details

property	value
status	complete
benchmark	size-11-alpha-3-num-11.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n124.star.cs.uiowa.edu
space	Waldmann_07_size11

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.5089571476 seconds
cpu usage	11.059419487
max memory	1.214861312E9

stage attributes

key	value
output-size	5402
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 1 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 85 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> b(x1) a(c(x1)) -> c(c(a(b(x1)))) b(b(x1)) -> a(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(x1) -> b(x1) c(a(x1)) -> b(a(c(c(x1)))) b(b(x1)) -> 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: A(x1) -> B(x1) C(a(x1)) -> B(a(c(c(x1)))) C(a(x1)) -> A(c(c(x1))) C(a(x1)) -> C(c(x1)) C(a(x1)) -> C(x1) B(b(x1)) -> A(x1) The TRS R consists of the following rules: a(x1) -> b(x1) c(a(x1)) -> b(a(c(c(x1)))) b(b(x1)) -> a(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 2 SCCs with 2 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) popout

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

job log

popout

actions

all output return to SRS Standard