details

property	value
status	complete
benchmark	ExSec11_1_Luc02a_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.28591 seconds
cpu usage	8.85409
user time	8.53437
system time	0.319723
max virtual memory	1.8501636E7
max residence set size	632240.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 50 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 326 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 1 ms] (12) 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)) a__half(0) -> 0 a__half(s(0)) -> 0 a__half(s(s(X))) -> s(a__half(mark(X))) a__half(dbl(X)) -> mark(X) 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(half(X)) -> a__half(mark(X)) 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) a__half(X) -> half(X) The set Q consists of the following terms: a__terms(x0) mark(terms(x0)) mark(sqr(x0)) mark(add(x0, x1)) mark(dbl(x0)) mark(first(x0, x1)) mark(half(x0)) mark(cons(x0, x1)) mark(recip(x0)) mark(s(x0)) mark(0) mark(nil) a__sqr(x0) a__add(x0, x1) a__dbl(x0) a__first(x0, x1) a__half(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__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) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Innermost