10.46/3.51 YES 10.85/3.57 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 10.85/3.57 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 10.85/3.57 10.85/3.57 10.85/3.57 Termination w.r.t. Q of the given QTRS could be proven: 10.85/3.57 10.85/3.57 (0) QTRS 10.85/3.57 (1) QTRS Reverse [EQUIVALENT, 0 ms] 10.85/3.57 (2) QTRS 10.85/3.57 (3) DependencyPairsProof [EQUIVALENT, 59 ms] 10.85/3.57 (4) QDP 10.85/3.57 (5) DependencyGraphProof [EQUIVALENT, 4 ms] 10.85/3.57 (6) QDP 10.85/3.57 (7) UsableRulesProof [EQUIVALENT, 2 ms] 10.85/3.57 (8) QDP 10.85/3.57 (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] 10.85/3.57 (10) YES 10.85/3.57 10.85/3.57 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (0) 10.85/3.57 Obligation: 10.85/3.57 Q restricted rewrite system: 10.85/3.57 The TRS R consists of the following rules: 10.85/3.57 10.85/3.57 4(2(4(x1))) -> 2(0(0(5(3(3(5(2(0(4(x1)))))))))) 10.85/3.57 4(4(2(4(2(x1))))) -> 2(0(5(2(1(4(0(2(0(1(x1)))))))))) 10.85/3.57 0(5(4(2(4(3(x1)))))) -> 5(1(5(5(3(5(3(0(0(0(x1)))))))))) 10.85/3.57 1(1(4(5(3(3(x1)))))) -> 1(3(1(1(3(0(1(2(2(1(x1)))))))))) 10.85/3.57 3(1(4(3(1(2(x1)))))) -> 0(0(1(1(4(2(3(0(0(3(x1)))))))))) 10.85/3.57 3(2(4(2(4(1(x1)))))) -> 0(2(1(1(1(5(3(1(3(3(x1)))))))))) 10.85/3.57 3(3(0(4(1(2(x1)))))) -> 3(5(1(2(0(2(0(5(3(1(x1)))))))))) 10.85/3.57 4(1(4(5(0(5(4(x1))))))) -> 4(1(5(3(1(0(5(3(1(0(x1)))))))))) 10.85/3.57 4(4(0(5(4(2(2(x1))))))) -> 4(0(4(3(4(4(4(5(4(1(x1)))))))))) 10.85/3.57 5(4(5(3(2(4(3(x1))))))) -> 2(5(5(5(0(4(5(0(1(4(x1)))))))))) 10.85/3.57 10.85/3.57 Q is empty. 10.85/3.57 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (1) QTRS Reverse (EQUIVALENT) 10.85/3.57 We applied the QTRS Reverse Processor [REVERSE]. 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (2) 10.85/3.57 Obligation: 10.85/3.57 Q restricted rewrite system: 10.85/3.57 The TRS R consists of the following rules: 10.85/3.57 10.85/3.57 4(2(4(x1))) -> 4(0(2(5(3(3(5(0(0(2(x1)))))))))) 10.85/3.57 2(4(2(4(4(x1))))) -> 1(0(2(0(4(1(2(5(0(2(x1)))))))))) 10.85/3.57 3(4(2(4(5(0(x1)))))) -> 0(0(0(3(5(3(5(5(1(5(x1)))))))))) 10.85/3.57 3(3(5(4(1(1(x1)))))) -> 1(2(2(1(0(3(1(1(3(1(x1)))))))))) 10.85/3.57 2(1(3(4(1(3(x1)))))) -> 3(0(0(3(2(4(1(1(0(0(x1)))))))))) 10.85/3.57 1(4(2(4(2(3(x1)))))) -> 3(3(1(3(5(1(1(1(2(0(x1)))))))))) 10.85/3.57 2(1(4(0(3(3(x1)))))) -> 1(3(5(0(2(0(2(1(5(3(x1)))))))))) 10.85/3.57 4(5(0(5(4(1(4(x1))))))) -> 0(1(3(5(0(1(3(5(1(4(x1)))))))))) 10.85/3.57 2(2(4(5(0(4(4(x1))))))) -> 1(4(5(4(4(4(3(4(0(4(x1)))))))))) 10.85/3.57 3(4(2(3(5(4(5(x1))))))) -> 4(1(0(5(4(0(5(5(5(2(x1)))))))))) 10.85/3.57 10.85/3.57 Q is empty. 