22.46/6.51 YES 23.08/6.69 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 23.08/6.69 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 23.08/6.69 23.08/6.69 23.08/6.69 Termination w.r.t. Q of the given QTRS could be proven: 23.08/6.69 23.08/6.69 (0) QTRS 23.08/6.69 (1) QTRS Reverse [EQUIVALENT, 0 ms] 23.08/6.69 (2) QTRS 23.08/6.69 (3) QTRSRRRProof [EQUIVALENT, 93 ms] 23.08/6.69 (4) QTRS 23.08/6.69 (5) Overlay + Local Confluence [EQUIVALENT, 36 ms] 23.08/6.69 (6) QTRS 23.08/6.69 (7) DependencyPairsProof [EQUIVALENT, 56 ms] 23.08/6.69 (8) QDP 23.08/6.69 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 23.08/6.69 (10) QDP 23.08/6.69 (11) UsableRulesProof [EQUIVALENT, 0 ms] 23.08/6.69 (12) QDP 23.08/6.69 (13) QReductionProof [EQUIVALENT, 0 ms] 23.08/6.69 (14) QDP 23.08/6.69 (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] 23.08/6.69 (16) YES 23.08/6.69 23.08/6.69 23.08/6.69 ---------------------------------------- 23.08/6.69 23.08/6.69 (0) 23.08/6.69 Obligation: 23.08/6.69 Q restricted rewrite system: 23.08/6.69 The TRS R consists of the following rules: 23.08/6.69 23.08/6.69 0(0(1(2(1(x1))))) -> 1(0(1(1(0(x1))))) 23.08/6.69 0(3(1(2(4(x1))))) -> 0(5(1(4(x1)))) 23.08/6.69 4(3(0(1(1(x1))))) -> 5(4(5(2(x1)))) 23.08/6.69 4(3(4(2(2(x1))))) -> 0(5(0(2(x1)))) 23.08/6.69 2(1(1(4(0(2(x1)))))) -> 1(1(5(3(5(x1))))) 23.08/6.69 2(2(4(1(3(4(2(x1))))))) -> 2(3(2(4(3(5(x1)))))) 23.08/6.69 4(0(5(4(2(4(0(x1))))))) -> 0(1(2(2(2(2(0(x1))))))) 23.08/6.69 4(2(4(2(5(1(0(1(5(x1))))))))) -> 5(0(3(1(0(2(2(5(x1)))))))) 23.08/6.69 5(2(4(5(0(3(0(2(3(x1))))))))) -> 5(3(2(1(0(1(3(4(0(x1))))))))) 23.08/6.69 1(4(1(3(2(3(3(1(2(1(x1)))))))))) -> 1(5(3(2(4(5(1(3(4(x1))))))))) 23.08/6.69 4(0(4(2(3(4(5(1(1(5(1(x1))))))))))) -> 5(4(3(2(2(3(2(1(2(4(0(x1))))))))))) 23.08/6.69 4(5(2(2(5(4(4(3(4(5(4(x1))))))))))) -> 5(2(3(2(5(0(0(0(5(4(x1)))))))))) 23.08/6.69 4(3(4(4(0(3(0(3(2(3(2(1(x1)))))))))))) -> 5(5(5(5(5(0(2(2(4(4(2(0(x1)))))))))))) 23.08/6.69 5(3(4(4(3(3(5(2(5(2(1(1(4(2(x1)))))))))))))) -> 5(4(0(5(2(5(5(3(1(0(3(3(5(x1))))))))))))) 23.08/6.69 0(5(1(0(3(3(2(5(5(4(0(5(5(5(2(x1))))))))))))))) -> 1(0(1(3(4(5(3(3(3(2(2(3(3(5(2(x1))))))))))))))) 23.08/6.69 4(5(3(2(1(1(5(2(2(3(4(3(2(3(1(x1))))))))))))))) -> 0(1(3(5(0(1(3(4(0(3(5(4(3(1(x1)))))))))))))) 23.08/6.69 5(0(1(0(1(1(5(1(1(5(5(2(1(1(0(x1))))))))))))))) -> 5(1(5(1(1(1(3(0(3(3(3(3(1(0(x1)))))))))))))) 23.08/6.69 5(3(0(4(4(1(1(5(3(4(1(1(2(3(2(x1))))))))))))))) -> 5(4(1(4(0(2(1(2(2(5(3(5(3(4(4(x1))))))))))))))) 23.08/6.69 2(4(1(0(2(3(2(3(5(3(1(2(3(1(1(4(x1)))))))))))))))) -> 2(2(2(1(4(5(0(1(0(3(1(3(5(1(2(x1))))))))))))))) 23.08/6.70 0(1(1(3(2(2(0(0(0(5(0(2(4(3(3(0(1(x1))))))))))))))))) -> 5(4(1(1(0(5(2(0(2(3(3(3(0(5(0(1(x1)))))))))))))))) 23.08/6.70 2(1(1(5(3(1(3(4(3(5(3(3(2(4(3(1(4(x1))))))))))))))))) -> 1(0(1(0(0(2(1(3(2(2(0(3(0(5(2(4(x1)))))))))))))))) 23.08/6.70 2(4(3(5(0(2(5(5(1(5(0(4(4(4(1(4(3(x1))))))))))))))))) -> 2(0(5(2(2(0(5(4(1(3(2(4(1(4(1(1(0(x1))))))))))))))))) 23.08/6.70 0(4(5(4(5(0(2(3(1(2(4(5(3(5(0(4(3(3(2(x1))))))))))))))))))) -> 1(0(2(4(5(5(2(2(4(2(1(1(4(0(1(2(3(0(2(5(x1)))))))))))))))))))) 23.08/6.70 4(5(1(0(2(0(5(4(5(4(4(2(5(5(2(3(5(4(2(3(x1)))))))))))))))))))) -> 0(5(5(2(0(5(2(4(2(5(2(5(2(0(1(5(2(3(3(0(x1)))))))))))))))))))) 23.08/6.70 3(1(2(4(3(4(3(2(0(3(2(3(4(3(4(5(4(3(4(1(1(x1))))))))))))))))))))) -> 3(4(0(0(2(4(5(0(0(4(3(5(4(3(0(3(2(2(1(1(x1)))))))))))))))))))) 23.08/6.70 23.08/6.70 Q is empty. 