51.49/13.99 YES 51.49/14.04 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 51.49/14.04 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 51.49/14.04 51.49/14.04 51.49/14.04 Termination w.r.t. Q of the given QTRS could be proven: 51.49/14.04 51.49/14.04 (0) QTRS 51.49/14.04 (1) QTRS Reverse [EQUIVALENT, 0 ms] 51.49/14.04 (2) QTRS 51.49/14.04 (3) QTRSRRRProof [EQUIVALENT, 118 ms] 51.49/14.04 (4) QTRS 51.49/14.04 (5) DependencyPairsProof [EQUIVALENT, 98 ms] 51.49/14.04 (6) QDP 51.49/14.04 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 51.49/14.04 (8) AND 51.49/14.04 (9) QDP 51.49/14.04 (10) MRRProof [EQUIVALENT, 626 ms] 51.49/14.04 (11) QDP 51.49/14.04 (12) PisEmptyProof [EQUIVALENT, 0 ms] 51.49/14.04 (13) YES 51.49/14.04 (14) QDP 51.49/14.04 (15) UsableRulesProof [EQUIVALENT, 0 ms] 51.49/14.04 (16) QDP 51.49/14.04 (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] 51.49/14.04 (18) YES 51.49/14.04 (19) QDP 51.49/14.04 (20) QDPSizeChangeProof [EQUIVALENT, 0 ms] 51.49/14.04 (21) YES 51.49/14.04 51.49/14.04 51.49/14.04 ---------------------------------------- 51.49/14.04 51.49/14.04 (0) 51.49/14.04 Obligation: 51.49/14.04 Q restricted rewrite system: 51.49/14.04 The TRS R consists of the following rules: 51.49/14.04 51.49/14.04 0(5(0(x1))) -> 3(4(3(3(2(0(4(4(4(0(x1)))))))))) 51.49/14.04 0(2(5(0(x1)))) -> 0(2(1(1(5(2(1(4(1(3(x1)))))))))) 51.49/14.04 0(5(0(5(x1)))) -> 3(2(0(2(0(4(4(1(5(4(x1)))))))))) 51.84/14.04 2(5(5(3(x1)))) -> 4(3(0(2(1(0(1(3(4(3(x1)))))))))) 51.84/14.04 5(5(5(0(x1)))) -> 4(1(2(1(2(3(5(0(1(3(x1)))))))))) 51.84/14.04 0(5(1(5(0(x1))))) -> 3(4(0(1(4(5(2(2(3(1(x1)))))))))) 51.84/14.04 3(0(0(5(3(x1))))) -> 1(3(4(3(5(2(4(1(3(3(x1)))))))))) 51.84/14.04 3(5(5(0(0(x1))))) -> 1(4(1(0(0(4(4(0(4(1(x1)))))))))) 51.84/14.04 0(2(5(1(5(0(x1)))))) -> 2(1(0(1(5(2(4(0(2(0(x1)))))))))) 51.84/14.04 0(2(5(5(1(4(x1)))))) -> 4(0(1(1(1(1(0(4(1(5(x1)))))))))) 51.84/14.04 0(5(3(1(2(5(x1)))))) -> 0(2(3(1(1(2(4(4(5(5(x1)))))))))) 51.84/14.04 0(5(5(1(5(1(x1)))))) -> 0(5(1(1(3(3(4(2(1(0(x1)))))))))) 51.84/14.04 4(5(5(4(2(0(x1)))))) -> 2(4(0(1(3(4(4(4(1(0(x1)))))))))) 51.84/14.04 5(0(0(3(5(2(x1)))))) -> 4(4(2(3(0(1(2(0(5(2(x1)))))))))) 51.84/14.04 5(1(5(0(2(5(x1)))))) -> 5(0(0(1(4(2(3(2(1(5(x1)))))))))) 51.84/14.04 5(2(0(2(5(5(x1)))))) -> 5(5(4(4(4(5(4(4(1(4(x1)))))))))) 51.84/14.04 5(5(0(2(5(0(x1)))))) -> 2(0(5(0(2(1(0(0(3(0(x1)))))))))) 51.84/14.04 5(5(0(3(4(5(x1)))))) -> 2(0(5(5(2(1(3(2(3(2(x1)))))))))) 51.84/14.04 5(5(3(5(0(5(x1)))))) -> 5(4(4(3(5(1(3(3(4(5(x1)))))))))) 51.84/14.04 0(4(4(0(0(5(1(x1))))))) -> 1(3(2(0(4(1(5(1(1(2(x1)))))))))) 51.84/14.04 0(4(4(2(5(5(5(x1))))))) -> 0(2(4(5(5(4(2(0(1(1(x1)))))))))) 51.84/14.04 1(0(2(5(2(0(0(x1))))))) -> 3(1(4(4(0(3(0(1(2(2(x1)))))))))) 51.84/14.04 1(2(0(4(2(5(0(x1))))))) -> 4(2(4(0(3(2(2(4(1(0(x1)))))))))) 51.84/14.04 1(2(5(5(0(3(3(x1))))))) -> 3(4(1(2(0(3(3(1(0(3(x1)))))))))) 51.84/14.04 1(5(5(3(3(3(4(x1))))))) -> 1(5(1(0(0(2(2(2(3(5(x1)))))))))) 51.84/14.04 2(5(4(5(2(5(1(x1))))))) -> 4(3(2(1(4(2(2(4(5(2(x1)))))))))) 51.84/14.04 3(2(3(5(1(5(2(x1))))))) -> 2(0(3(2(3(2(1(5(5(1(x1)))))))))) 51.84/14.04 3(3(4(2(5(5(2(x1))))))) -> 1(2(3(3(4(4(1(4(0(1(x1)))))))))) 51.84/14.04 3(5(0(5(5(5(0(x1))))))) -> 0(0(3(0(3(5(0(3(2(0(x1)))))))))) 51.84/14.04 4(3(1(2(5(2(4(x1))))))) -> 2(3(1(1(4(3(4(4(2(4(x1)))))))))) 51.84/14.04 4(5(5(3(1(0(5(x1))))))) -> 1(1(5(2(0(3(3(3(2(1(x1)))))))))) 51.84/14.04 5(0(5(3(1(0(5(x1))))))) -> 5(2(4(4(2(1(3(5(1(5(x1)))))))))) 51.84/14.04 5(0(5(3(5(1(5(x1))))))) -> 5(1(1(2(4(0(0(3(2(5(x1)))))))))) 51.84/14.04 5(1(5(3(3(0(5(x1))))))) -> 4(4(3(2(2(2(5(0(1(1(x1)))))))))) 51.84/14.04 5(2(0(2(5(3(3(x1))))))) -> 5(2(0(5(1(1(3(2(0(3(x1)))))))))) 51.84/14.04 51.84/14.04 Q is empty. 