details

property	value
status	complete
benchmark	#4.30a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	20.2298 seconds
cpu usage	20.187
user time	17.2677
system time	2.9193
max virtual memory	597920.0
max residence set size	6100.0

stage attributes

key	value
starexec-result	YES

output

YES Problem 1: (VAR v_NonEmpty:S 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) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(s(x:S),s(y:S)),s(y:S))) ) Problem 1: Innermost Equivalent 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) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(s(x:S),s(y:S)),s(y:S))) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: LE(s(x:S),s(y:S)) -> LE(x:S,y:S) MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) QUOT(s(x:S),s(y:S)) -> MINUS(s(x:S),s(y:S)) QUOT(s(x:S),s(y:S)) -> QUOT(minus(s(x:S),s(y:S)),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) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(s(x:S),s(y:S)),s(y:S))) Problem 1: SCC Processor: -> Pairs: LE(s(x:S),s(y:S)) -> LE(x:S,y:S) MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) QUOT(s(x:S),s(y:S)) -> MINUS(s(x:S),s(y:S)) QUOT(s(x:S),s(y:S)) -> QUOT(minus(s(x:S),s(y:S)),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) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(s(x:S),s(y:S)),s(y:S))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: MINUS(s(x:S),s(y:S)) -> MINUS(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) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(s(x:S),s(y:S)),s(y:S))) ->->Cycle: ->->-> Pairs: QUOT(s(x:S),s(y:S)) -> QUOT(minus(s(x:S),s(y:S)),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) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(s(x:S),s(y:S)),s(y:S))) ->->Cycle: ->->-> Pairs: 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) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(s(x:S),s(y:S)),s(y:S))) The problem is decomposed in 3 subproblems. popout

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actions

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