details

property	value
status	complete
benchmark	5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	11.4257 seconds
cpu usage	41.9618
user time	40.1242
system time	1.83763
max virtual memory	5.8396468E7
max residence set size	4216004.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 26 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 189 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 65 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 271 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 103 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 770 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(x1))) -> a(a(b(b(x1)))) b(b(b(b(x1)))) -> a(b(a(a(x1)))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(a(x1))) -> A(a(b(b(x1)))) A(a(a(x1))) -> A(b(b(x1))) A(a(a(x1))) -> B(b(x1)) A(a(a(x1))) -> B(x1) B(b(b(b(x1)))) -> A(b(a(a(x1)))) B(b(b(b(x1)))) -> B(a(a(x1))) B(b(b(b(x1)))) -> A(a(x1)) B(b(b(b(x1)))) -> A(x1) The TRS R consists of the following rules: a(a(a(x1))) -> a(a(b(b(x1)))) b(b(b(b(x1)))) -> a(b(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(a(x1))) -> A(b(b(x1))) A(a(a(x1))) -> A(a(b(b(x1)))) A(a(a(x1))) -> B(b(x1)) B(b(b(b(x1)))) -> A(b(a(a(x1)))) A(a(a(x1))) -> B(x1) B(b(b(b(x1)))) -> A(a(x1)) B(b(b(b(x1)))) -> A(x1) The TRS R consists of the following rules: a(a(a(x1))) -> a(a(b(b(x1)))) b(b(b(b(x1)))) -> a(b(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. popout

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