51.84/14.04 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (1) QTRS Reverse (EQUIVALENT) 51.84/14.04 We applied the QTRS Reverse Processor [REVERSE]. 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (2) 51.84/14.04 Obligation: 51.84/14.04 Q restricted rewrite system: 51.84/14.04 The TRS R consists of the following rules: 51.84/14.04 51.84/14.04 0(5(0(x1))) -> 0(4(4(4(0(2(3(3(4(3(x1)))))))))) 51.84/14.04 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.04 5(0(5(0(x1)))) -> 4(5(1(4(4(0(2(0(2(3(x1)))))))))) 51.84/14.04 3(5(5(2(x1)))) -> 3(4(3(1(0(1(2(0(3(4(x1)))))))))) 51.84/14.04 0(5(5(5(x1)))) -> 3(1(0(5(3(2(1(2(1(4(x1)))))))))) 51.84/14.04 0(5(1(5(0(x1))))) -> 1(3(2(2(5(4(1(0(4(3(x1)))))))))) 51.84/14.04 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.04 0(0(5(5(3(x1))))) -> 1(4(0(4(4(0(0(1(4(1(x1)))))))))) 51.84/14.04 0(5(1(5(2(0(x1)))))) -> 0(2(0(4(2(5(1(0(1(2(x1)))))))))) 51.84/14.04 4(1(5(5(2(0(x1)))))) -> 5(1(4(0(1(1(1(1(0(4(x1)))))))))) 51.84/14.04 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.04 1(5(1(5(5(0(x1)))))) -> 0(1(2(4(3(3(1(1(5(0(x1)))))))))) 51.84/14.04 0(2(4(5(5(4(x1)))))) -> 0(1(4(4(4(3(1(0(4(2(x1)))))))))) 51.84/14.04 2(5(3(0(0(5(x1)))))) -> 2(5(0(2(1(0(3(2(4(4(x1)))))))))) 51.84/14.04 5(2(0(5(1(5(x1)))))) -> 5(1(2(3(2(4(1(0(0(5(x1)))))))))) 51.84/14.04 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.04 0(5(2(0(5(5(x1)))))) -> 0(3(0(0(1(2(0(5(0(2(x1)))))))))) 51.84/14.04 5(4(3(0(5(5(x1)))))) -> 2(3(2(3(1(2(5(5(0(2(x1)))))))))) 51.84/14.04 5(0(5(3(5(5(x1)))))) -> 5(4(3(3(1(5(3(4(4(5(x1)))))))))) 51.84/14.04 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.04 5(5(5(2(4(4(0(x1))))))) -> 1(1(0(2(4(5(5(4(2(0(x1)))))))))) 51.84/14.04 0(0(2(5(2(0(1(x1))))))) -> 2(2(1(0(3(0(4(4(1(3(x1)))))))))) 51.84/14.04 0(5(2(4(0(2(1(x1))))))) -> 0(1(4(2(2(3(0(4(2(4(x1)))))))))) 51.84/14.04 3(3(0(5(5(2(1(x1))))))) -> 3(0(1(3(3(0(2(1(4(3(x1)))))))))) 51.84/14.04 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.04 1(5(2(5(4(5(2(x1))))))) -> 2(5(4(2(2(4(1(2(3(4(x1)))))))))) 51.84/14.04 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.04 2(5(5(2(4(3(3(x1))))))) -> 1(0(4(1(4(4(3(3(2(1(x1)))))))))) 51.84/14.04 0(5(5(5(0(5(3(x1))))))) -> 0(2(3(0(5(3(0(3(0(0(x1)))))))))) 51.84/14.04 4(2(5(2(1(3(4(x1))))))) -> 4(2(4(4(3(4(1(1(3(2(x1)))))))))) 51.84/14.04 5(0(1(3(5(5(4(x1))))))) -> 1(2(3(3(3(0(2(5(1(1(x1)))))))))) 51.84/14.04 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.04 5(1(5(3(5(0(5(x1))))))) -> 5(2(3(0(0(4(2(1(1(5(x1)))))))))) 51.84/14.04 5(0(3(3(5(1(5(x1))))))) -> 1(1(0(5(2(2(2(3(4(4(x1)))))))))) 51.84/14.04 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.04 51.84/14.04 Q is empty. 51.84/14.04 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (3) QTRSRRRProof (EQUIVALENT) 51.84/14.04 Used ordering: 51.84/14.04 Polynomial interpretation [POLO]: 51.84/14.04 51.84/14.04 POL(0(x_1)) = x_1 51.84/14.04 POL(1(x_1)) = x_1 51.84/14.04 POL(2(x_1)) = x_1 51.84/14.04 POL(3(x_1)) = x_1 51.84/14.04 POL(4(x_1)) = x_1 51.84/14.04 POL(5(x_1)) = 1 + x_1 51.84/14.04 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 51.84/14.04 51.84/14.04 0(5(0(x1))) -> 0(4(4(4(0(2(3(3(4(3(x1)))))))))) 51.84/14.04 5(0(5(0(x1)))) -> 4(5(1(4(4(0(2(0(2(3(x1)))))))))) 51.84/14.04 3(5(5(2(x1)))) -> 3(4(3(1(0(1(2(0(3(4(x1)))))))))) 51.84/14.04 0(5(5(5(x1)))) -> 3(1(0(5(3(2(1(2(1(4(x1)))))))))) 51.84/14.04 0(5(1(5(0(x1))))) -> 1(3(2(2(5(4(1(0(4(3(x1)))))))))) 51.84/14.04 0(0(5(5(3(x1))))) -> 1(4(0(4(4(0(0(1(4(1(x1)))))))))) 51.84/14.04 0(5(1(5(2(0(x1)))))) -> 0(2(0(4(2(5(1(0(1(2(x1)))))))))) 51.84/14.04 4(1(5(5(2(0(x1)))))) -> 5(1(4(0(1(1(1(1(0(4(x1)))))))))) 51.84/14.04 1(5(1(5(5(0(x1)))))) -> 0(1(2(4(3(3(1(1(5(0(x1)))))))))) 51.84/14.04 0(2(4(5(5(4(x1)))))) -> 0(1(4(4(4(3(1(0(4(2(x1)))))))))) 51.84/14.04 