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SRS Stand 10685 pair #381717452
details
property
value
status
complete
benchmark
06.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
Mixed_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
11.0184690952 seconds
cpu usage
40.495626036
max memory
4.211699712E9
stage attributes
key
value
output-size
10436
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) RootLabelingProof [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT,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) QDPOrderProof [EQUIVALENT, 106 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 66 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 57 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(a(x1)))) -> a(b(b(x1))) b(b(x1)) -> b(a(a(a(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: a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{a_1}(x1))) a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{b_1}(x1))) b_{b_1}(b_{a_1}(x1)) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) b_{b_1}(b_{b_1}(x1)) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_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: A_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> B_{B_1}(b_{a_1}(x1)) A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(x1)) A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(x1) B_{B_1}(b_{a_1}(x1)) -> A_{A_1}(a_{a_1}(a_{a_1}(x1))) B_{B_1}(b_{a_1}(x1)) -> A_{A_1}(a_{a_1}(x1)) B_{B_1}(b_{a_1}(x1)) -> A_{A_1}(x1) B_{B_1}(b_{b_1}(x1)) -> A_{A_1}(a_{a_1}(a_{b_1}(x1))) B_{B_1}(b_{b_1}(x1)) -> A_{A_1}(a_{b_1}(x1)) The TRS R consists of the following rules: a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{a_1}(x1))) a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{b_1}(x1))) b_{b_1}(b_{a_1}(x1)) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) b_{b_1}(b_{b_1}(x1)) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(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.
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