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SRS Standard pair #487519157
details
property
value
status
complete
benchmark
secr3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n032.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.35183405876 seconds
cpu usage
13.977744353
max memory
1.511555072E9
stage attributes
key
value
output-size
8060
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 1 ms]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) MRRProof [EQUIVALENT, 23 ms] (11) QDP (12) DependencyGraphProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPOrderProof [EQUIVALENT, 49 ms] (15) QDP (16) PisEmptyProof [EQUIVALENT, 0 ms] (17) YES (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(x1))) -> b(c(x1)) b(b(b(x1))) -> c(b(x1)) c(x1) -> a(b(x1)) c(d(x1)) -> d(c(b(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(b(a(x1))) -> c(b(x1)) b(b(b(x1))) -> b(c(x1)) c(x1) -> b(a(x1)) d(c(x1)) -> a(b(c(d(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(a(x1))) -> C(b(x1)) A(b(a(x1))) -> B(x1) B(b(b(x1))) -> B(c(x1)) B(b(b(x1))) -> C(x1) C(x1) -> B(a(x1)) C(x1) -> A(x1) D(c(x1)) -> A(b(c(d(x1)))) D(c(x1)) -> B(c(d(x1))) D(c(x1)) -> C(d(x1)) D(c(x1)) -> D(x1) The TRS R consists of the following rules: a(b(a(x1))) -> c(b(x1)) b(b(b(x1))) -> b(c(x1)) c(x1) -> b(a(x1)) d(c(x1)) -> a(b(c(d(x1)))) Q is empty.
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