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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 27 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 39 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(x1))) -> b(a(b(c(a(x1))))) b(a(x1)) -> a(b(b(x1))) b(c(a(x1))) -> c(a(b(x1))) 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: A(a(b(x1))) -> B(a(b(c(a(x1))))) A(a(b(x1))) -> A(b(c(a(x1)))) A(a(b(x1))) -> B(c(a(x1))) A(a(b(x1))) -> A(x1) B(a(x1)) -> A(b(b(x1))) B(a(x1)) -> B(b(x1)) B(a(x1)) -> B(x1) B(c(a(x1))) -> A(b(x1)) B(c(a(x1))) -> B(x1) The TRS R consists of the following rules: a(a(b(x1))) -> b(a(b(c(a(x1))))) b(a(x1)) -> a(b(b(x1))) b(c(a(x1))) -> c(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(a(b(x1))) -> B(c(a(x1))) A(a(b(x1))) -> A(x1) B(a(x1)) -> B(b(x1)) B(a(x1)) -> B(x1) B(c(a(x1))) -> B(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = 1 + x_1 POL(B(x_1)) = x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = x_1 POL(c(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(a(x1)) -> a(b(b(x1))) a(a(b(x1))) -> b(a(b(c(a(x1))))) b(c(a(x1))) -> c(a(b(x1))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(b(x1))) -> B(a(b(c(a(x1))))) A(a(b(x1))) -> A(b(c(a(x1)))) B(a(x1)) -> A(b(b(x1))) B(c(a(x1))) -> A(b(x1)) The TRS R consists of the following rules: a(a(b(x1))) -> b(a(b(c(a(x1))))) b(a(x1)) -> a(b(b(x1))) b(c(a(x1))) -> c(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(c(a(x1))) -> A(b(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = 0 POL(B(x_1)) = x_1 POL(a(x_1)) = 0 POL(b(x_1)) = x_1 POL(c(x_1)) = 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(a(x1)) -> a(b(b(x1))) a(a(b(x1))) -> b(a(b(c(a(x1))))) b(c(a(x1))) -> c(a(b(x1))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(b(x1))) -> B(a(b(c(a(x1))))) A(a(b(x1))) -> A(b(c(a(x1)))) B(a(x1)) -> A(b(b(x1))) The TRS R consists of the following rules: a(a(b(x1))) -> b(a(b(c(a(x1))))) b(a(x1)) -> a(b(b(x1))) b(c(a(x1))) -> c(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(a(b(x1))) -> A(b(c(a(x1)))) B(a(x1)) -> A(b(b(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A_1(x_1) ) = 2x_1 POL( B_1(x_1) ) = 2x_1 POL( b_1(x_1) ) = x_1 POL( a_1(x_1) ) = 2x_1 + 2 POL( c_1(x_1) ) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(a(x1)) -> a(b(b(x1))) a(a(b(x1))) -> b(a(b(c(a(x1))))) b(c(a(x1))) -> c(a(b(x1))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(b(x1))) -> B(a(b(c(a(x1))))) The TRS R consists of the following rules: a(a(b(x1))) -> b(a(b(c(a(x1))))) b(a(x1)) -> a(b(b(x1))) b(c(a(x1))) -> c(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (10) TRUE