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SRS Standard pair #487510745
details
property
value
status
complete
benchmark
random-158.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n128.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.58897709846 seconds
cpu usage
11.313605804
max memory
1.315037184E9
stage attributes
key
value
output-size
6425
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) RootLabelingProof [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 5 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 58 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(x1)))) -> a(b(a(b(x1)))) a(a(b(a(x1)))) -> a(b(a(a(x1)))) b(a(b(b(x1)))) -> b(b(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_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1)))) -> b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1)))) -> b_{b_1}(b_{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_{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}(a_{a_1}(a_{a_1}(x1)))) -> B_{A_1}(x1) A_{A_1}(a_{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}(b_{a_1}(a_{a_1}(x1)))) -> B_{A_1}(a_{a_1}(a_{a_1}(x1))) A_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> A_{A_1}(a_{a_1}(x1)) A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> B_{A_1}(a_{a_1}(a_{b_1}(x1))) A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> A_{A_1}(a_{b_1}(x1)) B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1)))) -> B_{A_1}(a_{a_1}(a_{a_1}(x1))) B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1)))) -> A_{A_1}(a_{a_1}(x1)) B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1)))) -> A_{A_1}(x1) B_{A_1}(a_{b_1}(b_{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}(b_{b_1}(x1)))) -> A_{A_1}(a_{b_1}(x1)) The TRS R consists of the following rules: a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1)))) -> b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1)))) -> b_{b_1}(b_{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].
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