details

property	value
status	complete
benchmark	wiehe06.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n100.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.60450410843 seconds
cpu usage	7.160399407
max memory	4.11394048E8

stage attributes

key	value
output-size	13558
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 23 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 51 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 48 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES (17) QDP (18) QDPOrderProof [EQUIVALENT, 0 ms] (19) QDP (20) PisEmptyProof [EQUIVALENT, 0 ms] (21) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))), x) times(x, 1) -> x times(x, 0) -> 0 plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) 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: TIMES(x, plus(y, 1)) -> PLUS(times(x, plus(y, times(1, 0))), x) TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, times(1, 0))) TIMES(x, plus(y, 1)) -> PLUS(y, times(1, 0)) TIMES(x, plus(y, 1)) -> TIMES(1, 0) PLUS(s(x), s(y)) -> PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) PLUS(s(x), s(y)) -> IF(gt(x, y), x, y) PLUS(s(x), s(y)) -> GT(x, y) PLUS(s(x), s(y)) -> IF(not(gt(x, y)), id(x), id(y)) PLUS(s(x), s(y)) -> NOT(gt(x, y)) PLUS(s(x), s(y)) -> ID(x) PLUS(s(x), s(y)) -> ID(y) PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x)) PLUS(s(x), x) -> IF(gt(x, x), id(x), id(x)) PLUS(s(x), x) -> GT(x, x) PLUS(s(x), x) -> ID(x) PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y))) PLUS(id(x), s(y)) -> IF(gt(s(y), y), y, s(y)) PLUS(id(x), s(y)) -> GT(s(y), y) NOT(x) -> IF(x, false, true) GT(s(x), s(y)) -> GT(x, y) The TRS R consists of the following rules: times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))), x) times(x, 1) -> x times(x, 0) -> 0 plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) popout

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

job log

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actions

all output return to TRS Stand 20472