details

property	value
status	complete
benchmark	Ex6_Luc98_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n146.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.20324 seconds
cpu usage	5.32305
user time	5.05612
system time	0.266931
max virtual memory	1.8411988E7
max residence set size	365488.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPQMonotonicMRRProof [EQUIVALENT, 41 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 4 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 1 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__first(0, X) -> nil a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) a__from(X) -> cons(mark(X), from(s(X))) mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) mark(from(X)) -> a__from(mark(X)) mark(0) -> 0 mark(nil) -> nil mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__first(X1, X2) -> first(X1, X2) a__from(X) -> from(X) The set Q consists of the following terms: a__from(x0) mark(first(x0, x1)) mark(from(x0)) mark(0) mark(nil) mark(s(x0)) mark(cons(x0, x1)) a__first(x0, x1) ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: A__FIRST(s(X), cons(Y, Z)) -> MARK(Y) A__FROM(X) -> MARK(X) MARK(first(X1, X2)) -> A__FIRST(mark(X1), mark(X2)) MARK(first(X1, X2)) -> MARK(X1) MARK(first(X1, X2)) -> MARK(X2) MARK(from(X)) -> A__FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__first(0, X) -> nil a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) a__from(X) -> cons(mark(X), from(s(X))) mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) mark(from(X)) -> a__from(mark(X)) mark(0) -> 0 mark(nil) -> nil mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__first(X1, X2) -> first(X1, X2) a__from(X) -> from(X) The set Q consists of the following terms: a__from(x0) mark(first(x0, x1)) mark(from(x0)) mark(0) mark(nil) mark(s(x0)) mark(cons(x0, x1)) a__first(x0, x1) We have to consider all minimal (P,Q,R)-chains. popout

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job log

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actions

all output return to TRS Innermost