3.60/1.65	WORST_CASE(NON_POLY, ?)
3.60/1.67	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
3.60/1.67	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.60/1.67	
3.60/1.67	
3.60/1.67	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.60/1.67	
3.60/1.67	(0) CpxTRS
3.60/1.67	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.60/1.67	(2) TRS for Loop Detection
3.60/1.67	(3) InfiniteLowerBoundProof [FINISHED, 42 ms]
3.60/1.67	(4) BOUNDS(INF, INF)
3.60/1.67	
3.60/1.67	
3.60/1.67	----------------------------------------
3.60/1.67	
3.60/1.67	(0)
3.60/1.67	Obligation:
3.60/1.67	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.60/1.67	
3.60/1.67	
3.60/1.67	The TRS R consists of the following rules:
3.60/1.67	
3.60/1.67	   pairNs -> cons(0, n__incr(oddNs))
3.60/1.67	   oddNs -> incr(pairNs)
3.60/1.67	   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
3.60/1.67	   take(0, XS) -> nil
3.60/1.67	   take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
3.60/1.67	   zip(nil, XS) -> nil
3.60/1.67	   zip(X, nil) -> nil
3.60/1.67	   zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
3.60/1.67	   tail(cons(X, XS)) -> activate(XS)
3.60/1.67	   repItems(nil) -> nil
3.60/1.67	   repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS))))
3.60/1.67	   incr(X) -> n__incr(X)
3.60/1.67	   take(X1, X2) -> n__take(X1, X2)
3.60/1.67	   zip(X1, X2) -> n__zip(X1, X2)
3.60/1.67	   cons(X1, X2) -> n__cons(X1, X2)
3.60/1.67	   repItems(X) -> n__repItems(X)
3.60/1.67	   activate(n__incr(X)) -> incr(X)
3.60/1.67	   activate(n__take(X1, X2)) -> take(X1, X2)
3.60/1.67	   activate(n__zip(X1, X2)) -> zip(X1, X2)
3.60/1.67	   activate(n__cons(X1, X2)) -> cons(X1, X2)
3.60/1.67	   activate(n__repItems(X)) -> repItems(X)
3.60/1.67	   activate(X) -> X
3.60/1.67	
3.60/1.67	S is empty.
3.60/1.67	Rewrite Strategy: FULL
3.60/1.67	----------------------------------------
3.60/1.67	
3.60/1.67	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.60/1.67	Transformed a relative TRS into a decreasing-loop problem.
3.60/1.67	----------------------------------------
3.60/1.67	
3.60/1.67	(2)
3.60/1.67	Obligation:
3.60/1.67	Analyzing the following TRS for decreasing loops:
3.60/1.67	
3.60/1.67	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.60/1.67	
3.60/1.67	
3.60/1.67	The TRS R consists of the following rules:
3.60/1.67	
3.60/1.67	   pairNs -> cons(0, n__incr(oddNs))
3.60/1.67	   oddNs -> incr(pairNs)
3.60/1.67	   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
3.60/1.67	   take(0, XS) -> nil
3.60/1.67	   take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
3.60/1.67	   zip(nil, XS) -> nil
3.60/1.67	   zip(X, nil) -> nil
3.60/1.67	   zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
3.60/1.67	   tail(cons(X, XS)) -> activate(XS)
3.60/1.67	   repItems(nil) -> nil
3.60/1.67	   repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS))))
3.60/1.67	   incr(X) -> n__incr(X)
3.60/1.67	   take(X1, X2) -> n__take(X1, X2)
3.60/1.67	   zip(X1, X2) -> n__zip(X1, X2)
3.60/1.67	   cons(X1, X2) -> n__cons(X1, X2)
3.60/1.67	   repItems(X) -> n__repItems(X)
3.60/1.67	   activate(n__incr(X)) -> incr(X)
3.60/1.67	   activate(n__take(X1, X2)) -> take(X1, X2)
3.60/1.67	   activate(n__zip(X1, X2)) -> zip(X1, X2)
3.60/1.67	   activate(n__cons(X1, X2)) -> cons(X1, X2)
3.60/1.67	   activate(n__repItems(X)) -> repItems(X)
3.60/1.67	   activate(X) -> X
3.60/1.67	
3.60/1.67	S is empty.
3.60/1.67	Rewrite Strategy: FULL
3.60/1.67	----------------------------------------
3.60/1.67	
3.60/1.67	(3) InfiniteLowerBoundProof (FINISHED)
3.60/1.67	The following loop proves infinite runtime complexity:
3.60/1.67	
3.60/1.67	The rewrite sequence
3.60/1.67	
3.60/1.67	oddNs ->^+ incr(cons(0, n__incr(oddNs)))
3.60/1.67	
3.60/1.67	gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0].
3.60/1.67	
3.60/1.67	The pumping substitution is [ ].
3.60/1.67	
3.60/1.67	The result substitution is [ ].
3.60/1.67	
3.60/1.67	
3.60/1.67	
3.60/1.67	
3.60/1.67	----------------------------------------
3.60/1.67	
3.60/1.67	(4)
3.60/1.67	BOUNDS(INF, INF)
3.74/1.70	EOF