details

property	value
status	complete
benchmark	214091.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n020.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.728525877 seconds
cpu usage	19.63130915
max memory	2.064863232E9

stage attributes

key	value
output-size	16303
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 188 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(x1))) -> 0(2(1(0(x1)))) 0(1(2(x1))) -> 1(0(2(3(x1)))) 0(1(2(x1))) -> 0(2(4(1(5(x1))))) 0(1(2(x1))) -> 0(3(2(1(0(x1))))) 0(1(2(x1))) -> 1(0(3(2(3(x1))))) 0(1(2(x1))) -> 0(1(3(4(2(3(x1)))))) 0(5(2(x1))) -> 0(2(4(5(3(x1))))) 0(5(2(x1))) -> 5(4(2(3(0(4(x1)))))) 2(0(1(x1))) -> 3(0(2(1(x1)))) 2(0(1(x1))) -> 0(2(1(1(4(x1))))) 2(0(1(x1))) -> 0(3(2(4(1(x1))))) 2(0(1(x1))) -> 3(0(2(1(4(x1))))) 2(0(1(x1))) -> 0(2(2(3(4(1(x1)))))) 2(0(1(x1))) -> 0(3(2(3(1(1(x1)))))) 2(0(1(x1))) -> 4(0(4(2(1(4(x1)))))) 2(5(1(x1))) -> 0(2(1(5(1(x1))))) 2(5(1(x1))) -> 1(4(5(4(2(x1))))) 2(5(1(x1))) -> 5(0(2(1(4(x1))))) 2(5(1(x1))) -> 5(2(1(4(1(x1))))) 2(5(1(x1))) -> 1(5(0(2(4(1(x1)))))) 2(5(1(x1))) -> 5(2(1(1(1(1(x1)))))) 0(1(2(1(x1)))) -> 3(1(4(0(2(1(x1)))))) 0(1(3(1(x1)))) -> 5(0(3(1(1(x1))))) 0(1(3(1(x1)))) -> 1(0(3(4(2(1(x1)))))) 0(1(5(1(x1)))) -> 5(0(3(1(1(x1))))) 0(2(1(2(x1)))) -> 0(2(2(1(5(x1))))) 0(2(5(1(x1)))) -> 1(1(5(0(2(1(x1)))))) 0(5(3(1(x1)))) -> 0(1(4(4(3(5(x1)))))) 0(5(5(2(x1)))) -> 5(4(2(3(5(0(x1)))))) 2(0(1(2(x1)))) -> 0(2(3(2(1(1(x1)))))) 2(0(1(2(x1)))) -> 4(0(2(1(1(2(x1)))))) 2(0(4(1(x1)))) -> 3(0(2(4(1(x1))))) 2(0(5(1(x1)))) -> 5(4(2(1(0(x1))))) 2(2(5(1(x1)))) -> 3(2(2(4(5(1(x1)))))) 2(4(0(1(x1)))) -> 1(0(2(4(4(x1))))) 2(4(0(1(x1)))) -> 3(0(0(2(4(1(x1)))))) 2(4(0(1(x1)))) -> 5(4(0(2(1(1(x1)))))) 2(5(2(1(x1)))) -> 1(5(2(2(3(1(x1)))))) 2(5(4(1(x1)))) -> 4(5(2(1(4(4(x1)))))) 2(5(5(1(x1)))) -> 1(5(4(2(4(5(x1)))))) 2(5(5(2(x1)))) -> 5(5(2(3(2(x1))))) 0(1(3(0(1(x1))))) -> 0(3(1(0(1(1(x1)))))) 0(2(4(3(1(x1))))) -> 1(3(4(2(3(0(x1)))))) 0(2(4(3(1(x1))))) -> 4(0(3(2(1(0(x1)))))) 0(2(5(3(1(x1))))) -> 5(0(2(3(5(1(x1)))))) 2(0(5(4(1(x1))))) -> 0(4(5(3(2(1(x1)))))) 2(2(0(1(2(x1))))) -> 2(4(0(2(2(1(x1)))))) 2(4(0(5(1(x1))))) -> 1(4(5(0(4(2(x1)))))) 2(4(2(3(1(x1))))) -> 4(2(2(3(3(1(x1)))))) 2(5(2(0(1(x1))))) -> 0(2(4(1(5(2(x1)))))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS Stand 10685