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SRS Stand 10685 pair #381722152
details
property
value
status
complete
benchmark
3268.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n103.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
12.6407799721 seconds
cpu usage
46.71548644
max memory
4.362551296E9
stage attributes
key
value
output-size
92875
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) FlatCCProof [EQUIVALENT, 0 ms] (4) QTRS (5) RootLabelingProof [EQUIVALENT, 6 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 1483 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 5 ms] (10) QTRS (11) QTRSRRRProof [EQUIVALENT, 7 ms] (12) QTRS (13) QTRSRRRProof [EQUIVALENT, 0 ms] (14) QTRS (15) RisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(4(x1)) -> 3(1(1(2(2(4(x1)))))) 5(4(x1)) -> 4(2(3(1(1(1(x1)))))) 0(3(0(x1))) -> 2(1(1(0(2(0(x1)))))) 0(5(5(x1))) -> 1(0(1(3(4(2(x1)))))) 1(5(4(x1))) -> 0(2(5(2(0(4(x1)))))) 3(5(4(x1))) -> 4(1(3(4(2(3(x1)))))) 4(1(4(x1))) -> 3(3(2(2(3(1(x1)))))) 5(4(0(x1))) -> 2(4(0(4(4(0(x1)))))) 5(4(0(x1))) -> 5(1(5(2(1(0(x1)))))) 5(4(4(x1))) -> 4(1(1(3(2(4(x1)))))) 5(5(4(x1))) -> 3(4(4(1(2(2(x1)))))) 0(5(5(0(x1)))) -> 0(2(0(0(3(0(x1)))))) 0(5(5(4(x1)))) -> 0(1(3(4(3(4(x1)))))) 1(4(5(4(x1)))) -> 0(4(5(0(2(1(x1)))))) 1(4(5(5(x1)))) -> 0(0(1(3(4(1(x1)))))) 2(5(4(0(x1)))) -> 0(4(1(2(4(0(x1)))))) 4(3(0(5(x1)))) -> 3(3(2(3(5(5(x1)))))) 5(4(0(0(x1)))) -> 1(0(4(0(2(2(x1)))))) 5(4(0(2(x1)))) -> 3(0(4(5(0(2(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: 4(1(x1)) -> 4(2(2(1(1(3(x1)))))) 4(5(x1)) -> 1(1(1(3(2(4(x1)))))) 0(3(0(x1))) -> 0(2(0(1(1(2(x1)))))) 5(5(0(x1))) -> 2(4(3(1(0(1(x1)))))) 4(5(1(x1))) -> 4(0(2(5(2(0(x1)))))) 4(5(3(x1))) -> 3(2(4(3(1(4(x1)))))) 4(1(4(x1))) -> 1(3(2(2(3(3(x1)))))) 0(4(5(x1))) -> 0(4(4(0(4(2(x1)))))) 0(4(5(x1))) -> 0(1(2(5(1(5(x1)))))) 4(4(5(x1))) -> 4(2(3(1(1(4(x1)))))) 4(5(5(x1))) -> 2(2(1(4(4(3(x1)))))) 0(5(5(0(x1)))) -> 0(3(0(0(2(0(x1)))))) 4(5(5(0(x1)))) -> 4(3(4(3(1(0(x1)))))) 4(5(4(1(x1)))) -> 1(2(0(5(4(0(x1)))))) 5(5(4(1(x1)))) -> 1(4(3(1(0(0(x1)))))) 0(4(5(2(x1)))) -> 0(4(2(1(4(0(x1)))))) 5(0(3(4(x1)))) -> 5(5(3(2(3(3(x1)))))) 0(0(4(5(x1)))) -> 2(2(0(4(0(1(x1)))))) 2(0(4(5(x1)))) -> 2(0(5(4(0(3(x1)))))) Q is empty. ---------------------------------------- (3) FlatCCProof (EQUIVALENT) We used flat context closure [ROOTLAB] As Q is empty the flat context closure was sound AND complete.
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