11.36/3.87 YES 11.71/3.93 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 11.71/3.93 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 11.71/3.93 11.71/3.93 11.71/3.93 Termination w.r.t. Q of the given QTRS could be proven: 11.71/3.93 11.71/3.93 (0) QTRS 11.71/3.93 (1) QTRSRRRProof [EQUIVALENT, 122 ms] 11.71/3.93 (2) QTRS 11.71/3.93 (3) QTRSRRRProof [EQUIVALENT, 2 ms] 11.71/3.93 (4) QTRS 11.71/3.93 (5) DependencyPairsProof [EQUIVALENT, 17 ms] 11.71/3.93 (6) QDP 11.71/3.93 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 11.71/3.93 (8) TRUE 11.71/3.93 11.71/3.93 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (0) 11.71/3.93 Obligation: 11.71/3.93 Q restricted rewrite system: 11.71/3.93 The TRS R consists of the following rules: 11.71/3.93 11.71/3.93 0(0(1(2(x1)))) -> 2(3(0(x1))) 11.71/3.93 4(2(5(2(x1)))) -> 1(4(0(x1))) 11.71/3.93 4(5(3(4(x1)))) -> 4(4(2(4(x1)))) 11.71/3.93 1(3(5(5(2(x1))))) -> 1(3(2(3(x1)))) 11.71/3.93 5(2(2(1(2(x1))))) -> 0(0(2(3(x1)))) 11.71/3.93 2(2(5(3(2(2(x1)))))) -> 5(1(1(0(3(x1))))) 11.71/3.93 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 11.71/3.93 3(4(1(4(2(4(x1)))))) -> 1(3(3(3(x1)))) 11.71/3.93 3(5(2(2(4(5(x1)))))) -> 3(2(4(3(0(x1))))) 11.71/3.93 5(2(1(0(1(5(x1)))))) -> 5(4(2(4(5(1(x1)))))) 11.71/3.93 1(3(5(4(1(2(2(x1))))))) -> 3(3(3(4(4(0(x1)))))) 11.71/3.93 4(5(4(3(0(5(1(x1))))))) -> 4(3(3(5(4(1(x1)))))) 11.71/3.93 2(1(5(2(1(3(4(4(x1)))))))) -> 4(0(3(4(0(1(2(x1))))))) 11.71/3.93 5(4(0(2(2(4(0(4(x1)))))))) -> 3(1(5(1(3(0(4(x1))))))) 11.71/3.93 3(4(2(1(1(2(2(5(4(x1))))))))) -> 3(3(3(1(3(3(4(x1))))))) 11.71/3.93 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(x1))))))))) 11.71/3.93 5(2(1(3(1(5(2(5(4(4(x1)))))))))) -> 5(3(4(5(0(1(4(0(3(x1))))))))) 11.71/3.93 2(4(1(2(5(2(4(1(3(2(0(3(x1)))))))))))) -> 4(3(0(4(2(3(4(3(4(2(0(x1))))))))))) 11.71/3.93 0(2(3(5(4(2(2(1(0(3(3(5(0(x1))))))))))))) -> 3(3(1(2(3(0(4(0(0(0(2(0(x1)))))))))))) 11.71/3.93 2(1(0(2(1(4(0(0(2(0(0(0(5(2(x1)))))))))))))) -> 2(2(4(2(1(4(3(0(5(1(3(3(0(x1))))))))))))) 11.71/3.93 4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x1))))))))))))))) -> 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x1))))))))))))))) 11.71/3.93 5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x1))))))))))))))) -> 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x1)))))))))))))) 11.71/3.93 5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x1))))))))))))))) -> 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x1)))))))))))))) 11.71/3.93 5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x1))))))))))))))))))) -> 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x1)))))))))))))))))) 11.71/3.93 2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x1)))))))))))))))))))) -> 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x1)))))))))))))))))))) 11.71/3.93 11.71/3.93 Q is empty. 11.71/3.93 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (1) QTRSRRRProof (EQUIVALENT) 11.71/3.93 Used ordering: 11.71/3.93 Polynomial interpretation [POLO]: 11.71/3.93 11.71/3.93 POL(0(x_1)) = 22 + x_1 11.71/3.93 POL(1(x_1)) = 14 + x_1 11.71/3.93 POL(2(x_1)) = 19 + x_1 11.71/3.93 POL(3(x_1)) = 21 + x_1 11.71/3.93 POL(4(x_1)) = 14 + x_1 11.71/3.93 POL(5(x_1)) = 13 + x_1 11.71/3.93 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 11.71/3.93 11.71/3.93 0(0(1(2(x1)))) -> 2(3(0(x1))) 11.71/3.93 4(2(5(2(x1)))) -> 1(4(0(x1))) 11.71/3.93 4(5(3(4(x1)))) -> 4(4(2(4(x1)))) 11.71/3.93 1(3(5(5(2(x1))))) -> 1(3(2(3(x1)))) 11.71/3.93 2(2(5(3(2(2(x1)))))) -> 5(1(1(0(3(x1))))) 11.71/3.93 3(4(1(4(2(4(x1)))))) -> 1(3(3(3(x1)))) 11.71/3.93 3(5(2(2(4(5(x1)))))) -> 3(2(4(3(0(x1))))) 11.71/3.93 5(2(1(0(1(5(x1)))))) -> 5(4(2(4(5(1(x1)))))) 11.71/3.93 1(3(5(4(1(2(2(x1))))))) -> 3(3(3(4(4(0(x1)))))) 11.71/3.93 4(5(4(3(0(5(1(x1))))))) -> 4(3(3(5(4(1(x1)))))) 11.71/3.93 2(1(5(2(1(3(4(4(x1)))))))) -> 4(0(3(4(0(1(2(x1))))))) 11.71/3.93 