details

property	value
status	complete
benchmark	increase3.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.53469 seconds
cpu usage	6.44164
user time	6.11215
system time	0.329488
max virtual memory	1.8677612E7
max residence set size	366380.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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, 376 ms] (6) IDP (7) IDependencyGraphProof [EQUIVALENT, 0 ms] (8) IDP (9) IDPNonInfProof [SOUND, 59 ms] (10) IDP (11) IDependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (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 + ~ Add: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean -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(x, y, z) -> Cond_f(x > y && x > z, x, y, z) Cond_f(TRUE, x, y, z) -> f(x, y + 1, z) f(x, y, z) -> Cond_f1(x > y && x > z, x, y, z) Cond_f1(TRUE, x, y, z) -> f(x, y, z + 1) The set Q consists of the following terms: f(x0, x1, x2) Cond_f(TRUE, x0, x1, x2) Cond_f1(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 + ~ Add: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean -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(x, y, z) -> Cond_f(x > y && x > z, x, y, z) Cond_f(TRUE, x, y, z) -> f(x, y + 1, z) popout

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

job log

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actions

all output return to Integer TRS Innermost