details

property	value
status	complete
benchmark	secr6.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.96069 seconds
cpu usage	16.4275
user time	15.5028
system time	0.924671
max virtual memory	4.0573572E7
max residence set size	2908224.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 2 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 408 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(c(x1))) -> c(c(c(b(b(b(a(a(a(x1))))))))) c(b(x1)) -> a(a(a(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> x1 Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(b(a(x1))) -> a(a(a(b(b(b(c(c(c(x1))))))))) b(c(x1)) -> a(a(a(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> x1 Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(b(a(x1))) -> A(a(a(b(b(b(c(c(c(x1))))))))) C(b(a(x1))) -> A(a(b(b(b(c(c(c(x1)))))))) C(b(a(x1))) -> A(b(b(b(c(c(c(x1))))))) C(b(a(x1))) -> B(b(b(c(c(c(x1)))))) C(b(a(x1))) -> B(b(c(c(c(x1))))) C(b(a(x1))) -> B(c(c(c(x1)))) C(b(a(x1))) -> C(c(c(x1))) C(b(a(x1))) -> C(c(x1)) C(b(a(x1))) -> C(x1) B(c(x1)) -> A(a(a(x1))) B(c(x1)) -> A(a(x1)) B(c(x1)) -> A(x1) The TRS R consists of the following rules: c(b(a(x1))) -> a(a(a(b(b(b(c(c(c(x1))))))))) b(c(x1)) -> a(a(a(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> x1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 9 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: C(b(a(x1))) -> C(c(x1)) popout

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