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SRS_Standard 2019-03-29 03.29 pair #432294091
details
property
value
status
complete
benchmark
secr6.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n081.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
11.1223 seconds
cpu usage
30.8573
user time
29.4763
system time
1.38093
max virtual memory
6.1788088E7
max residence set size
4060048.0
stage attributes
key
value
starexec-result
YES
output
26.81/7.78 YES 30.80/11.07 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 30.80/11.07 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 30.80/11.07 30.80/11.07 30.80/11.07 Termination w.r.t. Q of the given QTRS could be proven: 30.80/11.07 30.80/11.07 (0) QTRS 30.80/11.07 (1) QTRS Reverse [EQUIVALENT, 0 ms] 30.80/11.07 (2) QTRS 30.80/11.07 (3) DependencyPairsProof [EQUIVALENT, 10 ms] 30.80/11.07 (4) QDP 30.80/11.07 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 30.80/11.07 (6) QDP 30.80/11.07 (7) QDPOrderProof [EQUIVALENT, 429 ms] 30.80/11.07 (8) QDP 30.80/11.07 (9) PisEmptyProof [EQUIVALENT, 0 ms] 30.80/11.07 (10) YES 30.80/11.07 30.80/11.07 30.80/11.07 ---------------------------------------- 30.80/11.07 30.80/11.07 (0) 30.80/11.07 Obligation: 30.80/11.07 Q restricted rewrite system: 30.80/11.07 The TRS R consists of the following rules: 30.80/11.07 30.80/11.07 a(b(c(x1))) -> c(c(c(b(b(b(a(a(a(x1))))))))) 30.80/11.07 c(b(x1)) -> a(a(a(x1))) 30.80/11.07 a(x1) -> x1 30.80/11.07 b(x1) -> x1 30.80/11.07 c(x1) -> x1 30.80/11.07 30.80/11.07 Q is empty. 30.80/11.07 30.80/11.07 ---------------------------------------- 30.80/11.07 30.80/11.07 (1) QTRS Reverse (EQUIVALENT) 30.80/11.07 We applied the QTRS Reverse Processor [REVERSE]. 30.80/11.07 ---------------------------------------- 30.80/11.07 30.80/11.07 (2) 30.80/11.07 Obligation: 30.80/11.07 Q restricted rewrite system: 30.80/11.07 The TRS R consists of the following rules: 30.80/11.07 30.80/11.07 c(b(a(x1))) -> a(a(a(b(b(b(c(c(c(x1))))))))) 30.80/11.07 b(c(x1)) -> a(a(a(x1))) 30.80/11.07 a(x1) -> x1 30.80/11.07 b(x1) -> x1 30.80/11.07 c(x1) -> x1 30.80/11.07 30.80/11.07 Q is empty. 30.80/11.07 30.80/11.07 ---------------------------------------- 30.80/11.07 30.80/11.07 (3) DependencyPairsProof (EQUIVALENT) 30.80/11.07 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 30.80/11.07 ---------------------------------------- 30.80/11.07 30.80/11.07 (4) 30.80/11.07 Obligation: 30.80/11.07 Q DP problem: 30.80/11.07 The TRS P consists of the following rules: 30.80/11.07 30.80/11.07 C(b(a(x1))) -> A(a(a(b(b(b(c(c(c(x1))))))))) 30.80/11.07 C(b(a(x1))) -> A(a(b(b(b(c(c(c(x1)))))))) 30.80/11.07 C(b(a(x1))) -> A(b(b(b(c(c(c(x1))))))) 30.80/11.07 C(b(a(x1))) -> B(b(b(c(c(c(x1)))))) 30.80/11.07 C(b(a(x1))) -> B(b(c(c(c(x1))))) 30.80/11.07 C(b(a(x1))) -> B(c(c(c(x1)))) 30.80/11.07 C(b(a(x1))) -> C(c(c(x1))) 30.80/11.07 C(b(a(x1))) -> C(c(x1)) 30.80/11.07 C(b(a(x1))) -> C(x1) 30.80/11.07 B(c(x1)) -> A(a(a(x1))) 30.80/11.07 B(c(x1)) -> A(a(x1)) 30.80/11.07 B(c(x1)) -> A(x1) 30.80/11.07 30.80/11.07 The TRS R consists of the following rules: 30.80/11.07 30.80/11.07 c(b(a(x1))) -> a(a(a(b(b(b(c(c(c(x1))))))))) 30.80/11.07 b(c(x1)) -> a(a(a(x1))) 30.80/11.07 a(x1) -> x1 30.80/11.07 b(x1) -> x1 30.80/11.07 c(x1) -> x1 30.80/11.07 30.80/11.07 Q is empty. 30.80/11.07 We have to consider all minimal (P,Q,R)-chains. 30.80/11.07 ---------------------------------------- 30.80/11.07 30.80/11.07 (5) DependencyGraphProof (EQUIVALENT) 30.80/11.07 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 9 less nodes. 30.80/11.07 ---------------------------------------- 30.80/11.07 30.80/11.07 (6) 30.80/11.07 Obligation: 30.80/11.07 Q DP problem: 30.80/11.07 The TRS P consists of the following rules: 30.80/11.07 30.80/11.07 C(b(a(x1))) -> C(c(x1))
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