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TRS Standard pair #487068069
details
property
value
status
complete
benchmark
2.44.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n074.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.0502291 seconds
cpu usage
0.0463
user time
0.023395
system time
0.022905
max virtual memory
113188.0
max residence set size
5496.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR v_NonEmpty:S x:S y:S z:S) (RULES =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v)) =(.(x:S,y:S),nil) -> ffalse =(nil,.(y:S,z:S)) -> ffalse =(nil,nil) -> ttrue del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S) f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S))) f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S)) ) Problem 1: Innermost Equivalent Processor: -> Rules: =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v)) =(.(x:S,y:S),nil) -> ffalse =(nil,.(y:S,z:S)) -> ffalse =(nil,nil) -> ttrue del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S) f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S))) f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S)) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: DEL(.(x:S,.(y:S,z:S))) -> =#(x:S,y:S) DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S) F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) -> Rules: =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v)) =(.(x:S,y:S),nil) -> ffalse =(nil,.(y:S,z:S)) -> ffalse =(nil,nil) -> ttrue del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S) f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S))) f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S)) Problem 1: SCC Processor: -> Pairs: DEL(.(x:S,.(y:S,z:S))) -> =#(x:S,y:S) DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S) F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) -> Rules: =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v)) =(.(x:S,y:S),nil) -> ffalse =(nil,.(y:S,z:S)) -> ffalse =(nil,nil) -> ttrue del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S) f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S))) f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S)) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S) F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) ->->-> Rules: =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v)) =(.(x:S,y:S),nil) -> ffalse =(nil,.(y:S,z:S)) -> ffalse =(nil,nil) -> ttrue del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S) f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S))) f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S)) Problem 1: Reduction Pairs Processor: -> Pairs: DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S) F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S)) -> Rules: =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v)) =(.(x:S,y:S),nil) -> ffalse =(nil,.(y:S,z:S)) -> ffalse =(nil,nil) -> ttrue del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S) f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S))) f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S)) -> Usable rules: =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v)) =(.(x:S,y:S),nil) -> ffalse =(nil,.(y:S,z:S)) -> ffalse =(nil,nil) -> ttrue ->Interpretation type: Linear ->Coefficients:
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