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SRS Standard pair #516976180
details
property
value
status
complete
benchmark
random-90.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n113.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
11.6998150349 seconds
cpu usage
43.096908964
max memory
2.008952832E9
stage attributes
key
value
output-size
6203
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 15 ms] (4) QDP (5) MRRProof [EQUIVALENT, 118 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 319 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 753 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(a(b(x1)))) -> a(b(a(a(x1)))) a(a(b(a(x1)))) -> b(b(b(b(x1)))) b(b(a(b(x1)))) -> a(a(a(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(a(x1)))) -> a(a(b(a(x1)))) a(b(a(a(x1)))) -> b(b(b(b(x1)))) b(a(b(b(x1)))) -> a(a(a(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: B(a(a(a(x1)))) -> A(a(b(a(x1)))) B(a(a(a(x1)))) -> A(b(a(x1))) B(a(a(a(x1)))) -> B(a(x1)) A(b(a(a(x1)))) -> B(b(b(b(x1)))) A(b(a(a(x1)))) -> B(b(b(x1))) A(b(a(a(x1)))) -> B(b(x1)) A(b(a(a(x1)))) -> B(x1) B(a(b(b(x1)))) -> A(a(a(a(x1)))) B(a(b(b(x1)))) -> A(a(a(x1))) B(a(b(b(x1)))) -> A(a(x1)) B(a(b(b(x1)))) -> A(x1) The TRS R consists of the following rules: b(a(a(a(x1)))) -> a(a(b(a(x1)))) a(b(a(a(x1)))) -> b(b(b(b(x1)))) b(a(b(b(x1)))) -> a(a(a(a(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. Strictly oriented dependency pairs: B(a(a(a(x1)))) -> A(b(a(x1))) B(a(a(a(x1)))) -> B(a(x1)) A(b(a(a(x1)))) -> B(b(b(x1))) A(b(a(a(x1)))) -> B(b(x1))
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