23.08/6.70 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (1) QTRS Reverse (EQUIVALENT) 23.08/6.70 We applied the QTRS Reverse Processor [REVERSE]. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (2) 23.08/6.70 Obligation: 23.08/6.70 Q restricted rewrite system: 23.08/6.70 The TRS R consists of the following rules: 23.08/6.70 23.08/6.70 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 23.08/6.70 4(2(1(3(0(x1))))) -> 4(1(5(0(x1)))) 23.08/6.70 1(1(0(3(4(x1))))) -> 2(5(4(5(x1)))) 23.08/6.70 2(2(4(3(4(x1))))) -> 2(0(5(0(x1)))) 23.08/6.70 2(0(4(1(1(2(x1)))))) -> 5(3(5(1(1(x1))))) 23.08/6.70 2(4(3(1(4(2(2(x1))))))) -> 5(3(4(2(3(2(x1)))))) 23.08/6.70 0(4(2(4(5(0(4(x1))))))) -> 0(2(2(2(2(1(0(x1))))))) 23.08/6.70 5(1(0(1(5(2(4(2(4(x1))))))))) -> 5(2(2(0(1(3(0(5(x1)))))))) 23.08/6.70 3(2(0(3(0(5(4(2(5(x1))))))))) -> 0(4(3(1(0(1(2(3(5(x1))))))))) 23.08/6.70 1(2(1(3(3(2(3(1(4(1(x1)))))))))) -> 4(3(1(5(4(2(3(5(1(x1))))))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 23.08/6.70 4(5(4(3(4(4(5(2(2(5(4(x1))))))))))) -> 4(5(0(0(0(5(2(3(2(5(x1)))))))))) 23.08/6.70 1(2(3(2(3(0(3(0(4(4(3(4(x1)))))))))))) -> 0(2(4(4(2(2(0(5(5(5(5(5(x1)))))))))))) 23.08/6.70 2(4(1(1(2(5(2(5(3(3(4(4(3(5(x1)))))))))))))) -> 5(3(3(0(1(3(5(5(2(5(0(4(5(x1))))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 23.08/6.70 1(3(2(3(4(3(2(2(5(1(1(2(3(5(4(x1))))))))))))))) -> 1(3(4(5(3(0(4(3(1(0(5(3(1(0(x1)))))))))))))) 23.08/6.70 0(1(1(2(5(5(1(1(5(1(1(0(1(0(5(x1))))))))))))))) -> 0(1(3(3(3(3(0(3(1(1(1(5(1(5(x1)))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 23.08/6.70 4(1(1(3(2(1(3(5(3(2(3(2(0(1(4(2(x1)))))))))))))))) -> 2(1(5(3(1(3(0(1(0(5(4(1(2(2(2(x1))))))))))))))) 23.08/6.70 1(0(3(3(4(2(0(5(0(0(0(2(2(3(1(1(0(x1))))))))))))))))) -> 1(0(5(0(3(3(3(2(0(2(5(0(1(1(4(5(x1)))))))))))))))) 23.08/6.70 4(1(3(4(2(3(3(5(3(4(3(1(3(5(1(1(2(x1))))))))))))))))) -> 4(2(5(0(3(0(2(2(3(1(2(0(0(1(0(1(x1)))))))))))))))) 23.08/6.70 3(4(1(4(4(4(0(5(1(5(5(2(0(5(3(4(2(x1))))))))))))))))) -> 0(1(1(4(1(4(2(3(1(4(5(0(2(2(5(0(2(x1))))))))))))))))) 23.08/6.70 2(3(3(4(0(5(3(5(4(2(1(3(2(0(5(4(5(4(0(x1))))))))))))))))))) -> 5(2(0(3(2(1(0(4(1(1(2(4(2(2(5(5(4(2(0(1(x1)))))))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) 23.08/6.70 1(1(4(3(4(5(4(3(4(3(2(3(0(2(3(4(3(4(2(1(3(x1))))))))))))))))))))) -> 1(1(2(2(3(0(3(4(5(3(4(0(0(5(4(2(0(0(4(3(x1)))))))))))))))))))) 23.08/6.70 23.08/6.70 Q is empty. 