details

property	value
status	complete
benchmark	Ex2_Luc02a_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	16.4191 seconds
cpu usage	8.95238
user time	8.50788
system time	0.444499
max virtual memory	1.8564796E7
max residence set size	439140.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 42 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 373 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N))) a__sqr(0) -> 0 a__sqr(s(X)) -> s(a__add(a__sqr(mark(X)), a__dbl(mark(X)))) a__dbl(0) -> 0 a__dbl(s(X)) -> s(s(a__dbl(mark(X)))) a__add(0, X) -> mark(X) a__add(s(X), Y) -> s(a__add(mark(X), mark(Y))) a__first(0, X) -> nil a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) mark(terms(X)) -> a__terms(mark(X)) mark(sqr(X)) -> a__sqr(mark(X)) mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) mark(dbl(X)) -> a__dbl(mark(X)) mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(recip(X)) -> recip(mark(X)) mark(s(X)) -> s(mark(X)) mark(0) -> 0 mark(nil) -> nil a__terms(X) -> terms(X) a__sqr(X) -> sqr(X) a__add(X1, X2) -> add(X1, X2) a__dbl(X) -> dbl(X) a__first(X1, X2) -> first(X1, X2) Q is empty. ---------------------------------------- (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__TERMS(N) -> A__SQR(mark(N)) A__TERMS(N) -> MARK(N) A__SQR(s(X)) -> A__ADD(a__sqr(mark(X)), a__dbl(mark(X))) A__SQR(s(X)) -> A__SQR(mark(X)) A__SQR(s(X)) -> MARK(X) A__SQR(s(X)) -> A__DBL(mark(X)) A__DBL(s(X)) -> A__DBL(mark(X)) A__DBL(s(X)) -> MARK(X) A__ADD(0, X) -> MARK(X) A__ADD(s(X), Y) -> A__ADD(mark(X), mark(Y)) A__ADD(s(X), Y) -> MARK(X) A__ADD(s(X), Y) -> MARK(Y) A__FIRST(s(X), cons(Y, Z)) -> MARK(Y) MARK(terms(X)) -> A__TERMS(mark(X)) MARK(terms(X)) -> MARK(X) MARK(sqr(X)) -> A__SQR(mark(X)) MARK(sqr(X)) -> MARK(X) MARK(add(X1, X2)) -> A__ADD(mark(X1), mark(X2)) MARK(add(X1, X2)) -> MARK(X1) MARK(add(X1, X2)) -> MARK(X2) MARK(dbl(X)) -> A__DBL(mark(X)) MARK(dbl(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(cons(X1, X2)) -> MARK(X1) MARK(recip(X)) -> MARK(X) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N))) a__sqr(0) -> 0 a__sqr(s(X)) -> s(a__add(a__sqr(mark(X)), a__dbl(mark(X)))) a__dbl(0) -> 0 popout

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job log

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actions

all output return to TRS Standard