details

property	value
status	complete
benchmark	append.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	CiME_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.74902 seconds
cpu usage	3.66776
user time	3.50307
system time	0.164683
max virtual memory	1.8408912E7
max residence set size	228464.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [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: is_empty(nil) -> true is_empty(cons(x, l)) -> false hd(cons(x, l)) -> x tl(cons(x, l)) -> l append(l1, l2) -> ifappend(l1, l2, l1) ifappend(l1, l2, nil) -> l2 ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is hd(cons(x, l)) -> x tl(cons(x, l)) -> l ifappend(l1, l2, nil) -> l2 ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) append(l1, l2) -> ifappend(l1, l2, l1) The TRS R 2 is is_empty(nil) -> true is_empty(cons(x, l)) -> false The signature Sigma is {is_empty_1, true, false} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: is_empty(nil) -> true is_empty(cons(x, l)) -> false hd(cons(x, l)) -> x tl(cons(x, l)) -> l append(l1, l2) -> ifappend(l1, l2, l1) ifappend(l1, l2, nil) -> l2 ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) The set Q consists of the following terms: is_empty(nil) is_empty(cons(x0, x1)) hd(cons(x0, x1)) tl(cons(x0, x1)) append(x0, x1) ifappend(x0, x1, nil) ifappend(x0, x1, cons(x2, x3)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: APPEND(l1, l2) -> IFAPPEND(l1, l2, l1) IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2) The TRS R consists of the following rules: is_empty(nil) -> true is_empty(cons(x, l)) -> false hd(cons(x, l)) -> x tl(cons(x, l)) -> l append(l1, l2) -> ifappend(l1, l2, l1) ifappend(l1, l2, nil) -> l2 popout

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all output return to TRS Standard