(x, y)) -> MINUS(y) MINUS(*(x, y)) -> MINUS(minus(minus(x))) MINUS(*(x, y)) -> MINUS(minus(x)) MINUS(*(x, y)) -> MINUS(x) MINUS(*(x, y)) -> MINUS(minus(minus(y))) MINUS(*(x, y)) -> MINUS(minus(y)) MINUS(*(x, y)) -> MINUS(y) The TRS R consists of the following rules: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) f(minus(x)) -> minus(minus(minus(f(x)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(+(x, y)) -> MINUS(minus(x)) MINUS(+(x, y)) -> MINUS(minus(minus(x))) MINUS(+(x, y)) -> MINUS(x) MINUS(+(x, y)) -> MINUS(minus(minus(y))) MINUS(+(x, y)) -> MINUS(minus(y)) MINUS(+(x, y)) -> MINUS(y) MINUS(*(x, y)) -> MINUS(minus(minus(x))) MINUS(*(x, y)) -> MINUS(minus(x)) MINUS(*(x, y)) -> MINUS(x) MINUS(*(x, y)) -> MINUS(minus(minus(y))) MINUS(*(x, y)) -> MINUS(minus(y)) MINUS(*(x, y)) -> MINUS(y) The TRS R consists of the following rules: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: MINUS(+(x, y)) -> MINUS(minus(x)) MINUS(+(x, y)) -> MINUS(minus(minus(x))) MINUS(+(x, y)) -> MINUS(x) MINUS(+(x, y)) -> MINUS(minus(minus(y))) MINUS(+(x, y)) -> MINUS(minus(y)) MINUS(+(x, y)) -> MINUS(y) MINUS(*(x, y)) -> MINUS(minus(minus(x))) MINUS(*(x, y)) -> MINUS(minus(x)) MINUS(*(x, y)) -> MINUS(x) MINUS(*(x, y)) -> MINUS(minus(minus(y))) MINUS(*(x, y)) -> MINUS(minus(y)) MINUS(*(x, y)) -> MINUS(y) Strictly oriented rules of the TRS R: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) Used ordering: Knuth-Bendix order [KBO] with precedence:minus_1 > *_2 > MINUS_1 > +_2 and weight map: minus_1=0 MINUS_1=1 +_2=0 *_2=0 The variable weight is 1 ---------------------------------------- (9) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: F(minus(x)) -> F(x) The TRS R consists of the following rules: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) f(minus(x)) -> minus(minus(minus(f(x)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: F(minus(x)) -> F(x) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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: *F(minus(x)) -> F(x) The graph contains the following edges 1 > 1 ---------------------------------------- (16) YES