3.40/1.62	WORST_CASE(NON_POLY, ?)
3.40/1.62	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.40/1.62	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.40/1.62	
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	(0) CpxTRS
3.40/1.62	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.40/1.62	(2) TRS for Loop Detection
3.40/1.62	(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
3.40/1.62	(4) BEST
3.40/1.62	    (5) proven lower bound
3.40/1.62	        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
3.40/1.62	        (7) BOUNDS(n^1, INF)
3.40/1.62	    (8) TRS for Loop Detection
3.40/1.62	        (9) DecreasingLoopProof [FINISHED, 0 ms]
3.40/1.62	        (10) BOUNDS(EXP, INF)
3.40/1.62	
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(0)
3.40/1.62	Obligation:
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   plus(0, x) -> x
3.40/1.62	   plus(s(x), y) -> s(plus(p(s(x)), y))
3.40/1.62	   times(0, y) -> 0
3.40/1.62	   times(s(x), y) -> plus(y, times(p(s(x)), y))
3.40/1.62	   exp(x, 0) -> s(0)
3.40/1.62	   exp(x, s(y)) -> times(x, exp(x, y))
3.40/1.62	   p(s(0)) -> 0
3.40/1.62	   p(s(s(x))) -> s(p(s(x)))
3.40/1.62	   tower(x, y) -> towerIter(x, y, s(0))
3.40/1.62	   towerIter(0, y, z) -> z
3.40/1.62	   towerIter(s(x), y, z) -> towerIter(p(s(x)), y, exp(y, z))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.40/1.62	Transformed a relative TRS into a decreasing-loop problem.
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(2)
3.40/1.62	Obligation:
3.40/1.62	Analyzing the following TRS for decreasing loops:
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   plus(0, x) -> x
3.40/1.62	   plus(s(x), y) -> s(plus(p(s(x)), y))
3.40/1.62	   times(0, y) -> 0
3.40/1.62	   times(s(x), y) -> plus(y, times(p(s(x)), y))
3.40/1.62	   exp(x, 0) -> s(0)
3.40/1.62	   exp(x, s(y)) -> times(x, exp(x, y))
3.40/1.62	   p(s(0)) -> 0
3.40/1.62	   p(s(s(x))) -> s(p(s(x)))
3.40/1.62	   tower(x, y) -> towerIter(x, y, s(0))
3.40/1.62	   towerIter(0, y, z) -> z
3.40/1.62	   towerIter(s(x), y, z) -> towerIter(p(s(x)), y, exp(y, z))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(3) DecreasingLoopProof (LOWER BOUND(ID))
3.40/1.62	The following loop(s) give(s) rise to the lower bound Omega(n^1):
3.40/1.62	
3.40/1.62	The rewrite sequence
3.40/1.62	
3.40/1.62	p(s(s(x))) ->^+ s(p(s(x)))
3.40/1.62	
3.40/1.62	gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
3.40/1.62	
3.40/1.62	The pumping substitution is [x / s(x)].
3.40/1.62	
3.40/1.62	The result substitution is [ ].
3.40/1.62	
3.40/1.62	
3.40/1.62	
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(4)
3.40/1.62	Complex Obligation (BEST)
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(5)
3.40/1.62	Obligation:
3.40/1.62	Proved the lower bound n^1 for the following obligation:
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   plus(0, x) -> x
3.40/1.62	   plus(s(x), y) -> s(plus(p(s(x)), y))
3.40/1.62	   times(0, y) -> 0
3.40/1.62	   times(s(x), y) -> plus(y, times(p(s(x)), y))
3.40/1.62	   exp(x, 0) -> s(0)
3.40/1.62	   exp(x, s(y)) -> times(x, exp(x, y))
3.40/1.62	   p(s(0)) -> 0
3.40/1.62	   p(s(s(x))) -> s(p(s(x)))
3.40/1.62	   tower(x, y) -> towerIter(x, y, s(0))
3.40/1.62	   towerIter(0, y, z) -> z
3.40/1.62	   towerIter(s(x), y, z) -> towerIter(p(s(x)), y, exp(y, z))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(6) LowerBoundPropagationProof (FINISHED)
3.40/1.62	Propagated lower bound.
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(7)
3.40/1.62	BOUNDS(n^1, INF)
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(8)
3.40/1.62	Obligation:
3.40/1.62	Analyzing the following TRS for decreasing loops:
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   plus(0, x) -> x
3.40/1.62	   plus(s(x), y) -> s(plus(p(s(x)), y))
3.40/1.62	   times(0, y) -> 0
3.40/1.62	   times(s(x), y) -> plus(y, times(p(s(x)), y))
3.40/1.62	   exp(x, 0) -> s(0)
3.40/1.62	   exp(x, s(y)) -> times(x, exp(x, y))
3.40/1.62	   p(s(0)) -> 0
3.40/1.62	   p(s(s(x))) -> s(p(s(x)))
3.40/1.62	   tower(x, y) -> towerIter(x, y, s(0))
3.40/1.62	   towerIter(0, y, z) -> z
3.40/1.62	   towerIter(s(x), y, z) -> towerIter(p(s(x)), y, exp(y, z))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(9) DecreasingLoopProof (FINISHED)
3.40/1.62	The following loop(s) give(s) rise to the lower bound EXP:
3.40/1.62	
3.40/1.62	The rewrite sequence
3.40/1.62	
3.40/1.62	exp(s(x1_0), s(y)) ->^+ plus(exp(s(x1_0), y), times(p(s(x1_0)), exp(s(x1_0), y)))
3.40/1.62	
3.40/1.62	gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
3.40/1.62	
3.40/1.62	The pumping substitution is [y / s(y)].
3.40/1.62	
3.40/1.62	The result substitution is [ ].
3.40/1.62	
3.40/1.62	
3.40/1.62	
3.40/1.62	The rewrite sequence
3.40/1.62	
3.40/1.62	exp(s(x1_0), s(y)) ->^+ plus(exp(s(x1_0), y), times(p(s(x1_0)), exp(s(x1_0), y)))
3.40/1.62	
3.40/1.62	gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1].
3.40/1.62	
3.40/1.62	The pumping substitution is [y / s(y)].
3.40/1.62	
3.40/1.62	The result substitution is [ ].
3.40/1.62	
3.40/1.62	
3.40/1.62	
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(10)
3.40/1.62	BOUNDS(EXP, INF)
3.40/1.66	EOF