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	n097.star.cs.uiowa.edu
space	COPS

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	0.0972390174866 seconds
cpu usage	0.023456145
max memory	2129920.0

stage attributes

key	value
output-size	6243
starexec-result	YES

output

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

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