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SRS Standard pair #516975748
details
property
value
status
complete
benchmark
secr2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n178.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
5.93049001694 seconds
cpu usage
20.221702494
max memory
1.341554688E9
stage attributes
key
value
output-size
9401
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 15 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 1 ms] (9) QDP (10) MRRProof [EQUIVALENT, 12 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 0 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) MRRProof [EQUIVALENT, 29 ms] (25) QDP (26) PisEmptyProof [EQUIVALENT, 0 ms] (27) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x1)) -> b(b(b(x1))) a(c(x1)) -> c(b(b(x1))) b(a(x1)) -> a(a(a(x1))) b(c(x1)) -> c(c(c(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> 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(x1)) -> b(b(b(x1))) c(a(x1)) -> b(b(c(x1))) a(b(x1)) -> a(a(a(x1))) c(b(x1)) -> c(c(c(x1))) a(x1) -> x1 b(x1) -> x1 c(x1) -> 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(x1)) -> B(b(b(x1))) B(a(x1)) -> B(b(x1)) B(a(x1)) -> B(x1) C(a(x1)) -> B(b(c(x1))) C(a(x1)) -> B(c(x1)) C(a(x1)) -> C(x1) A(b(x1)) -> A(a(a(x1))) A(b(x1)) -> A(a(x1))
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