Runtime_Complexity_Innermost_Rewriting 2019-04-01 06.40 pair #433313706

details

property	value
status	complete
benchmark	otto04.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n160.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.60285 seconds
cpu usage	3.27257
user time	3.13166
system time	0.140914
max virtual memory	1.8275288E7
max residence set size	213044.0

stage attributes

key	value
starexec-result	WORST_CASE(NON_POLY, ?)

output

3.15/1.56	WORST_CASE(NON_POLY, ?)
3.15/1.57	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.15/1.57	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.15/1.57	
3.15/1.57	
3.15/1.57	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.15/1.57	
3.15/1.57	(0) CpxTRS
3.15/1.57	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.15/1.57	(2) TRS for Loop Detection
3.15/1.57	(3) DecreasingLoopProof [FINISHED, 0 ms]
3.15/1.57	(4) BOUNDS(EXP, INF)
3.15/1.57	
3.15/1.57	
3.15/1.57	----------------------------------------
3.15/1.57	
3.15/1.57	(0)
3.15/1.57	Obligation:
3.15/1.57	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.15/1.57	
3.15/1.57	
3.15/1.57	The TRS R consists of the following rules:
3.15/1.57	
3.15/1.57	   lt(0, s(x)) -> true
3.15/1.57	   lt(x, 0) -> false
3.15/1.57	   lt(s(x), s(y)) -> lt(x, y)
3.15/1.57	   fibo(0) -> fib(0)
3.15/1.57	   fibo(s(0)) -> fib(s(0))
3.15/1.57	   fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
3.15/1.57	   fib(0) -> s(0)
3.15/1.57	   fib(s(0)) -> s(0)
3.15/1.57	   fib(s(s(x))) -> if(true, 0, s(s(x)), 0, 0)
3.15/1.57	   if(true, c, s(s(x)), a, b) -> if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
3.15/1.57	   if(false, c, s(s(x)), a, b) -> sum(fibo(a), fibo(b))
3.15/1.57	   sum(x, 0) -> x
3.15/1.57	   sum(x, s(y)) -> s(sum(x, y))
3.15/1.57	
3.15/1.57	S is empty.
3.15/1.57	Rewrite Strategy: INNERMOST
3.15/1.57	----------------------------------------
3.15/1.57	
3.15/1.57	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.15/1.57	Transformed a relative TRS into a decreasing-loop problem.
3.15/1.57	----------------------------------------
3.15/1.57	
3.15/1.57	(2)
3.15/1.57	Obligation:
3.15/1.57	Analyzing the following TRS for decreasing loops:
3.15/1.57	
3.15/1.57	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.15/1.57	
3.15/1.57	
3.15/1.57	The TRS R consists of the following rules:
3.15/1.57	
3.15/1.57	   lt(0, s(x)) -> true
3.15/1.57	   lt(x, 0) -> false
3.15/1.57	   lt(s(x), s(y)) -> lt(x, y)
3.15/1.57	   fibo(0) -> fib(0)
3.15/1.57	   fibo(s(0)) -> fib(s(0))
3.15/1.57	   fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
3.15/1.57	   fib(0) -> s(0)
3.15/1.57	   fib(s(0)) -> s(0)
3.15/1.57	   fib(s(s(x))) -> if(true, 0, s(s(x)), 0, 0)
3.15/1.57	   if(true, c, s(s(x)), a, b) -> if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
3.15/1.57	   if(false, c, s(s(x)), a, b) -> sum(fibo(a), fibo(b))
3.15/1.57	   sum(x, 0) -> x
3.15/1.57	   sum(x, s(y)) -> s(sum(x, y))
3.15/1.57	
3.15/1.57	S is empty.
3.15/1.57	Rewrite Strategy: INNERMOST
3.15/1.57	----------------------------------------
3.15/1.57	
3.15/1.57	(3) DecreasingLoopProof (FINISHED)
3.15/1.57	The following loop(s) give(s) rise to the lower bound EXP:
3.15/1.57	
3.15/1.57	The rewrite sequence
3.15/1.57	
3.15/1.57	fibo(s(s(x))) ->^+ sum(fibo(s(x)), fibo(x))
3.15/1.57	
3.15/1.57	gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
3.15/1.57	
3.15/1.57	The pumping substitution is [x / s(x)].
3.15/1.57	
3.15/1.57	The result substitution is [ ].
3.15/1.57	
3.15/1.57	
3.15/1.57	
3.15/1.57	The rewrite sequence
3.15/1.57	
3.15/1.57	fibo(s(s(x))) ->^+ sum(fibo(s(x)), fibo(x))
3.15/1.57	
3.15/1.57	gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
3.15/1.57	
3.15/1.57	The pumping substitution is [x / s(s(x))].
3.15/1.57	
3.15/1.57	The result substitution is [ ].
3.15/1.57	
3.15/1.57	
3.15/1.57	
3.15/1.57

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