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	n051.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.99479699135 seconds
cpu usage	7.656664963
max memory	4.69856256E8

stage attributes

key	value
output-size	12662
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 251 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)) 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) 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) A__HALF(s(s(X))) -> A__HALF(mark(X)) A__HALF(s(s(X))) -> MARK(X) A__HALF(dbl(X)) -> MARK(X) 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) popout

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

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all output return to TRS Stand 20472