10.85/3.57 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (3) DependencyPairsProof (EQUIVALENT) 10.85/3.57 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (4) 10.85/3.57 Obligation: 10.85/3.57 Q DP problem: 10.85/3.57 The TRS P consists of the following rules: 10.85/3.57 10.85/3.57 4^1(2(4(x1))) -> 4^1(0(2(5(3(3(5(0(0(2(x1)))))))))) 10.85/3.57 4^1(2(4(x1))) -> 2^1(5(3(3(5(0(0(2(x1)))))))) 10.85/3.57 4^1(2(4(x1))) -> 3^1(3(5(0(0(2(x1)))))) 10.85/3.57 4^1(2(4(x1))) -> 3^1(5(0(0(2(x1))))) 10.85/3.57 4^1(2(4(x1))) -> 2^1(x1) 10.85/3.57 2^1(4(2(4(4(x1))))) -> 1^1(0(2(0(4(1(2(5(0(2(x1)))))))))) 10.85/3.57 2^1(4(2(4(4(x1))))) -> 2^1(0(4(1(2(5(0(2(x1)))))))) 10.85/3.57 2^1(4(2(4(4(x1))))) -> 4^1(1(2(5(0(2(x1)))))) 10.85/3.57 2^1(4(2(4(4(x1))))) -> 1^1(2(5(0(2(x1))))) 10.85/3.57 2^1(4(2(4(4(x1))))) -> 2^1(5(0(2(x1)))) 10.85/3.57 2^1(4(2(4(4(x1))))) -> 2^1(x1) 10.85/3.57 3^1(4(2(4(5(0(x1)))))) -> 3^1(5(3(5(5(1(5(x1))))))) 10.85/3.57 3^1(4(2(4(5(0(x1)))))) -> 3^1(5(5(1(5(x1))))) 10.85/3.57 3^1(4(2(4(5(0(x1)))))) -> 1^1(5(x1)) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 1^1(2(2(1(0(3(1(1(3(1(x1)))))))))) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 2^1(2(1(0(3(1(1(3(1(x1))))))))) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 2^1(1(0(3(1(1(3(1(x1)))))))) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 1^1(0(3(1(1(3(1(x1))))))) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 3^1(1(1(3(1(x1))))) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 1^1(1(3(1(x1)))) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 1^1(3(1(x1))) 10.85/3.57 3^1(3(5(4(1(1(x1)))))) -> 3^1(1(x1)) 10.85/3.57 2^1(1(3(4(1(3(x1)))))) -> 3^1(0(0(3(2(4(1(1(0(0(x1)))))))))) 10.85/3.57 2^1(1(3(4(1(3(x1)))))) -> 3^1(2(4(1(1(0(0(x1))))))) 10.85/3.57 2^1(1(3(4(1(3(x1)))))) -> 2^1(4(1(1(0(0(x1)))))) 10.85/3.57 2^1(1(3(4(1(3(x1)))))) -> 4^1(1(1(0(0(x1))))) 10.85/3.57 2^1(1(3(4(1(3(x1)))))) -> 1^1(1(0(0(x1)))) 10.85/3.57 2^1(1(3(4(1(3(x1)))))) -> 1^1(0(0(x1))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 3^1(3(1(3(5(1(1(1(2(0(x1)))))))))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 