YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 61 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) RisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: i(0) -> 0 +(0, y) -> y +(x, 0) -> x i(i(x)) -> x +(i(x), x) -> 0 +(x, i(x)) -> 0 i(+(x, y)) -> +(i(x), i(y)) +(x, +(y, z)) -> +(+(x, y), z) +(+(x, i(y)), y) -> x +(+(x, y), i(y)) -> x Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(+(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(0) = 0 POL(i(x_1)) = 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: +(0, y) -> y +(x, 0) -> x +(i(x), x) -> 0 +(x, i(x)) -> 0 i(+(x, y)) -> +(i(x), i(y)) +(x, +(y, z)) -> +(+(x, y), z) +(+(x, i(y)), y) -> x +(+(x, y), i(y)) -> x ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: i(0) -> 0 i(i(x)) -> x Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:i_1 > 0 and weight map: 0=1 i_1=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: i(0) -> 0 i(i(x)) -> x ---------------------------------------- (4) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (5) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (6) YES