details

property	value
status	complete
benchmark	jambox3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	Secret_05_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	23.5446 seconds
cpu usage	89.7025
user time	86.9098
system time	2.79265
max virtual memory	6.102314E7
max residence set size	5263696.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 1943 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 777 ms] (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(c(x1))) -> b(c(b(a(x1)))) b(x1) -> d(a(x1)) b(a(c(d(x1)))) -> a(a(a(x1))) c(x1) -> x1 b(x1) -> c(d(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(a(a(x1))) -> a(b(c(b(x1)))) b(x1) -> a(d(x1)) d(c(a(b(x1)))) -> a(a(a(x1))) c(x1) -> x1 b(x1) -> d(c(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(a(a(x1))) -> B(c(b(x1))) C(a(a(x1))) -> C(b(x1)) C(a(a(x1))) -> B(x1) B(x1) -> D(x1) B(x1) -> D(c(x1)) B(x1) -> C(x1) The TRS R consists of the following rules: c(a(a(x1))) -> a(b(c(b(x1)))) b(x1) -> a(d(x1)) d(c(a(b(x1)))) -> a(a(a(x1))) c(x1) -> x1 b(x1) -> d(c(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 2 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: B(x1) -> C(x1) C(a(a(x1))) -> B(c(b(x1))) C(a(a(x1))) -> C(b(x1)) popout

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