3.13/1.62	WORST_CASE(NON_POLY, ?)
3.13/1.62	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.13/1.62	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.13/1.62	
3.13/1.62	
3.13/1.62	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.13/1.62	
3.13/1.62	(0) CpxTRS
3.13/1.62	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.13/1.62	(2) TRS for Loop Detection
3.13/1.62	(3) InfiniteLowerBoundProof [FINISHED, 0 ms]
3.13/1.62	(4) BOUNDS(INF, INF)
3.13/1.62	
3.13/1.62	
3.13/1.62	----------------------------------------
3.13/1.62	
3.13/1.62	(0)
3.13/1.62	Obligation:
3.13/1.62	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.13/1.62	
3.13/1.62	
3.13/1.62	The TRS R consists of the following rules:
3.13/1.62	
3.13/1.62	   f(x) -> f(f(x))
3.13/1.62	
3.13/1.62	S is empty.
3.13/1.62	Rewrite Strategy: INNERMOST
3.13/1.62	----------------------------------------
3.13/1.62	
3.13/1.62	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.13/1.62	Transformed a relative TRS into a decreasing-loop problem.
3.13/1.62	----------------------------------------
3.13/1.62	
3.13/1.62	(2)
3.13/1.62	Obligation:
3.13/1.62	Analyzing the following TRS for decreasing loops:
3.13/1.62	
3.13/1.62	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.13/1.62	
3.13/1.62	
3.13/1.62	The TRS R consists of the following rules:
3.13/1.62	
3.13/1.62	   f(x) -> f(f(x))
3.13/1.62	
3.13/1.62	S is empty.
3.13/1.62	Rewrite Strategy: INNERMOST
3.13/1.62	----------------------------------------
3.13/1.62	
3.13/1.62	(3) InfiniteLowerBoundProof (FINISHED)
3.13/1.62	The following loop proves infinite runtime complexity:
3.13/1.62	
3.13/1.62	The rewrite sequence
3.13/1.62	
3.13/1.62	f(x) ->^+ f(f(x))
3.13/1.62	
3.13/1.62	gives rise to a decreasing loop by considering the right hand sides subterm at position [].
3.13/1.62	
3.13/1.62	The pumping substitution is [ ].
3.13/1.62	
3.13/1.62	The result substitution is [x / f(x)].
3.13/1.62	
3.13/1.62	
3.13/1.62	
3.13/1.62	
3.13/1.62	----------------------------------------
3.13/1.62	
3.13/1.62	(4)
3.13/1.62	BOUNDS(INF, INF)
3.37/2.42	EOF