details

property	value
status	complete
benchmark	#4.30a.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	Strategy_removed_AG01

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	76.0401 seconds
cpu usage	299.263
user time	296.48
system time	2.78347
max virtual memory	5393116.0
max residence set size	238148.0

stage attributes

key	value
starexec-result	YES

output

YES Problem: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(s(x),s(y)),s(y))) Proof: DP Processor: DPs: minus#(s(x),s(y)) -> minus#(x,y) le#(s(x),s(y)) -> le#(x,y) quot#(s(x),s(y)) -> minus#(s(x),s(y)) quot#(s(x),s(y)) -> quot#(minus(s(x),s(y)),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(s(x),s(y)),s(y))) TDG Processor: DPs: minus#(s(x),s(y)) -> minus#(x,y) le#(s(x),s(y)) -> le#(x,y) quot#(s(x),s(y)) -> minus#(s(x),s(y)) quot#(s(x),s(y)) -> quot#(minus(s(x),s(y)),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(s(x),s(y)),s(y))) graph: quot#(s(x),s(y)) -> quot#(minus(s(x),s(y)),s(y)) -> quot#(s(x),s(y)) -> quot#(minus(s(x),s(y)),s(y)) quot#(s(x),s(y)) -> quot#(minus(s(x),s(y)),s(y)) -> quot#(s(x),s(y)) -> minus#(s(x),s(y)) quot#(s(x),s(y)) -> minus#(s(x),s(y)) -> minus#(s(x),s(y)) -> minus#(x,y) le#(s(x),s(y)) -> le#(x,y) -> le#(s(x),s(y)) -> le#(x,y) minus#(s(x),s(y)) -> minus#(x,y) -> minus#(s(x),s(y)) -> minus#(x,y) SCC Processor: #sccs: 3 #rules: 3 #arcs: 5/16 DPs: le#(s(x),s(y)) -> le#(x,y) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(s(x),s(y)),s(y))) Subterm Criterion Processor: simple projection: pi(le#) = 0 problem: DPs: TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(s(x),s(y)),s(y))) Qed DPs: quot#(s(x),s(y)) -> quot#(minus(s(x),s(y)),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(s(x),s(y)),s(y))) Extended Uncurrying Processor: application symbol: le symbol table: quot# ==> quot{0,#}/2 quot ==> quot0/2 false ==> false0/0 true ==> true0/0 s ==> s0/1 s1/2 minus ==> minus0/2 0 ==> 00/0 01/1 popout

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