3.70/1.75	WORST_CASE(NON_POLY, ?)
3.70/1.75	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.70/1.75	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.70/1.75	
3.70/1.75	
3.70/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.70/1.75	
3.70/1.75	(0) CpxTRS
3.70/1.75	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.70/1.75	(2) TRS for Loop Detection
3.70/1.75	(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
3.70/1.75	(4) BEST
3.70/1.75	    (5) proven lower bound
3.70/1.75	        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
3.70/1.75	        (7) BOUNDS(n^1, INF)
3.70/1.75	    (8) TRS for Loop Detection
3.70/1.75	        (9) InfiniteLowerBoundProof [FINISHED, 85 ms]
3.70/1.75	        (10) BOUNDS(INF, INF)
3.70/1.75	
3.70/1.75	
3.70/1.75	----------------------------------------
3.70/1.75	
3.70/1.75	(0)
3.70/1.75	Obligation:
3.70/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.70/1.75	
3.70/1.75	
3.70/1.75	The TRS R consists of the following rules:
3.70/1.75	
3.70/1.75	   nats -> cons(0, n__incr(n__nats))
3.70/1.75	   pairs -> cons(0, n__incr(n__odds))
3.70/1.75	   odds -> incr(pairs)
3.70/1.75	   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
3.70/1.75	   head(cons(X, XS)) -> X
3.70/1.75	   tail(cons(X, XS)) -> activate(XS)
3.70/1.75	   incr(X) -> n__incr(X)
3.70/1.75	   nats -> n__nats
3.70/1.75	   odds -> n__odds
3.70/1.75	   activate(n__incr(X)) -> incr(activate(X))
3.70/1.75	   activate(n__nats) -> nats
3.70/1.75	   activate(n__odds) -> odds
3.70/1.75	   activate(X) -> X
3.70/1.75	
3.70/1.75	S is empty.
3.70/1.75	Rewrite Strategy: FULL
3.70/1.75	----------------------------------------
3.70/1.75	
3.70/1.75	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.70/1.75	Transformed a relative TRS into a decreasing-loop problem.
3.70/1.75	----------------------------------------
3.70/1.75	
3.70/1.75	(2)
3.70/1.75	Obligation:
3.70/1.75	Analyzing the following TRS for decreasing loops:
3.70/1.75	
3.70/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.70/1.75	
3.70/1.75	
3.70/1.75	The TRS R consists of the following rules:
3.70/1.75	
3.70/1.75	   nats -> cons(0, n__incr(n__nats))
3.70/1.75	   pairs -> cons(0, n__incr(n__odds))
3.70/1.75	   odds -> incr(pairs)
3.70/1.75	   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
3.70/1.75	   head(cons(X, XS)) -> X
3.70/1.75	   tail(cons(X, XS)) -> activate(XS)
3.70/1.75	   incr(X) -> n__incr(X)
3.70/1.75	   nats -> n__nats
3.70/1.75	   odds -> n__odds
3.70/1.75	   activate(n__incr(X)) -> incr(activate(X))
3.70/1.75	   activate(n__nats) -> nats
3.70/1.75	   activate(n__odds) -> odds
3.70/1.75	   activate(X) -> X
3.70/1.75	
3.70/1.75	S is empty.
3.70/1.75	Rewrite Strategy: FULL
3.70/1.75	----------------------------------------
3.70/1.75	
3.70/1.75	(3) DecreasingLoopProof (LOWER BOUND(ID))
3.70/1.75	The following loop(s) give(s) rise to the lower bound Omega(n^1):
3.70/1.75	
3.70/1.75	The rewrite sequence
3.70/1.75	
3.70/1.75	activate(n__incr(X)) ->^+ incr(activate(X))
3.70/1.75	
3.70/1.75	gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
3.70/1.75	
3.70/1.75	The pumping substitution is [X / n__incr(X)].
3.70/1.75	
3.70/1.75	The result substitution is [ ].