5(4(0(2(2(4(0(4(x1)))))))) -> 3(1(5(1(3(0(4(x1))))))) 11.71/3.93 3(4(2(1(1(2(2(5(4(x1))))))))) -> 3(3(3(1(3(3(4(x1))))))) 11.71/3.93 2(4(1(2(5(2(4(1(3(2(0(3(x1)))))))))))) -> 4(3(0(4(2(3(4(3(4(2(0(x1))))))))))) 11.71/3.93 0(2(3(5(4(2(2(1(0(3(3(5(0(x1))))))))))))) -> 3(3(1(2(3(0(4(0(0(0(2(0(x1)))))))))))) 11.71/3.93 2(1(0(2(1(4(0(0(2(0(0(0(5(2(x1)))))))))))))) -> 2(2(4(2(1(4(3(0(5(1(3(3(0(x1))))))))))))) 11.71/3.93 4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x1))))))))))))))) -> 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x1))))))))))))))) 11.71/3.93 5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x1))))))))))))))) -> 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x1)))))))))))))) 11.71/3.93 5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x1))))))))))))))) -> 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x1)))))))))))))) 11.71/3.93 5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x1))))))))))))))))))) -> 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x1)))))))))))))))))) 11.71/3.93 2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x1)))))))))))))))))))) -> 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x1)))))))))))))))))))) 11.71/3.93 11.71/3.93 11.71/3.93 11.71/3.93 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (2) 11.71/3.93 Obligation: 11.71/3.93 Q restricted rewrite system: 11.71/3.93 The TRS R consists of the following rules: 11.71/3.93 11.71/3.93 5(2(2(1(2(x1))))) -> 0(0(2(3(x1)))) 11.71/3.93 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 11.71/3.93 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(x1))))))))) 11.71/3.93 5(2(1(3(1(5(2(5(4(4(x1)))))))))) -> 5(3(4(5(0(1(4(0(3(x1))))))))) 11.71/3.93 11.71/3.93 Q is empty. 11.71/3.93 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (3) QTRSRRRProof (EQUIVALENT) 11.71/3.93 Used ordering: 11.71/3.93 Polynomial interpretation [POLO]: 11.71/3.93 11.71/3.93 POL(0(x_1)) = x_1 11.71/3.93 POL(1(x_1)) = x_1 11.71/3.93 POL(2(x_1)) = x_1 11.71/3.93 POL(3(x_1)) = x_1 11.71/3.93 POL(4(x_1)) = x_1 11.71/3.93 POL(5(x_1)) = 1 + x_1 11.71/3.93 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 11.71/3.93 11.71/3.93 5(2(2(1(2(x1))))) -> 0(0(2(3(x1)))) 11.71/3.93 5(2(1(3(1(5(2(5(4(4(x1)))))))))) -> 5(3(4(5(0(1(4(0(3(x1))))))))) 11.71/3.93 11.71/3.93 11.71/3.93 11.71/3.93 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (4) 11.71/3.93 Obligation: 11.71/3.93 Q restricted rewrite system: 11.71/3.93 The TRS R consists of the following rules: 11.71/3.93 11.71/3.93 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 11.71/3.93 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(x1))))))))) 11.71/3.93 11.71/3.93 Q is empty. 11.71/3.93 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (5) DependencyPairsProof (EQUIVALENT) 11.71/3.93 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (6) 11.71/3.93 Obligation: 11.71/3.93 Q DP problem: 11.71/3.93 The TRS P consists of the following rules: 11.71/3.93 11.71/3.93 2^1(5(1(2(1(1(x1)))))) -> 5^1(2(1(2(4(1(x1)))))) 11.71/3.93 2^1(5(1(2(1(1(x1)))))) -> 2^1(1(2(4(1(x1))))) 11.71/3.93 2^1(5(1(2(1(1(x1)))))) -> 2^1(4(1(x1))) 11.71/3.93 5^1(4(4(5(0(1(4(5(4(x1))))))))) -> 5^1(5(0(4(1(4(5(4(x1)))))))) 11.71/3.93 5^1(4(4(5(0(1(4(5(4(x1))))))))) -> 5^1(0(4(1(4(5(4(x1))))))) 11.71/3.93 11.71/3.93 The TRS R consists of the following rules: 11.71/3.93 11.71/3.93 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 11.71/3.93 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(x1))))))))) 11.71/3.93 11.71/3.93 Q is empty. 11.71/3.93 We have to consider all minimal (P,Q,R)-chains. 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (7) DependencyGraphProof (EQUIVALENT) 11.71/3.93 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 5 less nodes. 11.71/3.93 ---------------------------------------- 11.71/3.93 11.71/3.93 (8) 11.71/3.93 TRUE 12.15/9.02 EOF