3^1(1(3(5(1(1(1(2(0(x1))))))))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 1^1(3(5(1(1(1(2(0(x1)))))))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 3^1(5(1(1(1(2(0(x1))))))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 1^1(1(1(2(0(x1))))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 1^1(1(2(0(x1)))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 1^1(2(0(x1))) 10.85/3.57 1^1(4(2(4(2(3(x1)))))) -> 2^1(0(x1)) 10.85/3.57 2^1(1(4(0(3(3(x1)))))) -> 1^1(3(5(0(2(0(2(1(5(3(x1)))))))))) 10.85/3.57 2^1(1(4(0(3(3(x1)))))) -> 3^1(5(0(2(0(2(1(5(3(x1))))))))) 10.85/3.57 2^1(1(4(0(3(3(x1)))))) -> 2^1(0(2(1(5(3(x1)))))) 10.85/3.57 2^1(1(4(0(3(3(x1)))))) -> 2^1(1(5(3(x1)))) 10.85/3.57 2^1(1(4(0(3(3(x1)))))) -> 1^1(5(3(x1))) 10.85/3.57 4^1(5(0(5(4(1(4(x1))))))) -> 1^1(3(5(0(1(3(5(1(4(x1))))))))) 10.85/3.57 4^1(5(0(5(4(1(4(x1))))))) -> 3^1(5(0(1(3(5(1(4(x1)))))))) 10.85/3.57 4^1(5(0(5(4(1(4(x1))))))) -> 1^1(3(5(1(4(x1))))) 10.85/3.57 4^1(5(0(5(4(1(4(x1))))))) -> 3^1(5(1(4(x1)))) 10.85/3.57 2^1(2(4(5(0(4(4(x1))))))) -> 1^1(4(5(4(4(4(3(4(0(4(x1)))))))))) 10.85/3.57 2^1(2(4(5(0(4(4(x1))))))) -> 4^1(5(4(4(4(3(4(0(4(x1))))))))) 10.85/3.57 2^1(2(4(5(0(4(4(x1))))))) -> 4^1(4(4(3(4(0(4(x1))))))) 10.85/3.57 2^1(2(4(5(0(4(4(x1))))))) -> 4^1(4(3(4(0(4(x1)))))) 10.85/3.57 2^1(2(4(5(0(4(4(x1))))))) -> 4^1(3(4(0(4(x1))))) 10.85/3.57 2^1(2(4(5(0(4(4(x1))))))) -> 3^1(4(0(4(x1)))) 10.85/3.57 2^1(2(4(5(0(4(4(x1))))))) -> 4^1(0(4(x1))) 10.85/3.57 3^1(4(2(3(5(4(5(x1))))))) -> 4^1(1(0(5(4(0(5(5(5(2(x1)))))))))) 10.85/3.57 3^1(4(2(3(5(4(5(x1))))))) -> 1^1(0(5(4(0(5(5(5(2(x1))))))))) 10.85/3.57 3^1(4(2(3(5(4(5(x1))))))) -> 4^1(0(5(5(5(2(x1)))))) 10.85/3.57 3^1(4(2(3(5(4(5(x1))))))) -> 2^1(x1) 10.85/3.57 10.85/3.57 The TRS R consists of the following rules: 10.85/3.57 10.85/3.57 4(2(4(x1))) -> 4(0(2(5(3(3(5(0(0(2(x1)))))))))) 10.85/3.57 2(4(2(4(4(x1))))) -> 1(0(2(0(4(1(2(5(0(2(x1)))))))))) 10.85/3.57 3(4(2(4(5(0(x1)))))) -> 0(0(0(3(5(3(5(5(1(5(x1)))))))))) 10.85/3.57 3(3(5(4(1(1(x1)))))) -> 1(2(2(1(0(3(1(1(3(1(x1)))))))))) 10.85/3.57 2(1(3(4(1(3(x1)))))) -> 3(0(0(3(2(4(1(1(0(0(x1)))))))))) 10.85/3.57 1(4(2(4(2(3(x1)))))) -> 3(3(1(3(5(1(1(1(2(0(x1)))))))))) 10.85/3.57 2(1(4(0(3(3(x1)))))) -> 1(3(5(0(2(0(2(1(5(3(x1)))))))))) 10.85/3.57 4(5(0(5(4(1(4(x1))))))) -> 0(1(3(5(0(1(3(5(1(4(x1)))))))))) 10.85/3.57 2(2(4(5(0(4(4(x1))))))) -> 1(4(5(4(4(4(3(4(0(4(x1)))))))))) 10.85/3.57 3(4(2(3(5(4(5(x1))))))) -> 4(1(0(5(4(0(5(5(5(2(x1)))))))))) 10.85/3.57 10.85/3.57 Q is empty. 