details

property	value
status	complete
benchmark	#3.26.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n007.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.07864999771 seconds
cpu usage	24.564069198
max memory	3.510423552E9

stage attributes

key	value
output-size	6721
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) FlatCCProof [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 34 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 168 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x) -> s(x) f(s(s(x))) -> s(f(f(x))) 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: f(f(x)) -> f(s(x)) s(f(x)) -> s(s(x)) f(f(s(s(x)))) -> f(s(f(f(x)))) s(f(s(s(x)))) -> s(s(f(f(x)))) 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: f_{f_1}(f_{f_1}(x)) -> f_{s_1}(s_{f_1}(x)) f_{f_1}(f_{s_1}(x)) -> f_{s_1}(s_{s_1}(x)) s_{f_1}(f_{f_1}(x)) -> s_{s_1}(s_{f_1}(x)) s_{f_1}(f_{s_1}(x)) -> s_{s_1}(s_{s_1}(x)) f_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x)))) f_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x)))) s_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x)))) s_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x)))) 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: F_{F_1}(f_{f_1}(x)) -> S_{F_1}(x) S_{F_1}(f_{f_1}(x)) -> S_{F_1}(x) F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> S_{F_1}(f_{f_1}(f_{f_1}(x))) F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(f_{f_1}(x)) F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(x) popout

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