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SRS Standard pair #487089310
details
property
value
status
complete
benchmark
jambox5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n144.star.cs.uiowa.edu
space
Secret_05_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
12.9407 seconds
cpu usage
47.2536
user time
45.039
system time
2.21459
max virtual memory
7.9171156E7
max residence set size
4600528.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms]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) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPOrderProof [EQUIVALENT, 26 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 155 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 214 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) QDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) QDP (23) QDPSizeChangeProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(x1)) -> a(a(d(x1))) a(c(x1)) -> b(b(x1)) d(a(b(x1))) -> b(d(d(c(x1)))) d(x1) -> a(x1) b(a(c(a(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: a(b(x1)) -> d(a(a(x1))) c(a(x1)) -> b(b(x1)) b(a(d(x1))) -> c(d(d(b(x1)))) d(x1) -> a(x1) a(c(a(b(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: A(b(x1)) -> D(a(a(x1))) A(b(x1)) -> A(a(x1)) A(b(x1)) -> A(x1) C(a(x1)) -> B(b(x1)) C(a(x1)) -> B(x1) B(a(d(x1))) -> C(d(d(b(x1)))) B(a(d(x1))) -> D(d(b(x1))) B(a(d(x1))) -> D(b(x1)) B(a(d(x1))) -> B(x1) D(x1) -> A(x1) The TRS R consists of the following rules: a(b(x1)) -> d(a(a(x1))) c(a(x1)) -> b(b(x1)) b(a(d(x1))) -> c(d(d(b(x1)))) d(x1) -> a(x1) a(c(a(b(x1)))) -> x1 Q is empty. We have to consider all minimal (P,Q,R)-chains.
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