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SRS Standard pair #487085716
details
property
value
status
complete
benchmark
07.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Bouchare_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.59085 seconds
cpu usage
10.5078
user time
9.97899
system time
0.528825
max virtual memory
5.9400412E7
max residence set size
1307296.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) RootLabelingProof [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 21 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 9 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(x1)) -> b(b(x1)) b(a(b(x1))) -> b(a(a(x1))) a(a(a(x1))) -> a(b(b(x1))) Q is empty. ---------------------------------------- (1) RootLabelingProof (EQUIVALENT) We used plain root labeling [ROOTLAB] with the following heuristic: LabelAll: All function symbols get labeled As Q is empty the root labeling was sound AND complete. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(b_{b_1}(x1)) b_{a_1}(a_{a_1}(x1)) -> b_{b_1}(b_{a_1}(x1)) b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{b_1}(x1))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(x1))) a_{a_1}(a_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{b_1}(x1))) a_{a_1}(a_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(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: B_{A_1}(a_{a_1}(x1)) -> B_{A_1}(x1) B_{A_1}(a_{b_1}(b_{b_1}(x1))) -> B_{A_1}(a_{a_1}(a_{b_1}(x1))) B_{A_1}(a_{b_1}(b_{b_1}(x1))) -> A_{A_1}(a_{b_1}(x1)) B_{A_1}(a_{b_1}(b_{a_1}(x1))) -> B_{A_1}(a_{a_1}(a_{a_1}(x1))) B_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(a_{a_1}(x1)) B_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(x1) A_{A_1}(a_{a_1}(a_{a_1}(x1))) -> B_{A_1}(x1) The TRS R consists of the following rules: b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(b_{b_1}(x1)) b_{a_1}(a_{a_1}(x1)) -> b_{b_1}(b_{a_1}(x1)) b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{b_1}(x1))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(x1))) a_{a_1}(a_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{b_1}(x1))) a_{a_1}(a_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (6) Obligation: Q DP problem:
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