details

property	value
status	complete
benchmark	increase1.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n040.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.86329 seconds
cpu usage	3.99494
user time	3.80063
system time	0.194309
max virtual memory	1.854114E7
max residence set size	248444.0

stage attributes

key	value
starexec-result	YES

output

3.89/1.82 YES 3.89/1.83 proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs 3.89/1.83 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.89/1.83 3.89/1.83 3.89/1.83 Termination of the given ITRS could be proven: 3.89/1.83 3.89/1.83 (0) ITRS 3.89/1.83 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.89/1.83 (2) IDP 3.89/1.83 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.89/1.83 (4) IDP 3.89/1.83 (5) IDPNonInfProof [SOUND, 154 ms] 3.89/1.83 (6) IDP 3.89/1.83 (7) IDependencyGraphProof [EQUIVALENT, 0 ms] 3.89/1.83 (8) TRUE 3.89/1.83 3.89/1.83 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (0) 3.89/1.83 Obligation: 3.89/1.83 ITRS problem: 3.89/1.83 3.89/1.83 The following function symbols are pre-defined: 3.89/1.83 <<< 3.89/1.83 & ~ Bwand: (Integer, Integer) -> Integer 3.89/1.83 >= ~ Ge: (Integer, Integer) -> Boolean 3.89/1.83 | ~ Bwor: (Integer, Integer) -> Integer 3.89/1.83 / ~ Div: (Integer, Integer) -> Integer 3.89/1.83 != ~ Neq: (Integer, Integer) -> Boolean 3.89/1.83 && ~ Land: (Boolean, Boolean) -> Boolean 3.89/1.83 ! ~ Lnot: (Boolean) -> Boolean 3.89/1.83 = ~ Eq: (Integer, Integer) -> Boolean 3.89/1.83 <= ~ Le: (Integer, Integer) -> Boolean 3.89/1.83 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.89/1.83 % ~ Mod: (Integer, Integer) -> Integer 3.89/1.83 + ~ Add: (Integer, Integer) -> Integer 3.89/1.83 > ~ Gt: (Integer, Integer) -> Boolean 3.89/1.83 -1 ~ UnaryMinus: (Integer) -> Integer 3.89/1.83 < ~ Lt: (Integer, Integer) -> Boolean 3.89/1.83 || ~ Lor: (Boolean, Boolean) -> Boolean 3.89/1.83 - ~ Sub: (Integer, Integer) -> Integer 3.89/1.83 ~ ~ Bwnot: (Integer) -> Integer 3.89/1.83 * ~ Mul: (Integer, Integer) -> Integer 3.89/1.83 >>> 3.89/1.83 3.89/1.83 The TRS R consists of the following rules: 3.89/1.83 f(x, y) -> Cond_f(x > y, x, y) 3.89/1.83 Cond_f(TRUE, x, y) -> f(x, y + 1) 3.89/1.83 The set Q consists of the following terms: 3.89/1.83 f(x0, x1) 3.89/1.83 Cond_f(TRUE, x0, x1) 3.89/1.83 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (1) ITRStoIDPProof (EQUIVALENT) 3.89/1.83 Added dependency pairs 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (2) 3.89/1.83 Obligation: 3.89/1.83 IDP problem: 3.89/1.83 The following function symbols are pre-defined: 3.89/1.83 <<< 3.89/1.83 & ~ Bwand: (Integer, Integer) -> Integer 3.89/1.83 >= ~ Ge: (Integer, Integer) -> Boolean 3.89/1.83 | ~ Bwor: (Integer, Integer) -> Integer 3.89/1.83 / ~ Div: (Integer, Integer) -> Integer 3.89/1.83 != ~ Neq: (Integer, Integer) -> Boolean 3.89/1.83 && ~ Land: (Boolean, Boolean) -> Boolean 3.89/1.83 ! ~ Lnot: (Boolean) -> Boolean 3.89/1.83 = ~ Eq: (Integer, Integer) -> Boolean 3.89/1.83 <= ~ Le: (Integer, Integer) -> Boolean 3.89/1.83 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.89/1.83 % ~ Mod: (Integer, Integer) -> Integer 3.89/1.83 + ~ Add: (Integer, Integer) -> Integer 3.89/1.83 > ~ Gt: (Integer, Integer) -> Boolean 3.89/1.83 -1 ~ UnaryMinus: (Integer) -> Integer 3.89/1.83 < ~ Lt: (Integer, Integer) -> Boolean 3.89/1.83 || ~ Lor: (Boolean, Boolean) -> Boolean 3.89/1.83 - ~ Sub: (Integer, Integer) -> Integer 3.89/1.83 ~ ~ Bwnot: (Integer) -> Integer 3.89/1.83 * ~ Mul: (Integer, Integer) -> Integer 3.89/1.83 >>> 3.89/1.83 3.89/1.83 3.89/1.83 The following domains are used: 3.89/1.83 Integer 3.89/1.83 3.89/1.83 The ITRS R consists of the following rules: 3.89/1.83 f(x, y) -> Cond_f(x > y, x, y) 3.89/1.83 Cond_f(TRUE, x, y) -> f(x, y + 1) 3.89/1.83 3.89/1.83 The integer pair graph contains the following rules and edges: 3.89/1.83 (0): F(x[0], y[0]) -> COND_F(x[0] > y[0], x[0], y[0]) 3.89/1.83 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], y[1] + 1) 3.89/1.83 3.89/1.83 (0) -> (1), if (x[0] > y[0] & x[0] ->^* x[1] & y[0] ->^* y[1]) 3.89/1.83 (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0]) popout

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all output return to Integer_TRS_Innermost 2019-03-21 04.53