details

property	value
status	complete
benchmark	ExIntrod_GM01_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.70401 seconds
cpu usage	6.66049
user time	6.33448
system time	0.326011
max virtual memory	1.8545036E7
max residence set size	468480.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 54 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 19 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 17 ms] (8) QTRS (9) DependencyPairsProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) MRRProof [EQUIVALENT, 18 ms] (14) QDP (15) MRRProof [EQUIVALENT, 25 ms] (16) QDP (17) MRRProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 41 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__incr(nil) -> nil a__incr(cons(X, L)) -> cons(s(mark(X)), incr(L)) a__adx(nil) -> nil a__adx(cons(X, L)) -> a__incr(cons(mark(X), adx(L))) a__nats -> a__adx(a__zeros) a__zeros -> cons(0, zeros) a__head(cons(X, L)) -> mark(X) a__tail(cons(X, L)) -> mark(L) mark(incr(X)) -> a__incr(mark(X)) mark(adx(X)) -> a__adx(mark(X)) mark(nats) -> a__nats mark(zeros) -> a__zeros mark(head(X)) -> a__head(mark(X)) mark(tail(X)) -> a__tail(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) mark(0) -> 0 a__incr(X) -> incr(X) a__adx(X) -> adx(X) a__nats -> nats a__zeros -> zeros 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__adx(x_1)) = x_1 POL(a__head(x_1)) = 2*x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 1 POL(a__tail(x_1)) = 2*x_1 POL(a__zeros) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(head(x_1)) = 2*x_1 POL(incr(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nats) = 1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 2*x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__nats -> a__adx(a__zeros) ---------------------------------------- popout

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

job log

popout

actions

all output return to TRS Standard