3.70/1.75	
3.70/1.75	
3.70/1.75	
3.70/1.75	
3.70/1.75	----------------------------------------
3.70/1.75	
3.70/1.75	(4)
3.70/1.75	Complex Obligation (BEST)
3.70/1.75	
3.70/1.75	----------------------------------------
3.70/1.75	
3.70/1.75	(5)
3.70/1.75	Obligation:
3.70/1.75	Proved the lower bound n^1 for the following obligation:
3.70/1.75	
3.70/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.70/1.75	
3.70/1.75	
3.70/1.75	The TRS R consists of the following rules:
3.70/1.75	
3.70/1.75	   nats -> cons(0, n__incr(n__nats))
3.70/1.75	   pairs -> cons(0, n__incr(n__odds))
3.70/1.75	   odds -> incr(pairs)
3.70/1.75	   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
3.70/1.75	   head(cons(X, XS)) -> X
3.70/1.75	   tail(cons(X, XS)) -> activate(XS)
3.70/1.75	   incr(X) -> n__incr(X)
3.92/1.75	   nats -> n__nats
3.92/1.75	   odds -> n__odds
3.92/1.75	   activate(n__incr(X)) -> incr(activate(X))
3.92/1.75	   activate(n__nats) -> nats
3.92/1.75	   activate(n__odds) -> odds
3.92/1.75	   activate(X) -> X
3.92/1.75	
3.92/1.75	S is empty.
3.92/1.75	Rewrite Strategy: FULL
3.92/1.75	----------------------------------------
3.92/1.75	
3.92/1.75	(6) LowerBoundPropagationProof (FINISHED)
3.92/1.75	Propagated lower bound.
3.92/1.75	----------------------------------------
3.92/1.75	
3.92/1.75	(7)
3.92/1.75	BOUNDS(n^1, INF)
3.92/1.75	
3.92/1.75	----------------------------------------
3.92/1.75	
3.92/1.75	(8)
3.92/1.75	Obligation:
3.92/1.75	Analyzing the following TRS for decreasing loops:
3.92/1.75	
3.92/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).
3.92/1.75	
3.92/1.75	
3.92/1.75	The TRS R consists of the following rules:
3.92/1.75	
3.92/1.75	   nats -> cons(0, n__incr(n__nats))
3.92/1.75	   pairs -> cons(0, n__incr(n__odds))
3.92/1.75	   odds -> incr(pairs)
3.92/1.75	   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
3.92/1.75	   head(cons(X, XS)) -> X
3.92/1.75	   tail(cons(X, XS)) -> activate(XS)
3.92/1.75	   incr(X) -> n__incr(X)
3.92/1.75	   nats -> n__nats
3.92/1.75	   odds -> n__odds
3.92/1.75	   activate(n__incr(X)) -> incr(activate(X))
3.92/1.75	   activate(n__nats) -> nats
3.92/1.75	   activate(n__odds) -> odds
3.92/1.75	   activate(X) -> X
3.92/1.75	
3.92/1.75	S is empty.
3.92/1.75	Rewrite Strategy: FULL
3.92/1.75	----------------------------------------
3.92/1.75	
3.92/1.75	(9) InfiniteLowerBoundProof (FINISHED)
3.92/1.75	The following loop proves infinite runtime complexity:
3.92/1.75	
3.92/1.75	The rewrite sequence
3.92/1.75	
3.92/1.75	incr(cons(X, n__incr(n__nats))) ->^+ cons(s(X), n__incr(incr(cons(0, n__incr(n__nats)))))
3.92/1.75	
3.92/1.75	gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0].
3.92/1.75	
3.92/1.75	The pumping substitution is [ ].
3.92/1.75	
3.92/1.75	The result substitution is [X / 0].
3.92/1.75	
3.92/1.75	
3.92/1.75	
3.92/1.75	
3.92/1.75	----------------------------------------
3.92/1.75	
3.92/1.75	(10)
3.92/1.75	BOUNDS(INF, INF)
3.95/2.23	EOF