details

property	value
status	complete
benchmark	logarquot.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n189.star.cs.uiowa.edu
space	Rubio_04

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	0.0588819 seconds
cpu usage	0.057438
user time	0.021887
system time	0.035551
max virtual memory	113188.0
max residence set size	5568.0

stage attributes

key	value
starexec-result	YES

output

YES Problem 1: (VAR v_NonEmpty:S X:S Y:S) (RULES log(s(0)) -> 0 log(s(s(X:S))) -> s(log(s(quot(X:S,s(s(0)))))) min(s(X:S),s(Y:S)) -> min(X:S,Y:S) min(X:S,0) -> X:S quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(min(X:S,Y:S),s(Y:S))) ) Problem 1: Innermost Equivalent Processor: -> Rules: log(s(0)) -> 0 log(s(s(X:S))) -> s(log(s(quot(X:S,s(s(0)))))) min(s(X:S),s(Y:S)) -> min(X:S,Y:S) min(X:S,0) -> X:S quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(min(X: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: LOG(s(s(X:S))) -> LOG(s(quot(X:S,s(s(0))))) LOG(s(s(X:S))) -> QUOT(X:S,s(s(0))) MIN(s(X:S),s(Y:S)) -> MIN(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> MIN(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> QUOT(min(X:S,Y:S),s(Y:S)) -> Rules: log(s(0)) -> 0 log(s(s(X:S))) -> s(log(s(quot(X:S,s(s(0)))))) min(s(X:S),s(Y:S)) -> min(X:S,Y:S) min(X:S,0) -> X:S quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(min(X:S,Y:S),s(Y:S))) Problem 1: SCC Processor: -> Pairs: LOG(s(s(X:S))) -> LOG(s(quot(X:S,s(s(0))))) LOG(s(s(X:S))) -> QUOT(X:S,s(s(0))) MIN(s(X:S),s(Y:S)) -> MIN(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> MIN(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> QUOT(min(X:S,Y:S),s(Y:S)) -> Rules: log(s(0)) -> 0 log(s(s(X:S))) -> s(log(s(quot(X:S,s(s(0)))))) min(s(X:S),s(Y:S)) -> min(X:S,Y:S) min(X:S,0) -> X:S quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(min(X:S,Y:S),s(Y:S))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: MIN(s(X:S),s(Y:S)) -> MIN(X:S,Y:S) ->->-> Rules: log(s(0)) -> 0 log(s(s(X:S))) -> s(log(s(quot(X:S,s(s(0)))))) min(s(X:S),s(Y:S)) -> min(X:S,Y:S) min(X:S,0) -> X:S quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(min(X:S,Y:S),s(Y:S))) ->->Cycle: ->->-> Pairs: QUOT(s(X:S),s(Y:S)) -> QUOT(min(X:S,Y:S),s(Y:S)) ->->-> Rules: log(s(0)) -> 0 log(s(s(X:S))) -> s(log(s(quot(X:S,s(s(0)))))) min(s(X:S),s(Y:S)) -> min(X:S,Y:S) min(X:S,0) -> X:S quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(min(X:S,Y:S),s(Y:S))) ->->Cycle: ->->-> Pairs: LOG(s(s(X:S))) -> LOG(s(quot(X:S,s(s(0))))) ->->-> Rules: log(s(0)) -> 0 log(s(s(X:S))) -> s(log(s(quot(X:S,s(s(0)))))) min(s(X:S),s(Y:S)) -> min(X:S,Y:S) min(X:S,0) -> X:S quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(min(X:S,Y:S),s(Y:S))) The problem is decomposed in 3 subproblems. Problem 1.1: Subterm Processor: -> Pairs: MIN(s(X:S),s(Y:S)) -> MIN(X:S,Y:S) popout

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job log

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actions

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