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Integer_TRS_Innermost 2019-03-21 04.53 pair #429995146
details
property
value
status
complete
benchmark
unsatCond1.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n043.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.80941 seconds
cpu usage
3.6913
user time
3.50916
system time
0.182147
max virtual memory
1.8607704E7
max residence set size
239088.0
stage attributes
key
value
starexec-result
YES
output
3.63/1.77 YES 3.63/1.78 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 3.63/1.78 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.63/1.78 3.63/1.78 3.63/1.78 Termination of the given ITRS could be proven: 3.63/1.78 3.63/1.78 (0) ITRS 3.63/1.78 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.63/1.78 (2) IDP 3.63/1.78 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.63/1.78 (4) IDP 3.63/1.78 (5) IDPNonInfProof [SOUND, 100 ms] 3.63/1.78 (6) IDP 3.63/1.78 (7) PisEmptyProof [EQUIVALENT, 0 ms] 3.63/1.78 (8) YES 3.63/1.78 3.63/1.78 3.63/1.78 ---------------------------------------- 3.63/1.78 3.63/1.78 (0) 3.63/1.78 Obligation: 3.63/1.78 ITRS problem: 3.63/1.78 3.63/1.78 The following function symbols are pre-defined: 3.63/1.78 <<< 3.63/1.78 & ~ Bwand: (Integer, Integer) -> Integer 3.63/1.78 >= ~ Ge: (Integer, Integer) -> Boolean 3.63/1.78 | ~ Bwor: (Integer, Integer) -> Integer 3.63/1.78 / ~ Div: (Integer, Integer) -> Integer 3.63/1.78 != ~ Neq: (Integer, Integer) -> Boolean 3.63/1.78 && ~ Land: (Boolean, Boolean) -> Boolean 3.63/1.78 ! ~ Lnot: (Boolean) -> Boolean 3.63/1.78 = ~ Eq: (Integer, Integer) -> Boolean 3.63/1.78 <= ~ Le: (Integer, Integer) -> Boolean 3.63/1.78 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.63/1.78 % ~ Mod: (Integer, Integer) -> Integer 3.63/1.78 > ~ Gt: (Integer, Integer) -> Boolean 3.63/1.78 + ~ Add: (Integer, Integer) -> Integer 3.63/1.78 -1 ~ UnaryMinus: (Integer) -> Integer 3.63/1.78 < ~ Lt: (Integer, Integer) -> Boolean 3.63/1.78 || ~ Lor: (Boolean, Boolean) -> Boolean 3.63/1.78 - ~ Sub: (Integer, Integer) -> Integer 3.63/1.78 ~ ~ Bwnot: (Integer) -> Integer 3.63/1.78 * ~ Mul: (Integer, Integer) -> Integer 3.63/1.78 >>> 3.63/1.78 3.63/1.78 The TRS R consists of the following rules: 3.63/1.78 f(x) -> Cond_f(x > x, x) 3.63/1.78 Cond_f(TRUE, x) -> f(x) 3.63/1.78 The set Q consists of the following terms: 3.63/1.78 f(x0) 3.63/1.78 Cond_f(TRUE, x0) 3.63/1.78 3.63/1.78 ---------------------------------------- 3.63/1.78 3.63/1.78 (1) ITRStoIDPProof (EQUIVALENT) 3.63/1.78 Added dependency pairs 3.63/1.78 ---------------------------------------- 3.63/1.78 3.63/1.78 (2) 3.63/1.78 Obligation: 3.63/1.78 IDP problem: 3.63/1.78 The following function symbols are pre-defined: 3.63/1.78 <<< 3.63/1.78 & ~ Bwand: (Integer, Integer) -> Integer 3.63/1.78 >= ~ Ge: (Integer, Integer) -> Boolean 3.63/1.78 | ~ Bwor: (Integer, Integer) -> Integer 3.63/1.78 / ~ Div: (Integer, Integer) -> Integer 3.63/1.78 != ~ Neq: (Integer, Integer) -> Boolean 3.63/1.78 && ~ Land: (Boolean, Boolean) -> Boolean 3.63/1.78 ! ~ Lnot: (Boolean) -> Boolean 3.63/1.78 = ~ Eq: (Integer, Integer) -> Boolean 3.63/1.78 <= ~ Le: (Integer, Integer) -> Boolean 3.63/1.78 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.63/1.78 % ~ Mod: (Integer, Integer) -> Integer 3.63/1.78 > ~ Gt: (Integer, Integer) -> Boolean 3.63/1.78 + ~ Add: (Integer, Integer) -> Integer 3.63/1.78 -1 ~ UnaryMinus: (Integer) -> Integer 3.63/1.78 < ~ Lt: (Integer, Integer) -> Boolean 3.63/1.78 || ~ Lor: (Boolean, Boolean) -> Boolean 3.63/1.78 - ~ Sub: (Integer, Integer) -> Integer 3.63/1.78 ~ ~ Bwnot: (Integer) -> Integer 3.63/1.78 * ~ Mul: (Integer, Integer) -> Integer 3.63/1.78 >>> 3.63/1.78 3.63/1.78 3.63/1.78 The following domains are used: 3.63/1.78 Integer 3.63/1.78 3.63/1.78 The ITRS R consists of the following rules: 3.63/1.78 f(x) -> Cond_f(x > x, x) 3.63/1.78 Cond_f(TRUE, x) -> f(x) 3.63/1.78 3.63/1.78 The integer pair graph contains the following rules and edges: 3.63/1.78 (0): F(x[0]) -> COND_F(x[0] > x[0], x[0]) 3.63/1.78 (1): COND_F(TRUE, x[1]) -> F(x[1]) 3.63/1.78 3.63/1.78 (0) -> (1), if (x[0] > x[0] & x[0] ->^* x[1]) 3.63/1.78 (1) -> (0), if (x[1] ->^* x[0])
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