YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 73 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 8 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 48 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: i(0(x1)) -> p(s(p(s(0(p(s(p(s(x1))))))))) i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(0(x1)) -> p(s(p(p(s(s(0(p(s(p(s(x1))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(i(x_1)) = 1 + x_1 POL(j(x_1)) = 1 + x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: i(0(x1)) -> p(s(p(s(0(p(s(p(s(x1))))))))) j(0(x1)) -> p(s(p(p(s(s(0(p(s(p(s(x1))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) Q is empty. ---------------------------------------- (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: I(s(x1)) -> P(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) I(s(x1)) -> P(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))) I(s(x1)) -> J(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))) I(s(x1)) -> P(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))) I(s(x1)) -> P(s(p(p(p(p(s(s(s(s(x1)))))))))) I(s(x1)) -> P(p(p(p(s(s(s(s(x1)))))))) I(s(x1)) -> P(p(p(s(s(s(s(x1))))))) I(s(x1)) -> P(p(s(s(s(s(x1)))))) I(s(x1)) -> P(s(s(s(s(x1))))) J(s(x1)) -> P(p(s(s(i(p(s(p(s(x1))))))))) J(s(x1)) -> P(s(s(i(p(s(p(s(x1)))))))) J(s(x1)) -> I(p(s(p(s(x1))))) J(s(x1)) -> P(s(p(s(x1)))) J(s(x1)) -> P(s(x1)) P(p(s(x1))) -> P(x1) The TRS R consists of the following rules: i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) Q is empty. 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 12 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(s(x1))) -> P(x1) The TRS R consists of the following rules: i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(s(x1))) -> P(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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: *P(p(s(x1))) -> P(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: I(s(x1)) -> J(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))) J(s(x1)) -> I(p(s(p(s(x1))))) The TRS R consists of the following rules: i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: I(s(x1)) -> J(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))) J(s(x1)) -> I(p(s(p(s(x1))))) The TRS R consists of the following rules: i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) The set Q consists of the following terms: i(s(x0)) j(s(x0)) p(s(x0)) p(0(x0)) 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: I(s(x1)) -> J(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))) J(s(x1)) -> I(p(s(p(s(x1))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) The set Q consists of the following terms: i(s(x0)) j(s(x0)) p(s(x0)) p(0(x0)) 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]. i(s(x0)) j(s(x0)) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: I(s(x1)) -> J(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))) J(s(x1)) -> I(p(s(p(s(x1))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) The set Q consists of the following terms: p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. J(s(x1)) -> I(p(s(p(s(x1))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( I_1(x_1) ) = max{0, x_1 - 2} POL( J_1(x_1) ) = max{0, x_1 - 1} POL( p_1(x_1) ) = max{0, x_1 - 2} POL( s_1(x_1) ) = x_1 + 2 POL( 0_1(x_1) ) = max{0, -2} The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: I(s(x1)) -> J(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) The set Q consists of the following terms: p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (22) TRUE