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SRS Standard pair #487511411
details
property
value
status
complete
benchmark
dj.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n190.star.cs.uiowa.edu
space
Secret_07_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.22783207893 seconds
cpu usage
13.348443003
max memory
1.694912512E9
stage attributes
key
value
output-size
5575
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 13 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 10 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) MRRProof [EQUIVALENT, 16 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 1(1(x1)) -> 0(0(0(0(x1)))) 0(0(x1)) -> 0(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: 0(1(x1)) -> 1(0(0(0(x1)))) 1(0(x1)) -> 1(x1) 1(1(x1)) -> 0(0(0(0(x1)))) 0(0(x1)) -> 0(x1) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 1(1(x1)) -> 0(0(0(0(x1)))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(x1)) -> 1(0(0(0(x1)))) 1(0(x1)) -> 1(x1) 0(0(x1)) -> 0(x1) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ----------------------------------------
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