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SRS Stand 10685 pair #381718843
details
property
value
status
complete
benchmark
secr6.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n022.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
9.06600904465 seconds
cpu usage
32.789515889
max memory
4.241002496E9
stage attributes
key
value
output-size
4391
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 26 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 7 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 422 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(c(x1))) -> c(c(c(b(b(b(a(a(a(x1))))))))) c(b(x1)) -> a(a(a(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> 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: c(b(a(x1))) -> a(a(a(b(b(b(c(c(c(x1))))))))) b(c(x1)) -> a(a(a(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> 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: C(b(a(x1))) -> A(a(a(b(b(b(c(c(c(x1))))))))) C(b(a(x1))) -> A(a(b(b(b(c(c(c(x1)))))))) C(b(a(x1))) -> A(b(b(b(c(c(c(x1))))))) C(b(a(x1))) -> B(b(b(c(c(c(x1)))))) C(b(a(x1))) -> B(b(c(c(c(x1))))) C(b(a(x1))) -> B(c(c(c(x1)))) C(b(a(x1))) -> C(c(c(x1))) C(b(a(x1))) -> C(c(x1)) C(b(a(x1))) -> C(x1) B(c(x1)) -> A(a(a(x1))) B(c(x1)) -> A(a(x1)) B(c(x1)) -> A(x1) The TRS R consists of the following rules: c(b(a(x1))) -> a(a(a(b(b(b(c(c(c(x1))))))))) b(c(x1)) -> a(a(a(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> 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 9 less nodes. ----------------------------------------
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