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	n109.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	ttt2-1.17+nonreach
configuration	ttt2-1.17+nonreach
runtime (wallclock)	26.4357590675 seconds
cpu usage	104.240326612
max memory	6.56543744E8

stage attributes

key	value
output-size	9240
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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 popout

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job log

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actions

all output return to TRS Stand 20472