details

property	value
status	complete
benchmark	z089.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n028.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	6.24930810928 seconds
cpu usage	21.55878298
max memory	3.371339776E9

stage attributes

key	value
output-size	5699
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 7 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 30 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 70 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(b(a(x1)))) -> a(c(a(b(x1)))) a(c(x1)) -> c(c(a(x1))) c(c(c(x1))) -> b(c(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: a(b(b(a(x1)))) -> b(a(c(a(x1)))) c(a(x1)) -> a(c(c(x1))) c(c(c(x1))) -> b(c(b(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: A(b(b(a(x1)))) -> A(c(a(x1))) A(b(b(a(x1)))) -> C(a(x1)) C(a(x1)) -> A(c(c(x1))) C(a(x1)) -> C(c(x1)) C(a(x1)) -> C(x1) C(c(c(x1))) -> C(b(x1)) The TRS R consists of the following rules: a(b(b(a(x1)))) -> b(a(c(a(x1)))) c(a(x1)) -> a(c(c(x1))) c(c(c(x1))) -> b(c(b(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 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(b(a(x1)))) -> C(a(x1)) C(a(x1)) -> A(c(c(x1))) A(b(b(a(x1)))) -> A(c(a(x1))) popout

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

job log

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actions

all output return to SRS Stand 10685