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Integ TRS Inner 43557 pair #381740751
details
property
value
status
complete
benchmark
minsort.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n089.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.79973506927 seconds
cpu usage
6.932041546
max memory
3.78150912E8
stage attributes
key
value
output-size
29596
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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) AND (7) IDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) IDP (10) IDPtoQDPProof [SOUND, 35 ms] (11) QDP (12) QReductionProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) IDP (17) IDPtoQDPProof [SOUND, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 99 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) 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 > ~ 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: min(x, e) -> pair(x, e) min(x, ins(y, zs)) -> if_1(min(y, zs), x, y, zs) if_1(pair(m, zh), x, y, zs) -> Cond_if_1(m > x, pair(m, zh), x, y, zs) Cond_if_1(TRUE, pair(m, zh), x, y, zs) -> pair(x, ins(m, zh)) min(x, ins(y, zs)) -> if_2(min(y, zs), x, y, zs) if_2(pair(m, zh), x, y, zs) -> Cond_if_2(x >= m, pair(m, zh), x, y, zs) Cond_if_2(TRUE, pair(m, zh), x, y, zs) -> pair(m, ins(x, zh)) msort(e) -> nil msort(ins(x, ys)) -> if_3(min(x, ys), x, ys) if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) The set Q consists of the following terms: min(x0, e) min(x0, ins(x1, x2)) if_1(pair(x0, x1), x2, x3, x4) Cond_if_1(TRUE, pair(x0, x1), x2, x3, x4) if_2(pair(x0, x1), x2, x3, x4) Cond_if_2(TRUE, pair(x0, x1), x2, x3, x4) msort(e) msort(ins(x0, x1)) if_3(pair(x0, x1), x2, x3) ---------------------------------------- (1) ITRStoIDPProof (EQUIVALENT) Added dependency pairs ---------------------------------------- (2) Obligation: IDP problem:
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