3.10/1.49	WORST_CASE(NON_POLY, ?)
3.10/1.50	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.10/1.50	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.10/1.50	
3.10/1.50	
3.10/1.50	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.10/1.50	
3.10/1.50	(0) CpxTRS
3.10/1.50	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.10/1.50	(2) TRS for Loop Detection
3.10/1.50	(3) InfiniteLowerBoundProof [FINISHED, 0 ms]
3.10/1.50	(4) BOUNDS(INF, INF)
3.10/1.50	
3.10/1.50	
3.10/1.50	----------------------------------------
3.10/1.50	
3.10/1.50	(0)
3.10/1.50	Obligation:
3.10/1.50	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.10/1.50	
3.10/1.50	
3.10/1.50	The TRS R consists of the following rules:
3.10/1.50	
3.10/1.50	   f(X) -> g(h(f(X)))
3.10/1.50	
3.10/1.50	S is empty.
3.10/1.50	Rewrite Strategy: INNERMOST
3.10/1.50	----------------------------------------
3.10/1.50	
3.10/1.50	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.10/1.50	Transformed a relative TRS into a decreasing-loop problem.
3.10/1.50	----------------------------------------
3.10/1.50	
3.10/1.50	(2)
3.10/1.50	Obligation:
3.10/1.50	Analyzing the following TRS for decreasing loops:
3.10/1.50	
3.10/1.50	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.10/1.50	
3.10/1.50	
3.10/1.50	The TRS R consists of the following rules:
3.10/1.50	
3.10/1.50	   f(X) -> g(h(f(X)))
3.10/1.50	
3.10/1.50	S is empty.
3.10/1.50	Rewrite Strategy: INNERMOST
3.10/1.50	----------------------------------------
3.10/1.50	
3.10/1.50	(3) InfiniteLowerBoundProof (FINISHED)
3.10/1.50	The following loop proves infinite runtime complexity:
3.10/1.50	
3.10/1.50	The rewrite sequence
3.10/1.50	
3.10/1.50	f(X) ->^+ g(h(f(X)))
3.10/1.50	
3.10/1.50	gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
3.10/1.50	
3.10/1.50	The pumping substitution is [ ].
3.10/1.50	
3.10/1.50	The result substitution is [ ].
3.10/1.50	
3.10/1.50	
3.10/1.50	
3.10/1.50	
3.10/1.50	----------------------------------------
3.10/1.50	
3.10/1.50	(4)
3.10/1.50	BOUNDS(INF, INF)
3.21/1.53	EOF