details

property	value
status	complete
benchmark	Ex26_Luc03b_L.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n146.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.62675309181 seconds
cpu usage	3.497107127
max memory	2.21319168E8

stage attributes

key	value
output-size	2162
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 33 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: terms(N) -> cons(recip(sqr(N))) sqr(0) -> 0 sqr(s) -> s dbl(0) -> 0 dbl(s) -> s add(0, X) -> X add(s, Y) -> s first(0, X) -> nil first(s, cons(Y)) -> cons(Y) The set Q consists of the following terms: terms(x0) sqr(0) sqr(s) dbl(0) dbl(s) add(0, x0) add(s, x0) first(0, x0) first(s, cons(x0)) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:dbl_1 > first_2 > nil > add_2 > s > 0 > recip_1 > sqr_1 > terms_1 > cons_1 and weight map: 0=1 s=1 nil=2 terms_1=3 cons_1=1 recip_1=1 sqr_1=1 dbl_1=0 add_2=0 first_2=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: terms(N) -> cons(recip(sqr(N))) sqr(0) -> 0 sqr(s) -> s dbl(0) -> 0 dbl(s) -> s add(0, X) -> X add(s, Y) -> s first(0, X) -> nil first(s, cons(Y)) -> cons(Y) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. The set Q consists of the following terms: terms(x0) sqr(0) sqr(s) dbl(0) dbl(s) add(0, x0) add(s, x0) first(0, x0) popout

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

job log

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actions

all output return to TRS Innermost