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SRS Standard pair #487083322
details
property
value
status
complete
benchmark
uni-5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n148.star.cs.uiowa.edu
space
Waldmann_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.78672 seconds
cpu usage
11.7787
user time
11.1647
system time
0.61396
max virtual memory
1.9940332E7
max residence set size
1608720.0
stage attributes
key
value
starexec-result
YES
output
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 14 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 104 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 75 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(x1)) -> x1 b(b(x1)) -> c(c(c(c(x1)))) c(c(x1)) -> a(c(b(x1))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(x1)) -> C(c(c(c(x1)))) B(b(x1)) -> C(c(c(x1))) B(b(x1)) -> C(c(x1)) B(b(x1)) -> C(x1) C(c(x1)) -> A(c(b(x1))) C(c(x1)) -> C(b(x1)) C(c(x1)) -> B(x1) The TRS R consists of the following rules: a(a(x1)) -> x1 b(b(x1)) -> c(c(c(c(x1)))) c(c(x1)) -> a(c(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(x1)) -> C(b(x1)) C(c(x1)) -> B(x1) B(b(x1)) -> C(c(c(c(x1)))) B(b(x1)) -> C(c(c(x1))) B(b(x1)) -> C(c(x1)) B(b(x1)) -> C(x1) The TRS R consists of the following rules: a(a(x1)) -> x1 b(b(x1)) -> c(c(c(c(x1)))) c(c(x1)) -> a(c(b(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. B(b(x1)) -> C(c(c(c(x1)))) B(b(x1)) -> C(c(c(x1))) B(b(x1)) -> C(c(x1)) B(b(x1)) -> C(x1)
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