details

property	value
status	complete
benchmark	multum1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n122.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	49.4193 seconds
cpu usage	191.8
user time	189.0
system time	2.80052
max virtual memory	4.1061124E7
max residence set size	5819168.0

stage attributes

key	value
starexec-result	YES

output

191.41/49.31 YES 191.54/49.32 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 191.54/49.32 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 191.54/49.32 191.54/49.32 191.54/49.32 Termination w.r.t. Q of the given QTRS could be proven: 191.54/49.32 191.54/49.32 (0) QTRS 191.54/49.32 (1) QTRS Reverse [EQUIVALENT, 0 ms] 191.54/49.32 (2) QTRS 191.54/49.32 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 191.54/49.32 (4) QDP 191.54/49.32 (5) MRRProof [EQUIVALENT, 81 ms] 191.54/49.32 (6) QDP 191.54/49.32 (7) QDPOrderProof [EQUIVALENT, 4177 ms] 191.54/49.32 (8) QDP 191.54/49.32 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 191.54/49.32 (10) TRUE 191.54/49.32 191.54/49.32 191.54/49.32 ---------------------------------------- 191.54/49.32 191.54/49.32 (0) 191.54/49.32 Obligation: 191.54/49.32 Q restricted rewrite system: 191.54/49.32 The TRS R consists of the following rules: 191.54/49.32 191.54/49.32 a(a(b(x1))) -> b(a(a(a(x1)))) 191.54/49.32 b(a(b(a(x1)))) -> a(b(b(x1))) 191.54/49.32 191.54/49.32 Q is empty. 191.54/49.32 191.54/49.32 ---------------------------------------- 191.54/49.32 191.54/49.32 (1) QTRS Reverse (EQUIVALENT) 191.54/49.32 We applied the QTRS Reverse Processor [REVERSE]. 191.54/49.32 ---------------------------------------- 191.54/49.32 191.54/49.32 (2) 191.54/49.32 Obligation: 191.54/49.32 Q restricted rewrite system: 191.54/49.32 The TRS R consists of the following rules: 191.54/49.32 191.54/49.32 b(a(a(x1))) -> a(a(a(b(x1)))) 191.54/49.32 a(b(a(b(x1)))) -> b(b(a(x1))) 191.54/49.32 191.54/49.32 Q is empty. 191.54/49.32 191.54/49.32 ---------------------------------------- 191.54/49.32 191.54/49.32 (3) DependencyPairsProof (EQUIVALENT) 191.54/49.32 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 191.54/49.32 ---------------------------------------- 191.54/49.32 191.54/49.32 (4) 191.54/49.32 Obligation: 191.54/49.32 Q DP problem: 191.54/49.32 The TRS P consists of the following rules: 191.54/49.32 191.54/49.32 B(a(a(x1))) -> A(a(a(b(x1)))) 191.54/49.32 B(a(a(x1))) -> A(a(b(x1))) 191.54/49.32 B(a(a(x1))) -> A(b(x1)) 191.54/49.32 B(a(a(x1))) -> B(x1) 191.54/49.32 A(b(a(b(x1)))) -> B(b(a(x1))) 191.54/49.32 A(b(a(b(x1)))) -> B(a(x1)) 191.54/49.32 A(b(a(b(x1)))) -> A(x1) 191.54/49.32 191.54/49.32 The TRS R consists of the following rules: 191.54/49.32 191.54/49.32 b(a(a(x1))) -> a(a(a(b(x1)))) 191.54/49.32 a(b(a(b(x1)))) -> b(b(a(x1))) 191.54/49.32 191.54/49.32 Q is empty. 191.54/49.32 We have to consider all minimal (P,Q,R)-chains. 191.54/49.32 ---------------------------------------- 191.54/49.32 191.54/49.32 (5) MRRProof (EQUIVALENT) 191.54/49.32 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. 191.54/49.32 191.54/49.32 Strictly oriented dependency pairs: 191.54/49.32 191.54/49.32 A(b(a(b(x1)))) -> B(a(x1)) 191.54/49.32 A(b(a(b(x1)))) -> A(x1) 191.54/49.32 191.54/49.32 191.54/49.32 Used ordering: Polynomial interpretation [POLO]: 191.54/49.32 191.54/49.32 POL(A(x_1)) = x_1 191.54/49.32 POL(B(x_1)) = 1 + x_1 191.54/49.32 POL(a(x_1)) = x_1 191.54/49.32 POL(b(x_1)) = 1 + x_1 191.54/49.32 191.54/49.32 191.54/49.32 ---------------------------------------- 191.54/49.32 191.54/49.32 (6) 191.54/49.32 Obligation: 191.54/49.32 Q DP problem: 191.54/49.32 The TRS P consists of the following rules: 191.54/49.32 popout

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all output return to SRS_Standard 2019-03-29 03.29