details

property	value
status	complete
benchmark	03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	Bouchare_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.93405 seconds
cpu usage	12.6366
user time	12.0617
system time	0.574907
max virtual memory	5.9597704E7
max residence set size	1603036.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 18 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 176 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(x1)) -> a(a(a(x1))) a(a(b(x1))) -> b(x1) a(b(a(x1))) -> a(b(b(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: b(b(x1)) -> a(a(a(x1))) b(a(a(x1))) -> b(x1) a(b(a(x1))) -> b(b(a(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: B(b(x1)) -> A(a(a(x1))) B(b(x1)) -> A(a(x1)) B(b(x1)) -> A(x1) B(a(a(x1))) -> B(x1) A(b(a(x1))) -> B(b(a(x1))) The TRS R consists of the following rules: b(b(x1)) -> a(a(a(x1))) b(a(a(x1))) -> b(x1) a(b(a(x1))) -> b(b(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]. The following pairs can be oriented strictly and are deleted. A(b(a(x1))) -> B(b(a(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(B(x_1)) = [[-I]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[-I], [-I], [-I]] + [[0A, 0A, 0A], [1A, 0A, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(A(x_1)) = [[-I]] + [[0A, 0A, 0A]] * x_1 >>> popout

output may be truncated. 'popout' for the full output.

job log

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all output return to SRS Standard