details

property	value
status	complete
benchmark	terminate.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.09364 seconds
cpu usage	5.04777
user time	4.82908
system time	0.218692
max virtual memory	1.8607728E7
max residence set size	287264.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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, 291 ms] (6) IDP (7) IDependencyGraphProof [EQUIVALENT, 0 ms] (8) 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: eval(i, j, k) -> Cond_eval(i <= 100 && j <= k, i, j, k) Cond_eval(TRUE, i, j, k) -> eval(j, i + 1, k - 1) The set Q consists of the following terms: eval(x0, x1, x2) Cond_eval(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: eval(i, j, k) -> Cond_eval(i <= 100 && j <= k, i, j, k) Cond_eval(TRUE, i, j, k) -> eval(j, i + 1, k - 1) The integer pair graph contains the following rules and edges: (0): EVAL(i[0], j[0], k[0]) -> COND_EVAL(i[0] <= 100 && j[0] <= k[0], i[0], j[0], k[0]) (1): COND_EVAL(TRUE, i[1], j[1], k[1]) -> EVAL(j[1], i[1] + 1, k[1] - 1) (0) -> (1), if (i[0] <= 100 && j[0] <= k[0] & i[0] ->^* i[1] & j[0] ->^* j[1] & k[0] ->^* k[1]) (1) -> (0), if (j[1] ->^* i[0] & i[1] + 1 ->^* j[0] & k[1] - 1 ->^* k[0]) popout

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all output return to Integer TRS Innermost