details

property	value
status	complete
benchmark	006.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n072.star.cs.uiowa.edu
space	AotoYamada_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.97498 seconds
cpu usage	4.0205
user time	3.84324
system time	0.177263
max virtual memory	1.8343148E7
max residence set size	243996.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 4 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 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: app(app(mapbt, f), app(leaf, x)) -> app(leaf, app(f, x)) app(app(mapbt, f), app(app(app(branch, x), l), r)) -> app(app(app(branch, app(f, x)), app(app(mapbt, f), l)), app(app(mapbt, f), r)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(mapbt, f), app(leaf, x)) -> app(leaf, app(f, x)) app(app(mapbt, f), app(app(app(branch, x), l), r)) -> app(app(app(branch, app(f, x)), app(app(mapbt, f), l)), app(app(mapbt, f), r)) The set Q consists of the following terms: app(app(mapbt, x0), app(leaf, x1)) app(app(mapbt, x0), app(app(app(branch, x1), 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: APP(app(mapbt, f), app(leaf, x)) -> APP(leaf, app(f, x)) APP(app(mapbt, f), app(leaf, x)) -> APP(f, x) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(app(app(branch, app(f, x)), app(app(mapbt, f), l)), app(app(mapbt, f), r)) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(app(branch, app(f, x)), app(app(mapbt, f), l)) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(branch, app(f, x)) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(f, x) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(app(mapbt, f), l) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(app(mapbt, f), r) The TRS R consists of the following rules: app(app(mapbt, f), app(leaf, x)) -> app(leaf, app(f, x)) app(app(mapbt, f), app(app(app(branch, x), l), r)) -> app(app(app(branch, app(f, x)), app(app(mapbt, f), l)), app(app(mapbt, f), r)) The set Q consists of the following terms: app(app(mapbt, x0), app(leaf, x1)) app(app(mapbt, x0), app(app(app(branch, x1), x2), x3)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(f, x) APP(app(mapbt, f), app(leaf, x)) -> APP(f, x) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(app(mapbt, f), l) APP(app(mapbt, f), app(app(app(branch, x), l), r)) -> APP(app(mapbt, f), r) The TRS R consists of the following rules: popout

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

job log

popout

actions

all output return to TRS Standard