2(5(3(0(0(5(x1)))))) -> 2(5(0(2(1(0(3(2(4(4(x1)))))))))) 51.84/14.04 5(2(0(5(1(5(x1)))))) -> 5(1(2(3(2(4(1(0(0(5(x1)))))))))) 51.84/14.04 0(5(2(0(5(5(x1)))))) -> 0(3(0(0(1(2(0(5(0(2(x1)))))))))) 51.84/14.04 5(4(3(0(5(5(x1)))))) -> 2(3(2(3(1(2(5(5(0(2(x1)))))))))) 51.84/14.04 5(0(5(3(5(5(x1)))))) -> 5(4(3(3(1(5(3(4(4(5(x1)))))))))) 51.84/14.04 5(5(5(2(4(4(0(x1))))))) -> 1(1(0(2(4(5(5(4(2(0(x1)))))))))) 51.84/14.04 0(0(2(5(2(0(1(x1))))))) -> 2(2(1(0(3(0(4(4(1(3(x1)))))))))) 51.84/14.04 0(5(2(4(0(2(1(x1))))))) -> 0(1(4(2(2(3(0(4(2(4(x1)))))))))) 51.84/14.04 3(3(0(5(5(2(1(x1))))))) -> 3(0(1(3(3(0(2(1(4(3(x1)))))))))) 51.84/14.04 1(5(2(5(4(5(2(x1))))))) -> 2(5(4(2(2(4(1(2(3(4(x1)))))))))) 51.84/14.04 2(5(5(2(4(3(3(x1))))))) -> 1(0(4(1(4(4(3(3(2(1(x1)))))))))) 51.84/14.04 0(5(5(5(0(5(3(x1))))))) -> 0(2(3(0(5(3(0(3(0(0(x1)))))))))) 51.84/14.04 4(2(5(2(1(3(4(x1))))))) -> 4(2(4(4(3(4(1(1(3(2(x1)))))))))) 51.84/14.04 5(0(1(3(5(5(4(x1))))))) -> 1(2(3(3(3(0(2(5(1(1(x1)))))))))) 51.84/14.04 5(1(5(3(5(0(5(x1))))))) -> 5(2(3(0(0(4(2(1(1(5(x1)))))))))) 51.84/14.04 5(0(3(3(5(1(5(x1))))))) -> 1(1(0(5(2(2(2(3(4(4(x1)))))))))) 51.84/14.04 51.84/14.04 51.84/14.04 51.84/14.04 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (4) 51.84/14.04 Obligation: 51.84/14.04 Q restricted rewrite system: 51.84/14.04 The TRS R consists of the following rules: 51.84/14.04 51.84/14.04 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.04 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.04 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.04 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.04 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.04 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.04 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.04 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.04 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.04 51.84/14.04 Q is empty. 51.84/14.04 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (5) DependencyPairsProof (EQUIVALENT) 51.84/14.04 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (6) 51.84/14.04 Obligation: 51.84/14.04 Q DP problem: 51.84/14.04 The TRS P consists of the following rules: 51.84/14.04 51.84/14.04 0^1(5(2(0(x1)))) -> 3^1(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.04 0^1(5(2(0(x1)))) -> 1^1(4(1(2(5(1(1(2(0(x1))))))))) 51.84/14.04 0^1(5(2(0(x1)))) -> 4^1(1(2(5(1(1(2(0(x1)))))))) 51.84/14.04 0^1(5(2(0(x1)))) -> 1^1(2(5(1(1(2(0(x1))))))) 51.84/14.04 0^1(5(2(0(x1)))) -> 2^1(5(1(1(2(0(x1)))))) 51.84/14.04 0^1(5(2(0(x1)))) -> 5^1(1(1(2(0(x1))))) 51.84/14.04 0^1(5(2(0(x1)))) -> 1^1(1(2(0(x1)))) 51.84/14.04 0^1(5(2(0(x1)))) -> 1^1(2(0(x1))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 3^1(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 3^1(1(4(2(5(3(4(3(1(x1))))))))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 1^1(4(2(5(3(4(3(1(x1)))))))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 4^1(2(5(3(4(3(1(x1))))))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 2^1(5(3(4(3(1(x1)))))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 5^1(3(4(3(1(x1))))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 3^1(4(3(1(x1)))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 4^1(3(1(x1))) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 3^1(1(x1)) 51.84/14.04 3^1(5(0(0(3(x1))))) -> 1^1(x1) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 5^1(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 5^1(4(4(2(1(1(3(2(0(x1))))))))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 4^1(4(2(1(1(3(2(0(x1)))))))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 4^1(2(1(1(3(2(0(x1))))))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 2^1(1(1(3(2(0(x1)))))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 