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Integer TRS Innermost pair #487100514
details
property
value
status
complete
benchmark
sqrt_patrs.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.02318 seconds
cpu usage
4.65333
user time
4.44464
system time
0.208693
max virtual memory
1.8607728E7
max residence set size
280560.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5) IDependencyGraphProof [EQUIVALENT, 0 ms] (6) IDP (7) IDPNonInfProof [SOUND, 196 ms] (8) AND (9) IDP (10) IDependencyGraphProof [EQUIVALENT, 0 ms] (11) TRUE (12) IDP (13) IDependencyGraphProof [EQUIVALENT, 0 ms] (14) 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: sqrt(x) -> f(x, 0, 1, 1) f(x, z, u, w) -> Cond_f(x >= w && u >= 0, x, z, u, w) Cond_f(TRUE, x, z, u, w) -> f(x, z + 1, u + 2, w + u + 2) f(x, z, u, w) -> Cond_f1(w > x, x, z, u, w) Cond_f1(TRUE, x, z, u, w) -> z The set Q consists of the following terms: sqrt(x0) f(x0, x1, x2, x3) Cond_f(TRUE, x0, x1, x2, x3) Cond_f1(TRUE, x0, x1, x2, x3) ---------------------------------------- (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
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