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SRS Standard pair #487512599
details
property
value
status
complete
benchmark
size-12-alpha-3-num-4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n174.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.80640387535 seconds
cpu usage
15.514266663
max memory
2.02829824E9
stage attributes
key
value
output-size
6889
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 111 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 46 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) a(b(x1)) -> b(c(x1)) c(c(x1)) -> a(c(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: a(x1) -> x1 a(x1) -> b(x1) b(a(x1)) -> c(b(x1)) c(c(x1)) -> a(c(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(x1) -> B(x1) B(a(x1)) -> C(b(x1)) B(a(x1)) -> B(x1) C(c(x1)) -> A(c(a(x1))) C(c(x1)) -> C(a(x1)) C(c(x1)) -> A(x1) The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) b(a(x1)) -> c(b(x1)) c(c(x1)) -> a(c(a(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted.
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