details

property	value
status	complete
benchmark	361.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	COPS

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	101.666694164 seconds
cpu usage	101.561683095
max memory	1.16129792E8

stage attributes

key	value
output-size	5109
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO Problem 1: (VAR v_NonEmpty:S q:S r:S x:S y:S) (RULES div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S ) Problem 1: Valid CTRS Processor: -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S -> The system is a deterministic 3-CTRS. Problem 1: Dependency Pairs Processor: Conditional Termination Problem 1: -> Pairs: GREATER(s(x:S),s(y:S)) -> GREATER(x:S,y:S) LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S) M(s(x:S),s(y:S)) -> M(x:S,y:S) -> QPairs: Empty -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S Conditional Termination Problem 2: -> Pairs: DIV(x:S,y:S) -> DIV(m(x:S,y:S),y:S) | leq(y:S,x:S) ->* ttrue DIV(x:S,y:S) -> GREATER(y:S,x:S) DIV(x:S,y:S) -> LEQ(y:S,x:S) DIV(x:S,y:S) -> M(x:S,y:S) | leq(y:S,x:S) ->* ttrue -> QPairs: GREATER(s(x:S),s(y:S)) -> GREATER(x:S,y:S) LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S) M(s(x:S),s(y:S)) -> M(x:S,y:S) -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S Problem 1: SCC Processor: -> Pairs: DIV(x:S,y:S) -> DIV(m(x:S,y:S),y:S) | leq(y:S,x:S) ->* ttrue DIV(x:S,y:S) -> GREATER(y:S,x:S) DIV(x:S,y:S) -> LEQ(y:S,x:S) DIV(x:S,y:S) -> M(x:S,y:S) | leq(y:S,x:S) ->* ttrue -> QPairs: GREATER(s(x:S),s(y:S)) -> GREATER(x:S,y:S) LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S) M(s(x:S),s(y:S)) -> M(x:S,y:S) -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 popout

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job log

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actions

all output return to TRS Conditional