details

property	value
status	complete
benchmark	jw4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n102.star.cs.uiowa.edu
space	Waldmann_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	24.0835721493 seconds
cpu usage	92.501920137
max memory	5.541347328E9

stage attributes

key	value
output-size	13471
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 26 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 1990 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 519 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 10 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(b(b(x1)))) -> b(b(b(a(b(x1))))) b(a(a(b(b(x1))))) -> b(a(b(b(a(a(b(x1))))))) b(a(a(a(b(b(x1)))))) -> b(a(a(b(b(a(a(a(b(x1))))))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(b(x1)))) -> b(a(b(b(b(x1))))) b(b(a(a(b(x1))))) -> b(a(a(b(b(a(b(x1))))))) b(b(a(a(a(b(x1)))))) -> b(a(a(a(b(b(a(a(b(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: b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_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: B_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(b_{b_1}(x1))) B_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(x1)) B_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> B_{B_1}(b_{b_1}(b_{a_1}(x1))) B_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> B_{B_1}(b_{a_1}(x1)) B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> B_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) popout

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all output return to SRS Stand 10685