23.08/6.70 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (3) QTRSRRRProof (EQUIVALENT) 23.08/6.70 Used ordering: 23.08/6.70 Polynomial interpretation [POLO]: 23.08/6.70 23.08/6.70 POL(0(x_1)) = 9 + x_1 23.08/6.70 POL(1(x_1)) = 7 + x_1 23.08/6.70 POL(2(x_1)) = 7 + x_1 23.08/6.70 POL(3(x_1)) = 9 + x_1 23.08/6.70 POL(4(x_1)) = 8 + x_1 23.08/6.70 POL(5(x_1)) = 8 + x_1 23.08/6.70 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 23.08/6.70 23.08/6.70 4(2(1(3(0(x1))))) -> 4(1(5(0(x1)))) 23.08/6.70 1(1(0(3(4(x1))))) -> 2(5(4(5(x1)))) 23.08/6.70 2(2(4(3(4(x1))))) -> 2(0(5(0(x1)))) 23.08/6.70 2(0(4(1(1(2(x1)))))) -> 5(3(5(1(1(x1))))) 23.08/6.70 2(4(3(1(4(2(2(x1))))))) -> 5(3(4(2(3(2(x1)))))) 23.08/6.70 0(4(2(4(5(0(4(x1))))))) -> 0(2(2(2(2(1(0(x1))))))) 23.08/6.70 5(1(0(1(5(2(4(2(4(x1))))))))) -> 5(2(2(0(1(3(0(5(x1)))))))) 23.08/6.70 3(2(0(3(0(5(4(2(5(x1))))))))) -> 0(4(3(1(0(1(2(3(5(x1))))))))) 23.08/6.70 1(2(1(3(3(2(3(1(4(1(x1)))))))))) -> 4(3(1(5(4(2(3(5(1(x1))))))))) 23.08/6.70 4(5(4(3(4(4(5(2(2(5(4(x1))))))))))) -> 4(5(0(0(0(5(2(3(2(5(x1)))))))))) 23.08/6.70 1(2(3(2(3(0(3(0(4(4(3(4(x1)))))))))))) -> 0(2(4(4(2(2(0(5(5(5(5(5(x1)))))))))))) 23.08/6.70 2(4(1(1(2(5(2(5(3(3(4(4(3(5(x1)))))))))))))) -> 5(3(3(0(1(3(5(5(2(5(0(4(5(x1))))))))))))) 23.08/6.70 1(3(2(3(4(3(2(2(5(1(1(2(3(5(4(x1))))))))))))))) -> 1(3(4(5(3(0(4(3(1(0(5(3(1(0(x1)))))))))))))) 23.08/6.70 0(1(1(2(5(5(1(1(5(1(1(0(1(0(5(x1))))))))))))))) -> 0(1(3(3(3(3(0(3(1(1(1(5(1(5(x1)))))))))))))) 23.08/6.70 4(1(1(3(2(1(3(5(3(2(3(2(0(1(4(2(x1)))))))))))))))) -> 2(1(5(3(1(3(0(1(0(5(4(1(2(2(2(x1))))))))))))))) 23.08/6.70 1(0(3(3(4(2(0(5(0(0(0(2(2(3(1(1(0(x1))))))))))))))))) -> 1(0(5(0(3(3(3(2(0(2(5(0(1(1(4(5(x1)))))))))))))))) 23.08/6.70 4(1(3(4(2(3(3(5(3(4(3(1(3(5(1(1(2(x1))))))))))))))))) -> 4(2(5(0(3(0(2(2(3(1(2(0(0(1(0(1(x1)))))))))))))))) 23.08/6.70 3(4(1(4(4(4(0(5(1(5(5(2(0(5(3(4(2(x1))))))))))))))))) -> 0(1(1(4(1(4(2(3(1(4(5(0(2(2(5(0(2(x1))))))))))))))))) 23.08/6.70 2(3(3(4(0(5(3(5(4(2(1(3(2(0(5(4(5(4(0(x1))))))))))))))))))) -> 5(2(0(3(2(1(0(4(1(1(2(4(2(2(5(5(4(2(0(1(x1)))))))))))))))))))) 23.08/6.70 1(1(4(3(4(5(4(3(4(3(2(3(0(2(3(4(3(4(2(1(3(x1))))))))))))))))))))) -> 1(1(2(2(3(0(3(4(5(3(4(0(0(5(4(2(0(0(4(3(x1)))))))))))))))))))) 23.08/6.70 23.08/6.70 23.08/6.70 23.08/6.70 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (4) 23.08/6.70 Obligation: 23.08/6.70 Q restricted rewrite system: 23.08/6.70 The TRS R consists of the following rules: 23.08/6.70 23.08/6.70 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) 23.08/6.70 23.08/6.70 Q is empty. 23.08/6.70 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (5) Overlay + Local Confluence (EQUIVALENT) 23.08/6.70 The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (6) 23.08/6.70 Obligation: 23.08/6.70 Q restricted rewrite system: 23.08/6.70 The TRS R consists of the following rules: 23.08/6.70 23.08/6.70 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) 23.08/6.70 23.08/6.70 The set Q consists of the following terms: 23.08/6.70 23.08/6.70 1(2(1(0(0(x0))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) 23.08/6.70 23.08/6.70 