4.34/1.92	WORST_CASE(NON_POLY, ?)
4.43/1.93	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
4.43/1.93	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.43/1.93	
4.43/1.93	
4.43/1.93	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
4.43/1.93	
4.43/1.93	(0) CpxTRS
4.43/1.93	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
4.43/1.93	(2) TRS for Loop Detection
4.43/1.93	(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
4.43/1.93	(4) BEST
4.43/1.93	    (5) proven lower bound
4.43/1.93	        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
4.43/1.93	        (7) BOUNDS(n^1, INF)
4.43/1.93	    (8) TRS for Loop Detection
4.43/1.93	        (9) InfiniteLowerBoundProof [FINISHED, 108 ms]
4.43/1.93	        (10) BOUNDS(INF, INF)
4.43/1.93	
4.43/1.93	
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(0)
4.43/1.93	Obligation:
4.43/1.93	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
4.43/1.93	
4.43/1.93	
4.43/1.93	The TRS R consists of the following rules:
4.43/1.93	
4.43/1.93	   a__zeros -> cons(0, zeros)
4.43/1.93	   a__and(tt, X) -> mark(X)
4.43/1.93	   a__length(nil) -> 0
4.43/1.93	   a__length(cons(N, L)) -> s(a__length(mark(L)))
4.43/1.93	   mark(zeros) -> a__zeros
4.43/1.93	   mark(and(X1, X2)) -> a__and(mark(X1), X2)
4.43/1.93	   mark(length(X)) -> a__length(mark(X))
4.43/1.93	   mark(cons(X1, X2)) -> cons(mark(X1), X2)
4.43/1.93	   mark(0) -> 0
4.43/1.93	   mark(tt) -> tt
4.43/1.93	   mark(nil) -> nil
4.43/1.93	   mark(s(X)) -> s(mark(X))
4.43/1.93	   a__zeros -> zeros
4.43/1.93	   a__and(X1, X2) -> and(X1, X2)
4.43/1.93	   a__length(X) -> length(X)
4.43/1.93	
4.43/1.93	S is empty.
4.43/1.93	Rewrite Strategy: FULL
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
4.43/1.93	Transformed a relative TRS into a decreasing-loop problem.
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(2)
4.43/1.93	Obligation:
4.43/1.93	Analyzing the following TRS for decreasing loops:
4.43/1.93	
4.43/1.93	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
4.43/1.93	
4.43/1.93	
4.43/1.93	The TRS R consists of the following rules:
4.43/1.93	
4.43/1.93	   a__zeros -> cons(0, zeros)
4.43/1.93	   a__and(tt, X) -> mark(X)
4.43/1.93	   a__length(nil) -> 0
4.43/1.93	   a__length(cons(N, L)) -> s(a__length(mark(L)))
4.43/1.93	   mark(zeros) -> a__zeros
4.43/1.93	   mark(and(X1, X2)) -> a__and(mark(X1), X2)
4.43/1.93	   mark(length(X)) -> a__length(mark(X))
4.43/1.93	   mark(cons(X1, X2)) -> cons(mark(X1), X2)
4.43/1.93	   mark(0) -> 0
4.43/1.93	   mark(tt) -> tt
4.43/1.93	   mark(nil) -> nil
4.43/1.93	   mark(s(X)) -> s(mark(X))
4.43/1.93	   a__zeros -> zeros
4.43/1.93	   a__and(X1, X2) -> and(X1, X2)
4.43/1.93	   a__length(X) -> length(X)
4.43/1.93	
4.43/1.93	S is empty.
4.43/1.93	Rewrite Strategy: FULL
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(3) DecreasingLoopProof (LOWER BOUND(ID))
4.43/1.93	The following loop(s) give(s) rise to the lower bound Omega(n^1):
4.43/1.93	
4.43/1.93	The rewrite sequence
4.43/1.93	
4.43/1.93	mark(length(X)) ->^+ a__length(mark(X))
4.43/1.93	
4.43/1.93	gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
4.43/1.93	
4.43/1.93	The pumping substitution is [X / length(X)].
