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TRS Stand 20472 pair #381711920
details
property
value
status
complete
benchmark
t007.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n053.star.cs.uiowa.edu
space
HirokawaMiddeldorp_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.79921007156 seconds
cpu usage
4.343007584
max memory
3.35908864E8
stage attributes
key
value
output-size
3236
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) FlatCCProof [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4)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) QTRSRRRProof [EQUIVALENT, 4 ms] (6) QTRS (7) RisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a) -> f(b) g(b) -> g(a) f(x) -> g(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(a) -> f(b) g(b) -> g(a) f(f(x)) -> f(g(x)) g(f(x)) -> g(g(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_{a}(a) -> f_{b}(b) g_{b}(b) -> g_{a}(a) f_{f_1}(f_{f_1}(x)) -> f_{g_1}(g_{f_1}(x)) f_{f_1}(f_{a}(x)) -> f_{g_1}(g_{a}(x)) f_{f_1}(f_{b}(x)) -> f_{g_1}(g_{b}(x)) f_{f_1}(f_{g_1}(x)) -> f_{g_1}(g_{g_1}(x)) g_{f_1}(f_{f_1}(x)) -> g_{g_1}(g_{f_1}(x)) g_{f_1}(f_{a}(x)) -> g_{g_1}(g_{a}(x)) g_{f_1}(f_{b}(x)) -> g_{g_1}(g_{b}(x)) g_{f_1}(f_{g_1}(x)) -> g_{g_1}(g_{g_1}(x)) Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a) = 0 POL(b) = 1 POL(f_{a}(x_1)) = 2 + x_1 POL(f_{b}(x_1)) = x_1 POL(f_{f_1}(x_1)) = 1 + x_1 POL(f_{g_1}(x_1)) = x_1 POL(g_{a}(x_1)) = x_1 POL(g_{b}(x_1)) = x_1 POL(g_{f_1}(x_1)) = 1 + x_1
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