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TRS Stand 20472 pair #381710291
details
property
value
status
complete
benchmark
polycounter-5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
TCT_12
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.0218479633331 seconds
cpu usage
0.018353886
max memory
2031616.0
stage attributes
key
value
output-size
8025
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR x1 x2 x3 x4 x5) (RULES f(0,0,0,0,0) -> 0 f(0,0,0,0,s(x5)) -> f(x5,x5,x5,x5,x5) f(0,0,0,s(x4),x5) -> f(x4,x4,x4,x4,x5) f(0,0,s(x3),x4,x5) -> f(x3,x3,x3,x4,x5) f(0,s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5) f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5) ) Problem 1: Innermost Equivalent Processor: -> Rules: f(0,0,0,0,0) -> 0 f(0,0,0,0,s(x5)) -> f(x5,x5,x5,x5,x5) f(0,0,0,s(x4),x5) -> f(x4,x4,x4,x4,x5) f(0,0,s(x3),x4,x5) -> f(x3,x3,x3,x4,x5) f(0,s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5) f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: F(0,0,0,0,s(x5)) -> F(x5,x5,x5,x5,x5) F(0,0,0,s(x4),x5) -> F(x4,x4,x4,x4,x5) F(0,0,s(x3),x4,x5) -> F(x3,x3,x3,x4,x5) F(0,s(x2),x3,x4,x5) -> F(x2,x2,x3,x4,x5) F(s(x1),x2,x3,x4,x5) -> F(x1,x2,x3,x4,x5) -> Rules: f(0,0,0,0,0) -> 0 f(0,0,0,0,s(x5)) -> f(x5,x5,x5,x5,x5) f(0,0,0,s(x4),x5) -> f(x4,x4,x4,x4,x5) f(0,0,s(x3),x4,x5) -> f(x3,x3,x3,x4,x5) f(0,s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5) f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5) Problem 1: SCC Processor: -> Pairs: F(0,0,0,0,s(x5)) -> F(x5,x5,x5,x5,x5) F(0,0,0,s(x4),x5) -> F(x4,x4,x4,x4,x5) F(0,0,s(x3),x4,x5) -> F(x3,x3,x3,x4,x5) F(0,s(x2),x3,x4,x5) -> F(x2,x2,x3,x4,x5) F(s(x1),x2,x3,x4,x5) -> F(x1,x2,x3,x4,x5) -> Rules: f(0,0,0,0,0) -> 0 f(0,0,0,0,s(x5)) -> f(x5,x5,x5,x5,x5) f(0,0,0,s(x4),x5) -> f(x4,x4,x4,x4,x5) f(0,0,s(x3),x4,x5) -> f(x3,x3,x3,x4,x5) f(0,s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5) f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: F(0,0,0,0,s(x5)) -> F(x5,x5,x5,x5,x5) F(0,0,0,s(x4),x5) -> F(x4,x4,x4,x4,x5) F(0,0,s(x3),x4,x5) -> F(x3,x3,x3,x4,x5) F(0,s(x2),x3,x4,x5) -> F(x2,x2,x3,x4,x5) F(s(x1),x2,x3,x4,x5) -> F(x1,x2,x3,x4,x5) ->->-> Rules: f(0,0,0,0,0) -> 0 f(0,0,0,0,s(x5)) -> f(x5,x5,x5,x5,x5) f(0,0,0,s(x4),x5) -> f(x4,x4,x4,x4,x5) f(0,0,s(x3),x4,x5) -> f(x3,x3,x3,x4,x5) f(0,s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5) f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5) Problem 1: Subterm Processor: -> Pairs: F(0,0,0,0,s(x5)) -> F(x5,x5,x5,x5,x5) F(0,0,0,s(x4),x5) -> F(x4,x4,x4,x4,x5) F(0,0,s(x3),x4,x5) -> F(x3,x3,x3,x4,x5) F(0,s(x2),x3,x4,x5) -> F(x2,x2,x3,x4,x5) F(s(x1),x2,x3,x4,x5) -> F(x1,x2,x3,x4,x5) -> Rules: f(0,0,0,0,0) -> 0 f(0,0,0,0,s(x5)) -> f(x5,x5,x5,x5,x5) f(0,0,0,s(x4),x5) -> f(x4,x4,x4,x4,x5) f(0,0,s(x3),x4,x5) -> f(x3,x3,x3,x4,x5) f(0,s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5) f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5) ->Projection: pi(F) = 5
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