details

property	value
status	complete
benchmark	#3.8a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n087.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	1.08198714256 seconds
cpu usage	2.978447063
max memory	4.17472512E8

stage attributes

key	value
output-size	4510
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: 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))) log(s(0())) -> 0() log(s(s(x))) -> s(log(s(quot(x,s(s(0())))))) Proof: DP Processor: DPs: minus#(x,s(y)) -> minus#(x,y) minus#(x,s(y)) -> pred#(minus(x,y)) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) log#(s(s(x))) -> quot#(x,s(s(0()))) log#(s(s(x))) -> log#(s(quot(x,s(s(0()))))) TRS: 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))) log(s(0())) -> 0() log(s(s(x))) -> s(log(s(quot(x,s(s(0())))))) TDG Processor: DPs: minus#(x,s(y)) -> minus#(x,y) minus#(x,s(y)) -> pred#(minus(x,y)) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) log#(s(s(x))) -> quot#(x,s(s(0()))) log#(s(s(x))) -> log#(s(quot(x,s(s(0()))))) TRS: 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))) log(s(0())) -> 0() log(s(s(x))) -> s(log(s(quot(x,s(s(0())))))) graph: log#(s(s(x))) -> log#(s(quot(x,s(s(0()))))) -> log#(s(s(x))) -> log#(s(quot(x,s(s(0()))))) log#(s(s(x))) -> log#(s(quot(x,s(s(0()))))) -> log#(s(s(x))) -> quot#(x,s(s(0()))) log#(s(s(x))) -> quot#(x,s(s(0()))) -> quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) log#(s(s(x))) -> quot#(x,s(s(0()))) -> quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) -> quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) -> quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> pred#(minus(x,y)) quot#(s(x),s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> minus#(x,y) minus#(x,s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> pred#(minus(x,y)) minus#(x,s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> minus#(x,y) SCC Processor: #sccs: 3 #rules: 3 #arcs: 10/36 DPs: log#(s(s(x))) -> log#(s(quot(x,s(s(0()))))) TRS: 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))) log(s(0())) -> 0() log(s(s(x))) -> s(log(s(quot(x,s(s(0())))))) Subterm Criterion Processor: simple projection: pi(s) = [0,0] pi(pred) = 0 pi(minus) = 0 pi(quot) = 0 pi(log#) = [0,0] problem: DPs: TRS: 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))) log(s(0())) -> 0() popout

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

job log

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actions

all output return to TRS Standard