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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 7 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 65 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 26 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) a(a(x1)) -> a(b(a(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: B(a(a(x1))) -> A(b(c(x1))) B(a(a(x1))) -> B(c(x1)) B(a(a(x1))) -> C(x1) C(a(x1)) -> A(c(x1)) C(a(x1)) -> C(x1) C(b(x1)) -> B(a(x1)) C(b(x1)) -> A(x1) A(a(x1)) -> A(b(a(x1))) A(a(x1)) -> B(a(x1)) The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) a(a(x1)) -> a(b(a(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. B(a(a(x1))) -> A(b(c(x1))) B(a(a(x1))) -> B(c(x1)) B(a(a(x1))) -> C(x1) C(a(x1)) -> A(c(x1)) C(a(x1)) -> C(x1) C(b(x1)) -> A(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A_1(x_1) ) = max{0, 2x_1 - 2} POL( B_1(x_1) ) = max{0, 2x_1 - 2} POL( c_1(x_1) ) = x_1 + 2 POL( a_1(x_1) ) = x_1 + 2 POL( b_1(x_1) ) = x_1 POL( C_1(x_1) ) = 2x_1 + 2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) b(a(a(x1))) -> a(b(c(x1))) a(a(x1)) -> a(b(a(x1))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(b(x1)) -> B(a(x1)) A(a(x1)) -> A(b(a(x1))) A(a(x1)) -> B(a(x1)) The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) a(a(x1)) -> a(b(a(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 1 SCC with 2 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(x1)) -> A(b(a(x1))) The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) a(a(x1)) -> a(b(a(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(x1)) -> A(b(a(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A_1(x_1) ) = max{0, x_1 - 1} POL( a_1(x_1) ) = x_1 + 2 POL( b_1(x_1) ) = max{0, x_1 - 2} POL( c_1(x_1) ) = x_1 + 2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: a(a(x1)) -> a(b(a(x1))) b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) ---------------------------------------- (8) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) a(a(x1)) -> a(b(a(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (10) YES