details

property	value
status	complete
benchmark	A06.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.72064 seconds
cpu usage	6.96445
user time	6.56687
system time	0.397585
max virtual memory	1.867816E7
max residence set size	387332.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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, 723 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 + z, x, y + 1, z) f(TRUE, x, y, z) -> f(x > y + 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: Integer The ITRS R consists of the following rules: f(TRUE, x, y, z) -> f(x > y + z, x, y + 1, z) f(TRUE, x, y, z) -> f(x > y + z, x, y, z + 1) The integer pair graph contains the following rules and edges: (0): F(TRUE, x[0], y[0], z[0]) -> F(x[0] > y[0] + z[0], x[0], y[0] + 1, z[0]) (1): F(TRUE, x[1], y[1], z[1]) -> F(x[1] > y[1] + z[1], x[1], y[1], z[1] + 1) (0) -> (0), if (x[0] > y[0] + z[0] & x[0] ->^* x[0]' & y[0] + 1 ->^* y[0]' & z[0] ->^* z[0]') (0) -> (1), if (x[0] > y[0] + z[0] & x[0] ->^* x[1] & y[0] + 1 ->^* y[1] & z[0] ->^* z[1]) (1) -> (0), if (x[1] > y[1] + z[1] & x[1] ->^* x[0] & y[1] ->^* y[0] & z[1] + 1 ->^* z[0]) popout

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job log

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actions

all output return to Integer TRS Innermost