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SRS Standard pair #516975808
details
property
value
status
complete
benchmark
beans6.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n026.star.cs.uiowa.edu
space
Zantema_06
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
7.54798388481 seconds
cpu usage
26.886908593
max memory
1.425354752E9
stage attributes
key
value
output-size
5936
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 31 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 59 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 42 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) a(a(b(x1))) -> d(b(a(x1))) a(d(x1)) -> d(a(x1)) b(d(x1)) -> a(b(x1)) a(a(x1)) -> a(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: a(a(b(x1))) -> c(b(a(x1))) a(c(x1)) -> c(a(x1)) b(c(x1)) -> a(b(x1)) b(a(a(x1))) -> a(b(d(x1))) d(a(x1)) -> a(d(x1)) d(b(x1)) -> b(a(x1)) a(a(x1)) -> a(b(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(b(x1))) -> B(a(x1)) A(a(b(x1))) -> A(x1) A(c(x1)) -> A(x1) B(c(x1)) -> A(b(x1)) B(c(x1)) -> B(x1) B(a(a(x1))) -> A(b(d(x1))) B(a(a(x1))) -> B(d(x1)) B(a(a(x1))) -> D(x1) D(a(x1)) -> A(d(x1)) D(a(x1)) -> D(x1) D(b(x1)) -> B(a(x1)) D(b(x1)) -> A(x1) A(a(x1)) -> A(b(a(x1))) A(a(x1)) -> B(a(x1)) The TRS R consists of the following rules: a(a(b(x1))) -> c(b(a(x1))) a(c(x1)) -> c(a(x1)) b(c(x1)) -> a(b(x1)) b(a(a(x1))) -> a(b(d(x1))) d(a(x1)) -> a(d(x1)) d(b(x1)) -> b(a(x1))
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