details

property	value
status	complete
benchmark	A08.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n051.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.46557211876 seconds
cpu usage	6.415701688
max memory	3.50588928E8

stage attributes

key	value
output-size	42522
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5) IDPNonInfProof [SOUND, 593 ms] (6) IDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: ITRS problem: The following function symbols are pre-defined: <<< & ~ Bwand: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer != ~ Neq: (Integer, Integer) -> Boolean && ~ Land: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean = ~ Eq: (Integer, Integer) -> Boolean <= ~ Le: (Integer, Integer) -> Boolean ^ ~ Bwxor: (Integer, Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean + ~ Add: (Integer, Integer) -> Integer -1 ~ UnaryMinus: (Integer) -> Integer < ~ Lt: (Integer, Integer) -> Boolean || ~ Lor: (Boolean, Boolean) -> Boolean - ~ Sub: (Integer, Integer) -> Integer ~ ~ Bwnot: (Integer) -> Integer * ~ Mul: (Integer, Integer) -> Integer >>> The TRS R consists of the following rules: f(TRUE, x, y, z) -> f(x > y && x > z, x, y + 1, z) f(TRUE, x, y, z) -> f(x > y && x > z, x, y, z + 1) The set Q consists of the following terms: f(TRUE, x0, x1, x2) ---------------------------------------- (1) ITRStoIDPProof (EQUIVALENT) Added dependency pairs ---------------------------------------- (2) Obligation: IDP problem: The following function symbols are pre-defined: <<< & ~ Bwand: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer != ~ Neq: (Integer, Integer) -> Boolean && ~ Land: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean = ~ Eq: (Integer, Integer) -> Boolean <= ~ Le: (Integer, Integer) -> Boolean ^ ~ Bwxor: (Integer, Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean + ~ Add: (Integer, Integer) -> Integer -1 ~ UnaryMinus: (Integer) -> Integer < ~ Lt: (Integer, Integer) -> Boolean || ~ Lor: (Boolean, Boolean) -> Boolean - ~ Sub: (Integer, Integer) -> Integer ~ ~ Bwnot: (Integer) -> Integer * ~ Mul: (Integer, Integer) -> Integer >>> The following domains are used: Boolean, Integer The ITRS R consists of the following rules: f(TRUE, x, y, z) -> f(x > y && x > z, x, y + 1, z) f(TRUE, x, y, z) -> f(x > y && x > z, x, y, z + 1) The integer pair graph contains the following rules and edges: popout

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

job log

popout

actions

all output return to Integ TRS Inner 43557