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SRS_Standard 2019-03-29 03.29 pair #432291187
details
property
value
status
complete
benchmark
size-12-alpha-2-num-15.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n150.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
10.1617 seconds
cpu usage
36.298
user time
34.6313
system time
1.6667
max virtual memory
2.1064004E7
max residence set size
4453136.0
stage attributes
key
value
starexec-result
YES
output
35.88/10.05 YES 35.88/10.06 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 35.88/10.06 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 35.88/10.06 35.88/10.06 35.88/10.06 Termination w.r.t. Q of the given QTRS could be proven: 35.88/10.06 35.88/10.06 (0) QTRS 35.88/10.06 (1) FlatCCProof [EQUIVALENT, 0 ms] 35.88/10.06 (2) QTRS 35.88/10.06 (3) RootLabelingProof [EQUIVALENT, 1 ms] 35.88/10.06 (4) QTRS 35.88/10.06 (5) QTRSRRRProof [EQUIVALENT, 35 ms] 35.88/10.06 (6) QTRS 35.88/10.06 (7) DependencyPairsProof [EQUIVALENT, 7 ms] 35.88/10.06 (8) QDP 35.88/10.06 (9) DependencyGraphProof [EQUIVALENT, 4 ms] 35.88/10.06 (10) QDP 35.88/10.06 (11) QDPOrderProof [EQUIVALENT, 43 ms] 35.88/10.06 (12) QDP 35.88/10.06 (13) QDPOrderProof [EQUIVALENT, 24 ms] 35.88/10.06 (14) QDP 35.88/10.06 (15) DependencyGraphProof [EQUIVALENT, 0 ms] 35.88/10.06 (16) TRUE 35.88/10.06 35.88/10.06 35.88/10.06 ---------------------------------------- 35.88/10.06 35.88/10.06 (0) 35.88/10.06 Obligation: 35.88/10.06 Q restricted rewrite system: 35.88/10.06 The TRS R consists of the following rules: 35.88/10.06 35.88/10.06 a(x1) -> x1 35.88/10.06 a(a(x1)) -> a(b(a(b(x1)))) 35.88/10.06 b(b(b(b(x1)))) -> a(x1) 35.88/10.06 35.88/10.06 Q is empty. 35.88/10.06 35.88/10.06 ---------------------------------------- 35.88/10.06 35.88/10.06 (1) FlatCCProof (EQUIVALENT) 35.88/10.06 We used flat context closure [ROOTLAB] 35.88/10.06 As Q is empty the flat context closure was sound AND complete. 35.88/10.06 35.88/10.06 ---------------------------------------- 35.88/10.06 35.88/10.06 (2) 35.88/10.06 Obligation: 35.88/10.06 Q restricted rewrite system: 35.88/10.06 The TRS R consists of the following rules: 35.88/10.06 35.88/10.06 a(a(x1)) -> a(b(a(b(x1)))) 35.88/10.06 a(a(x1)) -> a(x1) 35.88/10.06 b(a(x1)) -> b(x1) 35.88/10.06 a(b(b(b(b(x1))))) -> a(a(x1)) 35.88/10.06 b(b(b(b(b(x1))))) -> b(a(x1)) 35.88/10.06 35.88/10.06 Q is empty. 35.88/10.06 35.88/10.06 ---------------------------------------- 35.88/10.06 35.88/10.06 (3) RootLabelingProof (EQUIVALENT) 35.88/10.06 We used plain root labeling [ROOTLAB] with the following heuristic: 35.88/10.06 LabelAll: All function symbols get labeled 35.88/10.06 35.88/10.06 As Q is empty the root labeling was sound AND complete. 35.88/10.06 35.88/10.06 ---------------------------------------- 35.88/10.06 35.88/10.06 (4) 35.88/10.06 Obligation: 35.88/10.06 Q restricted rewrite system: 35.88/10.06 The TRS R consists of the following rules: 35.88/10.06 35.88/10.06 a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) 35.88/10.06 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) 35.88/10.06 a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1) 35.88/10.06 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1) 35.88/10.06 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 35.88/10.06 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 35.88/10.06 a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1)) 35.88/10.06 a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> a_{a_1}(a_{b_1}(x1)) 35.88/10.06 b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(x1)) 35.88/10.06 b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{b_1}(x1)) 35.88/10.06 35.88/10.06 Q is empty. 35.88/10.06 35.88/10.06 ---------------------------------------- 35.88/10.06 35.88/10.06 (5) QTRSRRRProof (EQUIVALENT) 35.88/10.06 Used ordering: 35.88/10.06 Polynomial interpretation [POLO]: 35.88/10.06 35.88/10.06 POL(a_{a_1}(x_1)) = 2 + x_1 35.88/10.06 POL(a_{b_1}(x_1)) = x_1 35.88/10.06 POL(b_{a_1}(x_1)) = 1 + x_1 35.88/10.06 POL(b_{b_1}(x_1)) = 1 + x_1 35.88/10.06 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 35.88/10.06
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