details

property	value
status	complete
benchmark	aprove00.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n053.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.03004 seconds
cpu usage	12.7675
user time	12.0997
system time	0.66776
max virtual memory	4.0184584E7
max residence set size	1898064.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 11 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 4 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) MNOCProof [EQUIVALENT, 0 ms] (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 24 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(s(x1)) -> s(s(s(p(s(b(p(p(s(s(x1)))))))))) b(s(x1)) -> s(s(s(p(p(s(s(c(p(s(p(s(x1)))))))))))) c(s(x1)) -> p(s(p(s(a(p(s(p(s(x1))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> 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(s(x1)) -> P(s(b(p(p(s(s(x1))))))) A(s(x1)) -> B(p(p(s(s(x1))))) A(s(x1)) -> P(p(s(s(x1)))) A(s(x1)) -> P(s(s(x1))) B(s(x1)) -> P(p(s(s(c(p(s(p(s(x1))))))))) B(s(x1)) -> P(s(s(c(p(s(p(s(x1)))))))) B(s(x1)) -> C(p(s(p(s(x1))))) B(s(x1)) -> P(s(p(s(x1)))) B(s(x1)) -> P(s(x1)) C(s(x1)) -> P(s(p(s(a(p(s(p(s(x1))))))))) C(s(x1)) -> P(s(a(p(s(p(s(x1))))))) C(s(x1)) -> A(p(s(p(s(x1))))) C(s(x1)) -> P(s(p(s(x1)))) C(s(x1)) -> P(s(x1)) P(p(s(x1))) -> P(x1) The TRS R consists of the following rules: a(s(x1)) -> s(s(s(p(s(b(p(p(s(s(x1)))))))))) b(s(x1)) -> s(s(s(p(p(s(s(c(p(s(p(s(x1)))))))))))) c(s(x1)) -> p(s(p(s(a(p(s(p(s(x1))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> 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 2 SCCs with 11 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(s(x1))) -> P(x1) popout

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