details

property	value
status	complete
benchmark	#4.26.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	0.0957109 seconds
cpu usage	0.076161
user time	0.038857
system time	0.037304
max virtual memory	113188.0
max residence set size	5384.0

stage attributes

key	value
starexec-result	YES

output

YES Problem 1: (VAR v_NonEmpty:S x:S y:S) (RULES if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S)) if(ttrue,x:S,y:S) -> 0 le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S ) (STRATEGY INNERMOST) Problem 1: Dependency Pairs Processor: -> Pairs: IF(ffalse,x:S,y:S) -> MINUS(p(x:S),y:S) IF(ffalse,x:S,y:S) -> P(x:S) LE(s(x:S),s(y:S)) -> LE(x:S,y:S) MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S) MINUS(x:S,y:S) -> LE(x:S,y:S) -> Rules: if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S)) if(ttrue,x:S,y:S) -> 0 le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S Problem 1: SCC Processor: -> Pairs: IF(ffalse,x:S,y:S) -> MINUS(p(x:S),y:S) IF(ffalse,x:S,y:S) -> P(x:S) LE(s(x:S),s(y:S)) -> LE(x:S,y:S) MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S) MINUS(x:S,y:S) -> LE(x:S,y:S) -> Rules: if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S)) if(ttrue,x:S,y:S) -> 0 le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: LE(s(x:S),s(y:S)) -> LE(x:S,y:S) ->->-> Rules: if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S)) if(ttrue,x:S,y:S) -> 0 le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S ->->Cycle: ->->-> Pairs: IF(ffalse,x:S,y:S) -> MINUS(p(x:S),y:S) MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S) ->->-> Rules: if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S)) if(ttrue,x:S,y:S) -> 0 le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs: LE(s(x:S),s(y:S)) -> LE(x:S,y:S) -> Rules: if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S)) if(ttrue,x:S,y:S) -> 0 le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S ->Projection: pi(LE) = 1 popout

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

job log

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actions

all output return to TRS Innermost