details

property	value
status	complete
benchmark	#3.2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n177.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.7331 seconds
cpu usage	3.91196
user time	3.73095
system time	0.181009
max virtual memory	1.8408912E7
max residence set size	247240.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 0 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pred(s(x)) -> x minus(x, 0) -> x minus(x, s(y)) -> pred(minus(x, y)) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) 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: pred(s(x)) -> x minus(x, 0) -> x minus(x, s(y)) -> pred(minus(x, y)) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) The set Q consists of the following terms: pred(s(x0)) minus(x0, 0) minus(x0, s(x1)) quot(0, s(x0)) quot(s(x0), s(x1)) ---------------------------------------- (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: MINUS(x, s(y)) -> PRED(minus(x, y)) MINUS(x, s(y)) -> MINUS(x, y) QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y)) QUOT(s(x), s(y)) -> MINUS(x, y) The TRS R consists of the following rules: pred(s(x)) -> x minus(x, 0) -> x minus(x, s(y)) -> pred(minus(x, y)) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) The set Q consists of the following terms: pred(s(x0)) popout

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

job log

popout

actions

all output return to TRS Standard