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (7) DependencyPairsProof (EQUIVALENT) 23.08/6.70 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (8) 23.08/6.70 Obligation: 23.08/6.70 Q DP problem: 23.08/6.70 The TRS P consists of the following rules: 23.08/6.70 23.08/6.70 1^1(2(1(0(0(x1))))) -> 1^1(1(0(1(x1)))) 23.08/6.70 1^1(2(1(0(0(x1))))) -> 1^1(0(1(x1))) 23.08/6.70 1^1(2(1(0(0(x1))))) -> 1^1(x1) 23.08/6.70 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(1(2(3(2(2(3(4(5(x1))))))))) 23.08/6.70 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 1^1(2(3(2(2(3(4(5(x1)))))))) 23.08/6.70 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(3(2(2(3(4(5(x1))))))) 23.08/6.70 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 3^1(2(2(3(4(5(x1)))))) 23.08/6.70 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(2(3(4(5(x1))))) 23.08/6.70 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(3(4(5(x1)))) 23.08/6.70 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 3^1(4(5(x1))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2^1(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(2(2(3(3(3(5(4(3(1(0(1(x1)))))))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2^1(2(3(3(3(5(4(3(1(0(1(x1))))))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2^1(3(3(3(5(4(3(1(0(1(x1)))))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(3(3(5(4(3(1(0(1(x1))))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(3(5(4(3(1(0(1(x1)))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(5(4(3(1(0(1(x1))))))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(1(0(1(x1)))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 1^1(0(1(x1))) 23.08/6.70 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 1^1(x1) 23.08/6.70 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 3^1(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))) 23.08/6.70 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 3^1(5(2(2(1(2(0(4(1(4(5(x1))))))))))) 23.08/6.70 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 2^1(2(1(2(0(4(1(4(5(x1))))))))) 23.08/6.70 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 2^1(1(2(0(4(1(4(5(x1)))))))) 23.08/6.70 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 1^1(2(0(4(1(4(5(x1))))))) 23.08/6.70 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 2^1(0(4(1(4(5(x1)))))) 23.08/6.70 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 1^1(4(5(x1))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 3^1(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 3^1(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 1^1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(2(4(2(5(0(2(5(5(0(x1))))))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(4(2(5(0(2(5(5(0(x1))))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(0(2(5(5(0(x1))))))) 23.08/6.70 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(5(0(x1)))) 23.08/6.70 23.08/6.70 The TRS R consists of the following rules: 23.08/6.70 23.08/6.70 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) 23.08/6.70 23.08/6.70 The set Q consists of the following terms: 23.08/6.70 23.08/6.70 1(2(1(0(0(x0))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) 23.08/6.70 23.08/6.70 We have to consider all minimal (P,Q,R)-chains. