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SRS Standard pair #487513979
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
n125.star.cs.uiowa.edu
space
Bouchare_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.63314795494 seconds
cpu usage
15.428640758
max memory
1.732620288E9
stage attributes
key
value
output-size
8178
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) RootLabelingProof [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 8 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 53 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 4 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)
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