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SRS Stand 10685 pair #381722079
details
property
value
status
complete
benchmark
size-12-alpha-3-num-535.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n053.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
5.30466103554 seconds
cpu usage
17.546558316
max memory
2.585714688E9
stage attributes
key
value
output-size
5273
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Overlay + Local Confluence [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 155 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 71 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(x1))) -> c(a(c(a(a(x1))))) a(c(x1)) -> b(a(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(a(a(x1))) -> a(a(c(a(c(x1))))) c(a(x1)) -> a(b(x1)) Q is empty. ---------------------------------------- (3) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(x1))) -> a(a(c(a(c(x1))))) c(a(x1)) -> a(b(x1)) The set Q consists of the following terms: b(a(a(x0))) c(a(x0)) ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(x1))) -> C(a(c(x1))) B(a(a(x1))) -> C(x1) C(a(x1)) -> B(x1) The TRS R consists of the following rules: b(a(a(x1))) -> a(a(c(a(c(x1))))) c(a(x1)) -> a(b(x1)) The set Q consists of the following terms: b(a(a(x0))) c(a(x0))
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