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TRS Standard pair #487067388
details
property
value
status
complete
benchmark
enger-nonloop-while-lt.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n174.star.cs.uiowa.edu
space
EEG_IJCAR_12
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
11.839 seconds
cpu usage
36.7092
user time
35.7065
system time
1.00268
max virtual memory
1.8710724E7
max residence set size
3108812.0
stage attributes
key
value
starexec-result
NO
output
NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) MNOCProof [EQUIVALENT, 0 ms] (16) QDP (17) NonLoopProof [COMPLETE, 30 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: while(true, x, y) -> while(lt(x, y), s(x), s(y)) lt(s(x), s(y)) -> lt(x, y) lt(0, y) -> true Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is lt(s(x), s(y)) -> lt(x, y) lt(0, y) -> true The TRS R 2 is while(true, x, y) -> while(lt(x, y), s(x), s(y)) The signature Sigma is {while_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: while(true, x, y) -> while(lt(x, y), s(x), s(y)) lt(s(x), s(y)) -> lt(x, y) lt(0, y) -> true The set Q consists of the following terms: while(true, x0, x1) lt(s(x0), s(x1)) lt(0, x0) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: WHILE(true, x, y) -> WHILE(lt(x, y), s(x), s(y)) WHILE(true, x, y) -> LT(x, y) LT(s(x), s(y)) -> LT(x, y) The TRS R consists of the following rules: while(true, x, y) -> while(lt(x, y), s(x), s(y)) lt(s(x), s(y)) -> lt(x, y) lt(0, y) -> true The set Q consists of the following terms: while(true, x0, x1) lt(s(x0), s(x1)) lt(0, x0) We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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