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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 38 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPOrderProof [EQUIVALENT, 48 ms] (11) QDP (12) DependencyGraphProof [EQUIVALENT, 0 ms] (13) AND (14) QDP (15) QDPOrderProof [EQUIVALENT, 9 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) QDPOrderProof [EQUIVALENT, 9 ms] (21) QDP (22) PisEmptyProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) QDPOrderProof [EQUIVALENT, 41 ms] (26) QDP (27) PisEmptyProof [EQUIVALENT, 0 ms] (28) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(s(s(x1)))) -> s(s(a(a(x1)))) b(b(a(a(b(b(s(s(x1)))))))) -> a(a(b(b(s(s(a(a(x1)))))))) b(b(a(a(b(b(b(b(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) a(a(b(b(a(a(a(a(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: s(s(a(a(x1)))) -> a(a(s(s(x1)))) s(s(b(b(a(a(b(b(x1)))))))) -> a(a(s(s(b(b(a(a(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(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: S(s(a(a(x1)))) -> A(a(s(s(x1)))) S(s(a(a(x1)))) -> A(s(s(x1))) S(s(a(a(x1)))) -> S(s(x1)) S(s(a(a(x1)))) -> S(x1) S(s(b(b(a(a(b(b(x1)))))))) -> A(a(s(s(b(b(a(a(x1)))))))) S(s(b(b(a(a(b(b(x1)))))))) -> A(s(s(b(b(a(a(x1))))))) S(s(b(b(a(a(b(b(x1)))))))) -> S(s(b(b(a(a(x1)))))) S(s(b(b(a(a(b(b(x1)))))))) -> S(b(b(a(a(x1))))) S(s(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(x1)))) S(s(b(b(a(a(b(b(x1)))))))) -> B(a(a(x1))) S(s(b(b(a(a(b(b(x1)))))))) -> A(a(x1)) S(s(b(b(a(a(b(b(x1)))))))) -> A(x1) B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(b(b(a(a(x1)))))))) B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(b(b(a(a(x1))))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(b(b(a(a(x1)))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(b(b(a(a(x1))))) B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(x1)))) B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(x1))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(x1)) B(b(b(b(a(a(b(b(x1)))))))) -> A(x1) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(a(a(b(b(x1)))))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(a(a(b(b(x1))))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(a(a(b(b(x1)))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(a(a(b(b(x1))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(x1)))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(x1))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(x1)) A(a(a(a(b(b(a(a(x1)))))))) -> B(x1) The TRS R consists of the following rules: s(s(a(a(x1)))) -> a(a(s(s(x1)))) s(s(b(b(a(a(b(b(x1)))))))) -> a(a(s(s(b(b(a(a(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(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 9 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(b(b(a(a(x1))))))) B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(b(b(a(a(x1)))))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(b(b(a(a(x1)))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(a(a(b(b(x1)))))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(a(a(b(b(x1))))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(a(a(b(b(x1)))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(b(b(a(a(x1))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(a(a(b(b(x1))))) B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(x1)))) B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(x1))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(x1)) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(x1)))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(x1))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(x1)) B(b(b(b(a(a(b(b(x1)))))))) -> A(x1) A(a(a(a(b(b(a(a(x1)))))))) -> B(x1) The TRS R consists of the following rules: s(s(a(a(x1)))) -> a(a(s(s(x1)))) s(s(b(b(a(a(b(b(x1)))))))) -> a(a(s(s(b(b(a(a(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(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: B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(b(b(a(a(x1))))))) B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(b(b(a(a(x1)))))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(b(b(a(a(x1)))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(a(a(b(b(x1)))))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(a(a(b(b(x1))))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(a(a(b(b(x1)))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(b(b(a(a(x1))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(a(a(b(b(x1))))) B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(x1)))) B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(x1))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(x1)) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(x1)))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(x1))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(x1)) B(b(b(b(a(a(b(b(x1)))))))) -> A(x1) A(a(a(a(b(b(a(a(x1)))))))) -> B(x1) The TRS R consists of the following rules: b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(b(b(a(a(x1))))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(b(b(a(a(x1)))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(a(a(b(b(x1))))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(a(a(b(b(x1)))))) B(b(b(b(a(a(b(b(x1)))))))) -> A(b(b(a(a(x1))))) A(a(a(a(b(b(a(a(x1)))))))) -> B(a(a(b(b(x1))))) B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(x1)))) B(b(b(b(a(a(b(b(x1)))))))) -> B(a(a(x1))) B(b(b(b(a(a(b(b(x1)))))))) -> A(a(x1)) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(x1)))) A(a(a(a(b(b(a(a(x1)))))))) -> A(b(b(x1))) A(a(a(a(b(b(a(a(x1)))))))) -> B(b(x1)) B(b(b(b(a(a(b(b(x1)))))))) -> A(x1) A(a(a(a(b(b(a(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)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(b(b(a(a(x1)))))))) A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(a(a(b(b(x1)))))))) The TRS R consists of the following rules: b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs. ---------------------------------------- (13) Complex Obligation (AND) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(a(a(b(b(x1)))))))) The TRS R consists of the following rules: b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(a(a(a(b(b(a(a(x1)))))))) -> A(a(b(b(a(a(b(b(x1)))))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) ---------------------------------------- (16) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (18) YES ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(b(b(a(a(x1)))))))) The TRS R consists of the following rules: b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(b(b(b(a(a(b(b(x1)))))))) -> B(b(a(a(b(b(a(a(x1)))))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(B(x_1)) = x_1 POL(a(x_1)) = 0 POL(b(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) ---------------------------------------- (21) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (23) YES ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: S(s(a(a(x1)))) -> S(x1) S(s(a(a(x1)))) -> S(s(x1)) S(s(b(b(a(a(b(b(x1)))))))) -> S(s(b(b(a(a(x1)))))) The TRS R consists of the following rules: s(s(a(a(x1)))) -> a(a(s(s(x1)))) s(s(b(b(a(a(b(b(x1)))))))) -> a(a(s(s(b(b(a(a(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. S(s(a(a(x1)))) -> S(x1) S(s(a(a(x1)))) -> S(s(x1)) S(s(b(b(a(a(b(b(x1)))))))) -> S(s(b(b(a(a(x1)))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(S(x_1)) = x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 1 + x_1 POL(s(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: s(s(a(a(x1)))) -> a(a(s(s(x1)))) s(s(b(b(a(a(b(b(x1)))))))) -> a(a(s(s(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) ---------------------------------------- (26) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: s(s(a(a(x1)))) -> a(a(s(s(x1)))) s(s(b(b(a(a(b(b(x1)))))))) -> a(a(s(s(b(b(a(a(x1)))))))) b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1)))))))) a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1)))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (28) YES