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SRS_Standard 2019-03-29 03.29 pair #432289063
details
property
value
status
complete
benchmark
11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n104.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
5.27223 seconds
cpu usage
17.554
user time
16.5593
system time
0.994704
max virtual memory
2.0326816E7
max residence set size
2822944.0
stage attributes
key
value
starexec-result
YES
output
17.36/5.19 YES 17.46/5.22 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 17.46/5.22 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 17.46/5.22 17.46/5.22 17.46/5.22 Termination w.r.t. Q of the given QTRS could be proven: 17.46/5.22 17.46/5.22 (0) QTRS 17.46/5.22 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 17.46/5.22 (2) QDP 17.46/5.22 (3) MRRProof [EQUIVALENT, 50 ms] 17.46/5.22 (4) QDP 17.46/5.22 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 17.46/5.22 (6) QDP 17.46/5.22 (7) QDPOrderProof [EQUIVALENT, 70 ms] 17.46/5.22 (8) QDP 17.46/5.22 (9) PisEmptyProof [EQUIVALENT, 0 ms] 17.46/5.22 (10) YES 17.46/5.22 17.46/5.22 17.46/5.22 ---------------------------------------- 17.46/5.22 17.46/5.22 (0) 17.46/5.22 Obligation: 17.46/5.22 Q restricted rewrite system: 17.46/5.22 The TRS R consists of the following rules: 17.46/5.22 17.46/5.22 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 17.46/5.22 1(0(1(1(x1)))) -> 0(0(0(1(x1)))) 17.46/5.22 17.46/5.22 Q is empty. 17.46/5.22 17.46/5.22 ---------------------------------------- 17.46/5.22 17.46/5.22 (1) DependencyPairsProof (EQUIVALENT) 17.46/5.22 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 17.46/5.22 ---------------------------------------- 17.46/5.22 17.46/5.22 (2) 17.46/5.22 Obligation: 17.46/5.22 Q DP problem: 17.46/5.22 The TRS P consists of the following rules: 17.46/5.22 17.46/5.22 0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1)))) 17.46/5.22 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 17.46/5.22 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 17.46/5.22 0^1(0(0(0(x1)))) -> 1^1(x1) 17.46/5.22 1^1(0(1(1(x1)))) -> 0^1(0(0(1(x1)))) 17.46/5.22 1^1(0(1(1(x1)))) -> 0^1(0(1(x1))) 17.46/5.22 1^1(0(1(1(x1)))) -> 0^1(1(x1)) 17.46/5.22 17.46/5.22 The TRS R consists of the following rules: 17.46/5.22 17.46/5.22 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 17.46/5.22 1(0(1(1(x1)))) -> 0(0(0(1(x1)))) 17.46/5.22 17.46/5.22 Q is empty. 17.46/5.22 We have to consider all minimal (P,Q,R)-chains. 17.46/5.22 ---------------------------------------- 17.46/5.22 17.46/5.22 (3) MRRProof (EQUIVALENT) 17.46/5.22 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. 17.46/5.22 17.46/5.22 Strictly oriented dependency pairs: 17.46/5.22 17.46/5.22 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 17.46/5.22 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 17.46/5.22 0^1(0(0(0(x1)))) -> 1^1(x1) 17.46/5.22 1^1(0(1(1(x1)))) -> 0^1(0(1(x1))) 17.46/5.22 1^1(0(1(1(x1)))) -> 0^1(1(x1)) 17.46/5.22 17.46/5.22 17.46/5.22 Used ordering: Polynomial interpretation [POLO]: 17.46/5.22 17.46/5.22 POL(0(x_1)) = 1 + x_1 17.46/5.22 POL(0^1(x_1)) = x_1 17.46/5.22 POL(1(x_1)) = 1 + x_1 17.46/5.22 POL(1^1(x_1)) = x_1 17.46/5.22 17.46/5.22 17.46/5.22 ---------------------------------------- 17.46/5.22 17.46/5.22 (4) 17.46/5.22 Obligation: 17.46/5.22 Q DP problem: 17.46/5.22 The TRS P consists of the following rules: 17.46/5.22 17.46/5.22 0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1)))) 17.46/5.22 1^1(0(1(1(x1)))) -> 0^1(0(0(1(x1)))) 17.46/5.22 17.46/5.22 The TRS R consists of the following rules: 17.46/5.22 17.46/5.22 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 17.46/5.22 1(0(1(1(x1)))) -> 0(0(0(1(x1)))) 17.46/5.22 17.46/5.22 Q is empty. 17.46/5.22 We have to consider all minimal (P,Q,R)-chains. 17.46/5.22 ---------------------------------------- 17.46/5.22 17.46/5.22 (5) DependencyGraphProof (EQUIVALENT)
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