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SRS Stand 10685 pair #381718361
details
property
value
status
complete
benchmark
x09.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n082.star.cs.uiowa.edu
space
Secret_07_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
26.7850339413 seconds
cpu usage
101.105144295
max memory
5.625331712E9
stage attributes
key
value
output-size
11441
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 10 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 451 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 228 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 376 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 297 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 321 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 1455 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(b(b(x1))))) -> b(b(a(b(b(a(a(a(x1)))))))) b(b(a(x1))) -> x1 a(x1) -> b(b(b(x1))) a(x1) -> 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: b(b(a(a(a(x1))))) -> a(a(a(b(b(a(b(b(x1)))))))) a(b(b(x1))) -> x1 a(x1) -> b(b(b(x1))) a(x1) -> 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: B(b(a(a(a(x1))))) -> A(a(a(b(b(a(b(b(x1)))))))) B(b(a(a(a(x1))))) -> A(a(b(b(a(b(b(x1))))))) B(b(a(a(a(x1))))) -> A(b(b(a(b(b(x1)))))) B(b(a(a(a(x1))))) -> B(b(a(b(b(x1))))) B(b(a(a(a(x1))))) -> B(a(b(b(x1)))) B(b(a(a(a(x1))))) -> A(b(b(x1))) B(b(a(a(a(x1))))) -> B(b(x1)) B(b(a(a(a(x1))))) -> B(x1) A(x1) -> B(b(b(x1))) A(x1) -> B(b(x1)) A(x1) -> B(x1) The TRS R consists of the following rules: b(b(a(a(a(x1))))) -> a(a(a(b(b(a(b(b(x1)))))))) a(b(b(x1))) -> x1 a(x1) -> b(b(b(x1))) a(x1) -> b(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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