10.85/3.57 We have to consider all minimal (P,Q,R)-chains. 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (5) DependencyGraphProof (EQUIVALENT) 10.85/3.57 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 55 less nodes. 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (6) 10.85/3.57 Obligation: 10.85/3.57 Q DP problem: 10.85/3.57 The TRS P consists of the following rules: 10.85/3.57 10.85/3.57 2^1(4(2(4(4(x1))))) -> 2^1(x1) 10.85/3.57 10.85/3.57 The TRS R consists of the following rules: 10.85/3.57 10.85/3.57 4(2(4(x1))) -> 4(0(2(5(3(3(5(0(0(2(x1)))))))))) 10.85/3.57 2(4(2(4(4(x1))))) -> 1(0(2(0(4(1(2(5(0(2(x1)))))))))) 10.85/3.57 3(4(2(4(5(0(x1)))))) -> 0(0(0(3(5(3(5(5(1(5(x1)))))))))) 10.85/3.57 3(3(5(4(1(1(x1)))))) -> 1(2(2(1(0(3(1(1(3(1(x1)))))))))) 10.85/3.57 2(1(3(4(1(3(x1)))))) -> 3(0(0(3(2(4(1(1(0(0(x1)))))))))) 10.85/3.57 1(4(2(4(2(3(x1)))))) -> 3(3(1(3(5(1(1(1(2(0(x1)))))))))) 10.85/3.57 2(1(4(0(3(3(x1)))))) -> 1(3(5(0(2(0(2(1(5(3(x1)))))))))) 10.85/3.57 4(5(0(5(4(1(4(x1))))))) -> 0(1(3(5(0(1(3(5(1(4(x1)))))))))) 10.85/3.57 2(2(4(5(0(4(4(x1))))))) -> 1(4(5(4(4(4(3(4(0(4(x1)))))))))) 10.85/3.57 3(4(2(3(5(4(5(x1))))))) -> 4(1(0(5(4(0(5(5(5(2(x1)))))))))) 10.85/3.57 10.85/3.57 Q is empty. 10.85/3.57 We have to consider all minimal (P,Q,R)-chains. 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (7) UsableRulesProof (EQUIVALENT) 10.85/3.57 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. 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (8) 10.85/3.57 Obligation: 10.85/3.57 Q DP problem: 10.85/3.57 The TRS P consists of the following rules: 10.85/3.57 10.85/3.57 2^1(4(2(4(4(x1))))) -> 2^1(x1) 10.85/3.57 10.85/3.57 R is empty. 10.85/3.57 Q is empty. 10.85/3.57 We have to consider all minimal (P,Q,R)-chains. 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (9) QDPSizeChangeProof (EQUIVALENT) 10.85/3.57 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. 10.85/3.57 10.85/3.57 From the DPs we obtained the following set of size-change graphs: 10.85/3.57 *2^1(4(2(4(4(x1))))) -> 2^1(x1) 10.85/3.57 The graph contains the following edges 1 > 1 10.85/3.57 10.85/3.57 10.85/3.57 ---------------------------------------- 10.85/3.57 10.85/3.57 (10) 10.85/3.57 YES 11.21/3.70 EOF