1^1(1(3(2(0(x1))))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 1^1(3(2(0(x1)))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 3^1(2(0(x1))) 51.84/14.04 5^1(2(1(3(5(0(x1)))))) -> 2^1(0(x1)) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 4^1(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 1^1(4(4(5(4(4(4(5(5(x1))))))))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 4^1(4(5(4(4(4(5(5(x1)))))))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 4^1(5(4(4(4(5(5(x1))))))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 5^1(4(4(4(5(5(x1)))))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 4^1(4(4(5(5(x1))))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 4^1(4(5(5(x1)))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 4^1(5(5(x1))) 51.84/14.04 5^1(5(2(0(2(5(x1)))))) -> 5^1(5(x1)) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 2^1(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(1(5(1(4(0(2(3(1(x1))))))))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(5(1(4(0(2(3(1(x1)))))))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 5^1(1(4(0(2(3(1(x1))))))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(4(0(2(3(1(x1)))))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 4^1(0(2(3(1(x1))))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 0^1(2(3(1(x1)))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 2^1(3(1(x1))) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 3^1(1(x1)) 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 5^1(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 3^1(2(2(2(0(0(1(5(1(x1))))))))) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 2^1(2(2(0(0(1(5(1(x1)))))))) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 2^1(2(0(0(1(5(1(x1))))))) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 2^1(0(0(1(5(1(x1)))))) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 0^1(0(1(5(1(x1))))) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 0^1(1(5(1(x1)))) 51.84/14.04 4^1(3(3(3(5(5(1(x1))))))) -> 1^1(5(1(x1))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 1^1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 5^1(5(1(2(3(2(3(0(2(x1))))))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 5^1(1(2(3(2(3(0(2(x1)))))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 1^1(2(3(2(3(0(2(x1))))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(2(3(0(2(x1)))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 3^1(2(3(0(2(x1))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(0(2(x1)))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 3^1(0(2(x1))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 0^1(2(x1)) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(x1) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 5^1(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 1^1(5(3(1(2(4(4(2(5(x1))))))))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 5^1(3(1(2(4(4(2(5(x1)))))))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 3^1(1(2(4(4(2(5(x1))))))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 1^1(2(4(4(2(5(x1)))))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 2^1(4(4(2(5(x1))))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 4^1(4(2(5(x1)))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 4^1(2(5(x1))) 51.84/14.04 5^1(0(1(3(5(0(5(x1))))))) -> 2^1(5(x1)) 51.84/14.04 3^1(3(5(2(0(2(5(x1))))))) -> 3^1(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.04 3^1(3(5(2(0(2(5(x1))))))) -> 0^1(2(3(1(1(5(0(2(5(x1))))))))) 51.84/14.04 3^1(3(5(2(0(2(5(x1))))))) -> 2^1(3(1(1(5(0(2(5(x1)))))))) 51.84/14.04 3^1(3(5(2(0(2(5(x1))))))) -> 3^1(1(1(5(0(2(5(x1))))))) 51.84/14.04 3^1(3(5(2(0(2(5(x1))))))) -> 1^1(1(5(0(2(5(x1)))))) 51.84/14.04 