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (9) DependencyGraphProof (EQUIVALENT) 23.08/6.70 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 36 less nodes. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (10) 23.08/6.70 Obligation: 23.08/6.70 Q DP problem: 23.08/6.70 The TRS P consists of the following rules: 23.08/6.70 23.08/6.70 1^1(2(1(0(0(x1))))) -> 1^1(x1) 23.08/6.70 23.08/6.70 The TRS R consists of the following rules: 23.08/6.70 23.08/6.70 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) 23.08/6.70 23.08/6.70 The set Q consists of the following terms: 23.08/6.70 23.08/6.70 1(2(1(0(0(x0))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) 23.08/6.70 23.08/6.70 We have to consider all minimal (P,Q,R)-chains. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (11) UsableRulesProof (EQUIVALENT) 23.08/6.70 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. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (12) 23.08/6.70 Obligation: 23.08/6.70 Q DP problem: 23.08/6.70 The TRS P consists of the following rules: 23.08/6.70 23.08/6.70 1^1(2(1(0(0(x1))))) -> 1^1(x1) 23.08/6.70 23.08/6.70 R is empty. 23.08/6.70 The set Q consists of the following terms: 23.08/6.70 23.08/6.70 1(2(1(0(0(x0))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) 23.08/6.70 23.08/6.70 We have to consider all minimal (P,Q,R)-chains. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (13) QReductionProof (EQUIVALENT) 23.08/6.70 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 23.08/6.70 23.08/6.70 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) 23.08/6.70 23.08/6.70 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (14) 23.08/6.70 Obligation: 23.08/6.70 Q DP problem: 23.08/6.70 The TRS P consists of the following rules: 23.08/6.70 23.08/6.70 1^1(2(1(0(0(x1))))) -> 1^1(x1) 23.08/6.70 23.08/6.70 R is empty. 23.08/6.70 The set Q consists of the following terms: 23.08/6.70 23.08/6.70 1(2(1(0(0(x0))))) 23.08/6.70 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 23.08/6.70 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 23.08/6.70 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 23.08/6.70 23.08/6.70 We have to consider all minimal (P,Q,R)-chains. 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (15) QDPSizeChangeProof (EQUIVALENT) 23.08/6.70 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. 23.08/6.70 23.08/6.70 From the DPs we obtained the following set of size-change graphs: 23.08/6.70 *1^1(2(1(0(0(x1))))) -> 1^1(x1) 23.08/6.70 The graph contains the following edges 1 > 1 23.08/6.70 23.08/6.70 23.08/6.70 ---------------------------------------- 23.08/6.70 23.08/6.70 (16) 23.08/6.70 YES 24.98/8.97 EOF