details

property	value
status	complete
benchmark	Ex6_9_Luc02c_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.23471 seconds
cpu usage	8.92885
user time	8.58313
system time	0.345717
max virtual memory	1.8543288E7
max residence set size	495416.0

stage attributes

key	value
starexec-result	YES

output

YES 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 15 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 91 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 33 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__2nd(cons1(X, cons(Y, Z))) -> mark(Y) a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1))) a__from(X) -> cons(mark(X), from(s(X))) mark(2nd(X)) -> a__2nd(mark(X)) mark(from(X)) -> a__from(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2)) a__2nd(X) -> 2nd(X) a__from(X) -> from(X) The set Q consists of the following terms: a__from(x0) mark(2nd(x0)) mark(from(x0)) mark(cons(x0, x1)) mark(s(x0)) mark(cons1(x0, x1)) a__2nd(x0) ---------------------------------------- (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__2ND(cons1(X, cons(Y, Z))) -> MARK(Y) A__2ND(cons(X, X1)) -> A__2ND(cons1(mark(X), mark(X1))) A__2ND(cons(X, X1)) -> MARK(X) A__2ND(cons(X, X1)) -> MARK(X1) A__FROM(X) -> MARK(X) MARK(2nd(X)) -> A__2ND(mark(X)) MARK(2nd(X)) -> MARK(X) MARK(from(X)) -> A__FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) MARK(cons1(X1, X2)) -> MARK(X1) MARK(cons1(X1, X2)) -> MARK(X2) The TRS R consists of the following rules: a__2nd(cons1(X, cons(Y, Z))) -> mark(Y) a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1))) a__from(X) -> cons(mark(X), from(s(X))) mark(2nd(X)) -> a__2nd(mark(X)) mark(from(X)) -> a__from(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2)) a__2nd(X) -> 2nd(X) a__from(X) -> from(X) The set Q consists of the following terms: a__from(x0) mark(2nd(x0)) mark(from(x0)) popout

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actions

all output return to TRS Innermost