3^1(3(5(2(0(2(5(x1))))))) -> 1^1(5(0(2(5(x1))))) 51.84/14.04 3^1(3(5(2(0(2(5(x1))))))) -> 5^1(0(2(5(x1)))) 51.84/14.04 51.84/14.04 The TRS R consists of the following rules: 51.84/14.04 51.84/14.04 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.04 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.04 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.04 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.04 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.04 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.04 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.04 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.04 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.04 51.84/14.04 Q is empty. 51.84/14.04 We have to consider all minimal (P,Q,R)-chains. 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (7) DependencyGraphProof (EQUIVALENT) 51.84/14.04 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 75 less nodes. 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (8) 51.84/14.04 Complex Obligation (AND) 51.84/14.04 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (9) 51.84/14.04 Obligation: 51.84/14.04 Q DP problem: 51.84/14.04 The TRS P consists of the following rules: 51.84/14.04 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(0(2(x1)))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(2(3(0(2(x1)))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(x1) 51.84/14.04 51.84/14.04 The TRS R consists of the following rules: 51.84/14.04 51.84/14.04 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.04 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.04 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.04 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.04 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.04 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.04 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.04 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.04 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.04 51.84/14.04 Q is empty. 51.84/14.04 We have to consider all minimal (P,Q,R)-chains. 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (10) MRRProof (EQUIVALENT) 51.84/14.04 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 51.84/14.04 51.84/14.04 Strictly oriented dependency pairs: 51.84/14.04 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(0(2(x1)))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(2(3(0(2(x1)))))) 51.84/14.04 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(x1) 51.84/14.04 51.84/14.04 51.84/14.04 Used ordering: Polynomial interpretation [POLO]: 51.84/14.04 51.84/14.04 POL(0(x_1)) = x_1 51.84/14.04 POL(1(x_1)) = x_1 51.84/14.04 POL(2(x_1)) = x_1 51.84/14.04 POL(2^1(x_1)) = x_1 51.84/14.04 POL(3(x_1)) = x_1 51.84/14.04 POL(4(x_1)) = x_1 51.84/14.04 POL(5(x_1)) = 2 + 2*x_1 51.84/14.04 51.84/14.04 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (11) 51.84/14.04 Obligation: 51.84/14.04 Q DP problem: 51.84/14.04 P is empty. 51.84/14.04 The TRS R consists of the following rules: 51.84/14.04 51.84/14.04 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.04 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.04 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.04 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.04 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.04 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.04 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.04 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.04 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.04 51.84/14.04 Q is empty. 51.84/14.04 We have to consider all minimal (P,Q,R)-chains. 