details

property	value
status	complete
benchmark	Ex1_Luc04b_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.37311 seconds
cpu usage	5.97341
user time	5.6733
system time	0.300114
max virtual memory	1.8477752E7
max residence set size	416136.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 69 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 18 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 5 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 0 ms] (10) QTRS (11) DependencyPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) MRRProof [EQUIVALENT, 25 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 31 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) QDP (23) UsableRulesProof [EQUIVALENT, 0 ms] (24) QDP (25) QDPSizeChangeProof [EQUIVALENT, 0 ms] (26) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) a__head(cons(X, XS)) -> mark(X) a__tail(cons(X, XS)) -> mark(XS) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(head(X)) -> a__head(mark(X)) mark(tail(X)) -> a__tail(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) a__nats -> nats a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) a__head(X) -> head(X) a__tail(X) -> tail(X) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__head(x_1)) = 1 + x_1 POL(a__incr(x_1)) = 2*x_1 POL(a__nats) = 0 POL(a__odds) = 0 POL(a__pairs) = 0 POL(a__tail(x_1)) = 1 + 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(head(x_1)) = 1 + x_1 POL(incr(x_1)) = 2*x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 0 POL(nil) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 1 + 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__head(cons(X, XS)) -> mark(X) a__tail(cons(X, XS)) -> mark(XS) mark(head(X)) -> a__head(mark(X)) mark(tail(X)) -> a__tail(mark(X)) popout

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all output return to TRS Standard