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SRS Standard pair #487518743
details
property
value
status
complete
benchmark
aabacaaa-aaaaabacaabac.srs.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n186.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
172.320924997 seconds
cpu usage
684.188042217
max memory
1.5535702016E10
stage attributes
key
value
output-size
5919
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 24 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 8567 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 5586 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(a(c(a(a(a(x1)))))))) -> a(a(a(a(a(b(a(c(a(a(b(a(c(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(a(a(c(a(b(a(a(x1)))))))) -> c(a(b(a(a(c(a(b(a(a(a(a(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(a(a(c(a(b(a(a(x1)))))))) -> A(b(a(a(c(a(b(a(a(a(a(a(x1)))))))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(c(a(b(a(a(a(a(a(x1)))))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(c(a(b(a(a(a(a(a(x1))))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(b(a(a(a(a(a(x1))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(a(x1))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(x1)))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(x1))) The TRS R consists of the following rules: a(a(a(c(a(b(a(a(x1)))))))) -> c(a(b(a(a(c(a(b(a(a(a(a(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 1 SCC with 4 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(x1)))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(a(x1))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(x1))) The TRS R consists of the following rules: a(a(a(c(a(b(a(a(x1)))))))) -> c(a(b(a(a(c(a(b(a(a(a(a(a(x1))))))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains.
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