4.43/1.93	
4.43/1.93	The result substitution is [ ].
4.43/1.93	
4.43/1.93	
4.43/1.93	
4.43/1.93	
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(4)
4.43/1.93	Complex Obligation (BEST)
4.43/1.93	
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(5)
4.43/1.93	Obligation:
4.43/1.93	Proved the lower bound n^1 for the following obligation:
4.43/1.93	
4.43/1.93	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
4.43/1.93	
4.43/1.93	
4.43/1.93	The TRS R consists of the following rules:
4.43/1.93	
4.43/1.93	   a__zeros -> cons(0, zeros)
4.43/1.93	   a__and(tt, X) -> mark(X)
4.43/1.93	   a__length(nil) -> 0
4.43/1.93	   a__length(cons(N, L)) -> s(a__length(mark(L)))
4.43/1.93	   mark(zeros) -> a__zeros
4.43/1.93	   mark(and(X1, X2)) -> a__and(mark(X1), X2)
4.43/1.93	   mark(length(X)) -> a__length(mark(X))
4.43/1.93	   mark(cons(X1, X2)) -> cons(mark(X1), X2)
4.43/1.93	   mark(0) -> 0
4.43/1.93	   mark(tt) -> tt
4.43/1.93	   mark(nil) -> nil
4.43/1.93	   mark(s(X)) -> s(mark(X))
4.43/1.93	   a__zeros -> zeros
4.43/1.93	   a__and(X1, X2) -> and(X1, X2)
4.43/1.93	   a__length(X) -> length(X)
4.43/1.93	
4.43/1.93	S is empty.
4.43/1.93	Rewrite Strategy: FULL
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(6) LowerBoundPropagationProof (FINISHED)
4.43/1.93	Propagated lower bound.
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(7)
4.43/1.93	BOUNDS(n^1, INF)
4.43/1.93	
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(8)
4.43/1.93	Obligation:
4.43/1.93	Analyzing the following TRS for decreasing loops:
4.43/1.93	
4.43/1.93	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
4.43/1.93	
4.43/1.93	
4.43/1.93	The TRS R consists of the following rules:
4.43/1.93	
4.43/1.93	   a__zeros -> cons(0, zeros)
4.43/1.93	   a__and(tt, X) -> mark(X)
4.43/1.93	   a__length(nil) -> 0
4.43/1.93	   a__length(cons(N, L)) -> s(a__length(mark(L)))
4.43/1.93	   mark(zeros) -> a__zeros
4.43/1.93	   mark(and(X1, X2)) -> a__and(mark(X1), X2)
4.43/1.93	   mark(length(X)) -> a__length(mark(X))
4.43/1.93	   mark(cons(X1, X2)) -> cons(mark(X1), X2)
4.43/1.93	   mark(0) -> 0
4.43/1.93	   mark(tt) -> tt
4.43/1.93	   mark(nil) -> nil
4.43/1.93	   mark(s(X)) -> s(mark(X))
4.43/1.93	   a__zeros -> zeros
4.43/1.93	   a__and(X1, X2) -> and(X1, X2)
4.43/1.93	   a__length(X) -> length(X)
4.43/1.93	
4.43/1.93	S is empty.
4.43/1.93	Rewrite Strategy: FULL
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(9) InfiniteLowerBoundProof (FINISHED)
4.43/1.93	The following loop proves infinite runtime complexity:
4.43/1.93	
4.43/1.93	The rewrite sequence
4.43/1.93	
4.43/1.93	a__length(cons(N, zeros)) ->^+ s(a__length(cons(0, zeros)))
4.43/1.93	
4.43/1.93	gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
4.43/1.93	
4.43/1.93	The pumping substitution is [ ].
4.43/1.93	
4.43/1.93	The result substitution is [N / 0].
4.43/1.93	
4.43/1.93	
4.43/1.93	
4.43/1.93	
4.43/1.93	----------------------------------------
4.43/1.93	
4.43/1.93	(10)
4.43/1.93	BOUNDS(INF, INF)
4.47/2.49	EOF