TRS Context Sensitive pair #487523746

details

property	value
status	complete
benchmark	ExProp7_Luc06.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	CSR_04

run statistics

property	value
solver	muterm 5.18
configuration	default
runtime (wallclock)	0.112241983414 seconds
cpu usage	0.094080615
max memory	3145728.0

stage attributes

key	value
output-size	1776
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 X)
(STRATEGY CONTEXTSENSITIVE
(f 1)
(p 1)
(0)
(cons 1)
(s 1)
)
(RULES
f(0) -> cons(0,f(s(0)))
f(s(0)) -> f(p(s(0)))
p(s(X)) -> X
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 f(0) -> cons(0,f(s(0)))
 f(s(0)) -> f(p(s(0)))
 p(s(X)) -> X
-> The context-sensitive term rewriting system is an orthogonal system. Therefore, innermost cs-termination implies cs-termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(s(0)) -> F(p(s(0)))
 F(s(0)) -> P(s(0))
-> Rules:
 f(0) -> cons(0,f(s(0)))
 f(s(0)) -> f(p(s(0)))
 p(s(X)) -> X
-> Unhiding Rules:
 Empty

Problem 1: 

SCC Processor:
-> Pairs:
 F(s(0)) -> F(p(s(0)))
 F(s(0)) -> P(s(0))
-> Rules:
 f(0) -> cons(0,f(s(0)))
 f(s(0)) -> f(p(s(0)))
 p(s(X)) -> X
-> Unhiding rules:
 Empty
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(s(0)) -> F(p(s(0)))
->->-> Rules:
 f(0) -> cons(0,f(s(0)))
 f(s(0)) -> f(p(s(0)))
 p(s(X)) -> X
->->-> Unhiding rules:
 Empty

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 F(s(0)) -> F(p(s(0)))
-> Rules:
 f(0) -> cons(0,f(s(0)))
 f(s(0)) -> f(p(s(0)))
 p(s(X)) -> X
-> Unhiding rules:
 Empty
-> Usable rules:
 p(s(X)) -> X
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[p](X) = 1/2.X + 1/2
[0] = 0
[s](X) = 2.X + 2
[F](X) = 1/2.X

Problem 1:

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job log

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