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SRS Standard pair #516976216
details
property
value
status
complete
benchmark
random-101.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n090.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
8.86508393288 seconds
cpu usage
31.7446669
max memory
1.688465408E9
stage attributes
key
value
output-size
9444
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 11 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 2 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) DependencyGraphProof [EQUIVALENT, 0 ms] (11) QDP (12) MNOCProof [EQUIVALENT, 0 ms] (13) QDP (14) UsableRulesProof [EQUIVALENT, 0 ms] (15) QDP (16) QReductionProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) QDPOrderProof [EQUIVALENT, 84 ms] (22) QDP (23) QDPOrderProof [EQUIVALENT, 787 ms] (24) QDP (25) PisEmptyProof [EQUIVALENT, 0 ms] (26) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(b(a(x1)))) -> a(b(a(a(x1)))) a(a(a(a(x1)))) -> b(a(a(b(x1)))) a(a(a(b(x1)))) -> b(b(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: a(b(b(b(x1)))) -> a(a(b(a(x1)))) a(a(a(a(x1)))) -> b(a(a(b(x1)))) b(a(a(a(x1)))) -> b(b(b(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(b(x1)))) -> A(a(b(a(x1)))) A(b(b(b(x1)))) -> A(b(a(x1))) A(b(b(b(x1)))) -> B(a(x1)) A(b(b(b(x1)))) -> A(x1) A(a(a(a(x1)))) -> B(a(a(b(x1)))) A(a(a(a(x1)))) -> A(a(b(x1))) A(a(a(a(x1)))) -> A(b(x1)) A(a(a(a(x1)))) -> B(x1) B(a(a(a(x1)))) -> B(b(b(b(x1)))) B(a(a(a(x1)))) -> B(b(b(x1))) B(a(a(a(x1)))) -> B(b(x1)) B(a(a(a(x1)))) -> B(x1) The TRS R consists of the following rules: a(b(b(b(x1)))) -> a(a(b(a(x1)))) a(a(a(a(x1)))) -> b(a(a(b(x1))))
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