details

property	value
status	complete
benchmark	wiehe08.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n182.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.91864 seconds
cpu usage	6.86723
user time	6.59109
system time	0.276134
max virtual memory	1.85464E7
max residence set size	406516.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 86 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 19 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES (17) QDP (18) UsableRulesProof [EQUIVALENT, 0 ms] (19) QDP (20) QDPSizeChangeProof [EQUIVALENT, 0 ms] (21) YES (22) QDP (23) QDPOrderProof [EQUIVALENT, 0 ms] (24) QDP (25) PisEmptyProof [EQUIVALENT, 0 ms] (26) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(s(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) PLUS(s(x), s(y)) -> PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) PLUS(s(x), s(y)) -> IF(gt(x, y), x, y) PLUS(s(x), s(y)) -> GT(x, y) PLUS(s(x), s(y)) -> IF(not(gt(x, y)), id(x), id(y)) PLUS(s(x), s(y)) -> NOT(gt(x, y)) PLUS(s(x), s(y)) -> ID(x) PLUS(s(x), s(y)) -> ID(y) PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x)) PLUS(s(x), x) -> IF(gt(x, x), id(x), id(x)) PLUS(s(x), x) -> GT(x, x) PLUS(s(x), x) -> ID(x) PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y))) PLUS(id(x), s(y)) -> IF(gt(s(y), y), y, s(y)) PLUS(id(x), s(y)) -> GT(s(y), y) NOT(x) -> IF(x, false, true) GT(s(x), s(y)) -> GT(x, y) The TRS R consists of the following rules: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) popout

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

job log

popout

actions

all output return to TRS Standard