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	n172.star.cs.uiowa.edu
space	COPS

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	0.0202119 seconds
cpu usage	0.020698
user time	0.010779
system time	0.009919
max virtual memory	113176.0
max residence set size	6852.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.01 YES 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 0.00/0.01 (VAR v_NonEmpty:S x:S y:S ys:S zs1:S zs2:S) 0.00/0.01 (RULES 0.00/0.01 le(0,y:S) -> ttrue 0.00/0.01 le(s(x:S),0) -> ffalse 0.00/0.01 le(s(x:S),s(y:S)) -> le(x:S,y:S) 0.00/0.01 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 0.00/0.01 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 0.00/0.01 split(x:S,nil) -> tp2(nil,nil) 0.00/0.01 ) 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 Valid CTRS Processor: 0.00/0.01 -> Rules: 0.00/0.01 le(0,y:S) -> ttrue 0.00/0.01 le(s(x:S),0) -> ffalse 0.00/0.01 le(s(x:S),s(y:S)) -> le(x:S,y:S) 0.00/0.01 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 0.00/0.01 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 0.00/0.01 split(x:S,nil) -> tp2(nil,nil) 0.00/0.01 -> The system is a deterministic 3-CTRS. 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 0.00/0.01 Dependency Pairs Processor: 0.00/0.01 0.00/0.01 Conditional Termination Problem 1: 0.00/0.01 -> Pairs: 0.00/0.01 LE(s(x:S),s(y:S)) -> LE(x:S,y:S) 0.00/0.01 -> QPairs: 0.00/0.01 Empty 0.00/0.01 -> Rules: 0.00/0.01 le(0,y:S) -> ttrue 0.00/0.01 le(s(x:S),0) -> ffalse 0.00/0.01 le(s(x:S),s(y:S)) -> le(x:S,y:S) 0.00/0.01 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 0.00/0.01 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 0.00/0.01 split(x:S,nil) -> tp2(nil,nil) 0.00/0.01 0.00/0.01 Conditional Termination Problem 2: 0.00/0.01 -> Pairs: 0.00/0.01 SPLIT(x:S,cons(y:S,ys:S)) -> LE(x:S,y:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S) 0.00/0.01 SPLIT(x:S,cons(y:S,ys:S)) -> SPLIT(x:S,ys:S) 0.00/0.01 -> QPairs: 0.00/0.01 LE(s(x:S),s(y:S)) -> LE(x:S,y:S) 0.00/0.01 -> Rules: 0.00/0.01 le(0,y:S) -> ttrue 0.00/0.01 le(s(x:S),0) -> ffalse 0.00/0.01 le(s(x:S),s(y:S)) -> le(x:S,y:S) 0.00/0.01 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 0.00/0.01 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 0.00/0.01 split(x:S,nil) -> tp2(nil,nil) 0.00/0.01 0.00/0.01 0.00/0.01 The problem is decomposed in 2 subproblems. 0.00/0.01 0.00/0.01 Problem 1.1: 0.00/0.01 0.00/0.01 SCC Processor: 0.00/0.01 -> Pairs: 0.00/0.01 LE(s(x:S),s(y:S)) -> LE(x:S,y:S) 0.00/0.01 -> QPairs: 0.00/0.01 Empty 0.00/0.01 -> Rules: 0.00/0.01 le(0,y:S) -> ttrue 0.00/0.01 le(s(x:S),0) -> ffalse 0.00/0.01 le(s(x:S),s(y:S)) -> le(x:S,y:S) 0.00/0.01 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 0.00/0.01 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 0.00/0.01 split(x:S,nil) -> tp2(nil,nil) 0.00/0.01 ->Strongly Connected Components: 0.00/0.01 ->->Cycle: 0.00/0.01 ->->-> Pairs: 0.00/0.01 LE(s(x:S),s(y:S)) -> LE(x:S,y:S) 0.00/0.01 -> QPairs: 0.00/0.01 Empty 0.00/0.01 ->->-> Rules: 0.00/0.01 le(0,y:S) -> ttrue 0.00/0.01 le(s(x:S),0) -> ffalse 0.00/0.01 le(s(x:S),s(y:S)) -> le(x:S,y:S) 0.00/0.01 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 0.00/0.01 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 0.00/0.01 split(x:S,nil) -> tp2(nil,nil) 0.00/0.01 0.00/0.01 Problem 1.1: 0.00/0.01 0.00/0.01 Conditional Subterm Processor: 0.00/0.01 -> Pairs: 0.00/0.01 LE(s(x:S),s(y:S)) -> LE(x:S,y:S) 0.00/0.01 -> QPairs: 0.00/0.01 Empty 0.00/0.01 -> Rules: 0.00/0.01 le(0,y:S) -> ttrue 0.00/0.01 le(s(x:S),0) -> ffalse 0.00/0.01 le(s(x:S),s(y:S)) -> le(x:S,y:S) 0.00/0.01 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 0.00/0.01 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 popout

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all output return to TRS_Conditional 2019-03-28 22.59