details

property	value
status	complete
benchmark	322.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	COPS

run statistics

property	value
solver	muterm 5.18
configuration	default
runtime (wallclock)	0.017450094223 seconds
cpu usage	0.015070526
max memory	1597440.0

stage attributes

key	value
output-size	5221
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR x y ys zs1 zs2) (RULES le(0,y) -> true le(s(x),0) -> false le(s(x),s(y)) -> le(x,y) split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> false split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> true split(x,nil) -> tp2(nil,nil) ) Problem 1: Valid CTRS Processor: -> Rules: le(0,y) -> true le(s(x),0) -> false le(s(x),s(y)) -> le(x,y) split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> false split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> true split(x,nil) -> tp2(nil,nil) -> The system is a deterministic 3-CTRS. Problem 1: Dependency Pairs Processor: Conditional Termination Problem 1: -> Pairs: LE(s(x),s(y)) -> LE(x,y) -> QPairs: Empty -> Rules: le(0,y) -> true le(s(x),0) -> false le(s(x),s(y)) -> le(x,y) split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> false split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> true split(x,nil) -> tp2(nil,nil) Conditional Termination Problem 2: -> Pairs: SPLIT(x,cons(y,ys)) -> LE(x,y) | split(x,ys) -> tp2(zs1,zs2) SPLIT(x,cons(y,ys)) -> SPLIT(x,ys) -> QPairs: LE(s(x),s(y)) -> LE(x,y) -> Rules: le(0,y) -> true le(s(x),0) -> false le(s(x),s(y)) -> le(x,y) split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> false split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> true split(x,nil) -> tp2(nil,nil) The problem is decomposed in 2 subproblems. Problem 1.1: SCC Processor: -> Pairs: LE(s(x),s(y)) -> LE(x,y) -> QPairs: Empty -> Rules: le(0,y) -> true le(s(x),0) -> false le(s(x),s(y)) -> le(x,y) split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> false split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> true split(x,nil) -> tp2(nil,nil) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: LE(s(x),s(y)) -> LE(x,y) ->->-> Rules: le(0,y) -> true le(s(x),0) -> false le(s(x),s(y)) -> le(x,y) split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> false split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) -> tp2(zs1,zs2), le(x,y) -> true split(x,nil) -> tp2(nil,nil) Problem 1.1: Conditional Subterm Processor: -> Pairs: LE(s(x),s(y)) -> LE(x,y) -> QPairs: Empty -> Rules: le(0,y) -> true popout

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job log

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actions

all output return to TRS Condi 20667