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SRS Standard pair #516973354
details
property
value
status
complete
benchmark
02.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
26.2642598152 seconds
cpu usage
101.444166856
max memory
2.446237696E9
stage attributes
key
value
output-size
13961
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) FlatCCProof [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 47 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 461 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 5970 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 807 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(0(0(1(x1)))) 1(0(0(1(x1)))) -> 0(1(0(0(x1)))) Q is empty. ---------------------------------------- (1) FlatCCProof (EQUIVALENT) We used flat context closure [ROOTLAB] As Q is empty the flat context closure was sound AND complete. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(0(0(1(x1)))) 0(1(0(0(1(x1))))) -> 0(0(1(0(0(x1))))) 1(1(0(0(1(x1))))) -> 1(0(1(0(0(x1))))) Q is empty. ---------------------------------------- (3) 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. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0_{0_1}(0_{0_1}(0_{0_1}(0_{0_1}(x1)))) -> 0_{0_1}(0_{0_1}(0_{1_1}(1_{0_1}(x1)))) 0_{0_1}(0_{0_1}(0_{0_1}(0_{1_1}(x1)))) -> 0_{0_1}(0_{0_1}(0_{1_1}(1_{1_1}(x1)))) 0_{1_1}(1_{0_1}(0_{0_1}(0_{1_1}(1_{0_1}(x1))))) -> 0_{0_1}(0_{1_1}(1_{0_1}(0_{0_1}(0_{0_1}(x1))))) 0_{1_1}(1_{0_1}(0_{0_1}(0_{1_1}(1_{1_1}(x1))))) -> 0_{0_1}(0_{1_1}(1_{0_1}(0_{0_1}(0_{1_1}(x1))))) 1_{1_1}(1_{0_1}(0_{0_1}(0_{1_1}(1_{0_1}(x1))))) -> 1_{0_1}(0_{1_1}(1_{0_1}(0_{0_1}(0_{0_1}(x1))))) 1_{1_1}(1_{0_1}(0_{0_1}(0_{1_1}(1_{1_1}(x1))))) -> 1_{0_1}(0_{1_1}(1_{0_1}(0_{0_1}(0_{1_1}(x1))))) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 0_{0_1}^1(0_{0_1}(0_{0_1}(0_{0_1}(x1)))) -> 0_{0_1}^1(0_{0_1}(0_{1_1}(1_{0_1}(x1)))) 0_{0_1}^1(0_{0_1}(0_{0_1}(0_{0_1}(x1)))) -> 0_{0_1}^1(0_{1_1}(1_{0_1}(x1))) 0_{0_1}^1(0_{0_1}(0_{0_1}(0_{0_1}(x1)))) -> 0_{1_1}^1(1_{0_1}(x1)) 0_{0_1}^1(0_{0_1}(0_{0_1}(0_{1_1}(x1)))) -> 0_{0_1}^1(0_{0_1}(0_{1_1}(1_{1_1}(x1))))
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