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SRS Stand 10685 pair #381719046
details
property
value
status
complete
benchmark
z117.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n191.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
10.8683569431 seconds
cpu usage
39.863137469
max memory
4.08938496E9
stage attributes
key
value
output-size
6900
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 30 ms] (4) QDP (5) MRRProof [EQUIVALENT, 79 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 26 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 1035 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: d(a(x1)) -> b(d(x1)) b(x1) -> a(a(a(x1))) c(d(c(x1))) -> a(d(x1)) b(d(d(x1))) -> c(c(d(d(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(d(x1)) -> d(b(x1)) b(x1) -> a(a(a(x1))) c(d(c(x1))) -> d(a(x1)) d(d(b(x1))) -> c(d(d(c(c(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(d(x1)) -> D(b(x1)) A(d(x1)) -> B(x1) B(x1) -> A(a(a(x1))) B(x1) -> A(a(x1)) B(x1) -> A(x1) C(d(c(x1))) -> D(a(x1)) C(d(c(x1))) -> A(x1) D(d(b(x1))) -> C(d(d(c(c(x1))))) D(d(b(x1))) -> D(d(c(c(x1)))) D(d(b(x1))) -> D(c(c(x1))) D(d(b(x1))) -> C(c(x1)) D(d(b(x1))) -> C(x1) The TRS R consists of the following rules: a(d(x1)) -> d(b(x1)) b(x1) -> a(a(a(x1))) c(d(c(x1))) -> d(a(x1)) d(d(b(x1))) -> c(d(d(c(c(x1))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
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