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (12) PisEmptyProof (EQUIVALENT) 51.84/14.04 The TRS P is empty. Hence, there is no (P,Q,R) chain. 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (13) 51.84/14.04 YES 51.84/14.04 51.84/14.04 ---------------------------------------- 51.84/14.04 51.84/14.04 (14) 51.84/14.04 Obligation: 51.84/14.04 Q DP problem: 51.84/14.04 The TRS P consists of the following rules: 51.84/14.04 51.84/14.04 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) 51.84/14.04 51.84/14.04 The TRS R consists of the following rules: 51.84/14.04 51.84/14.04 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.04 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.04 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.04 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.04 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.05 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.05 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.05 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.05 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.05 51.84/14.05 Q is empty. 51.84/14.05 We have to consider all minimal (P,Q,R)-chains. 51.84/14.05 ---------------------------------------- 51.84/14.05 51.84/14.05 (15) UsableRulesProof (EQUIVALENT) 51.84/14.05 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. 51.84/14.05 ---------------------------------------- 51.84/14.05 51.84/14.05 (16) 51.84/14.05 Obligation: 51.84/14.05 Q DP problem: 51.84/14.05 The TRS P consists of the following rules: 51.84/14.05 51.84/14.05 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) 51.84/14.05 51.84/14.05 R is empty. 51.84/14.05 Q is empty. 51.84/14.05 We have to consider all minimal (P,Q,R)-chains. 51.84/14.05 ---------------------------------------- 51.84/14.05 51.84/14.05 (17) QDPSizeChangeProof (EQUIVALENT) 51.84/14.05 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. 51.84/14.05 51.84/14.05 From the DPs we obtained the following set of size-change graphs: 51.84/14.05 *1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) 51.84/14.05 The graph contains the following edges 1 > 1 51.84/14.05 51.84/14.05 51.84/14.05 ---------------------------------------- 51.84/14.05 51.84/14.05 (18) 51.84/14.05 YES 51.84/14.05 51.84/14.05 ---------------------------------------- 51.84/14.05 51.84/14.05 (19) 51.84/14.05 Obligation: 51.84/14.05 Q DP problem: 51.84/14.05 The TRS P consists of the following rules: 51.84/14.05 51.84/14.05 5^1(5(2(0(2(5(x1)))))) -> 5^1(5(x1)) 51.84/14.05 51.84/14.05 The TRS R consists of the following rules: 51.84/14.05 51.84/14.05 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 51.84/14.05 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 51.84/14.05 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 51.84/14.05 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 51.84/14.05 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 51.84/14.05 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 51.84/14.05 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 51.84/14.05 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 51.84/14.05 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 51.84/14.05 51.84/14.05 Q is empty. 51.84/14.05 We have to consider all minimal (P,Q,R)-chains. 51.84/14.05 ---------------------------------------- 51.84/14.05 51.84/14.05 (20) QDPSizeChangeProof (EQUIVALENT) 51.84/14.05 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. 51.84/14.05 51.84/14.05 From the DPs we obtained the following set of size-change graphs: 51.84/14.05 *5^1(5(2(0(2(5(x1)))))) -> 5^1(5(x1)) 51.84/14.05 The graph contains the following edges 1 > 1 51.84/14.05 51.84/14.05 51.84/14.05 ---------------------------------------- 51.84/14.05 51.84/14.05 (21) 51.84/14.05 YES 51.91/14.13 EOF