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TRS Standard pair #487067684
details
property
value
status
complete
benchmark
2.46.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
SK90
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.069195 seconds
cpu usage
0.076625
user time
0.037939
system time
0.038686
max virtual memory
113188.0
max residence set size
5504.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR v_NonEmpty:S x:S) (RULES a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) ) Problem 1: Dependency Pairs Processor: -> Pairs: A(a(x:S)) -> B(b(x:S)) A(a(x:S)) -> B(x:S) B(b(a(x:S))) -> A(b(b(x:S))) B(b(a(x:S))) -> B(b(x:S)) B(b(a(x:S))) -> B(x:S) -> Rules: a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) Problem 1: SCC Processor: -> Pairs: A(a(x:S)) -> B(b(x:S)) A(a(x:S)) -> B(x:S) B(b(a(x:S))) -> A(b(b(x:S))) B(b(a(x:S))) -> B(b(x:S)) B(b(a(x:S))) -> B(x:S) -> Rules: a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: A(a(x:S)) -> B(b(x:S)) A(a(x:S)) -> B(x:S) B(b(a(x:S))) -> A(b(b(x:S))) B(b(a(x:S))) -> B(b(x:S)) B(b(a(x:S))) -> B(x:S) ->->-> Rules: a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) Problem 1: Reduction Pair Processor: -> Pairs: A(a(x:S)) -> B(b(x:S)) A(a(x:S)) -> B(x:S) B(b(a(x:S))) -> A(b(b(x:S))) B(b(a(x:S))) -> B(b(x:S)) B(b(a(x:S))) -> B(x:S) -> Rules: a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) -> Usable rules: a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [a](X) = 2.X + 1 [b](X) = 2.X [A](X) = 2.X + 2 [B](X) = 2.X + 2 Problem 1: SCC Processor: -> Pairs: A(a(x:S)) -> B(x:S) B(b(a(x:S))) -> A(b(b(x:S))) B(b(a(x:S))) -> B(b(x:S)) B(b(a(x:S))) -> B(x:S) -> Rules: a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: A(a(x:S)) -> B(x:S) B(b(a(x:S))) -> A(b(b(x:S))) B(b(a(x:S))) -> B(b(x:S)) B(b(a(x:S))) -> B(x:S) ->->-> Rules: a(a(x:S)) -> b(b(x:S)) b(b(a(x:S))) -> a(b(b(x:S))) Problem 1:
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