3.52/1.64	WORST_CASE(NON_POLY, ?)
3.52/1.65	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.52/1.65	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.52/1.65	
3.52/1.65	
3.52/1.65	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.52/1.65	
3.52/1.65	(0) CpxTRS
3.52/1.65	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.52/1.65	(2) TRS for Loop Detection
3.52/1.65	(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
3.52/1.65	(4) BEST
3.52/1.65	    (5) proven lower bound
3.52/1.65	        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
3.52/1.65	        (7) BOUNDS(n^1, INF)
3.52/1.65	    (8) TRS for Loop Detection
3.52/1.65	        (9) DecreasingLoopProof [FINISHED, 34 ms]
3.52/1.65	        (10) BOUNDS(EXP, INF)
3.52/1.65	
3.52/1.65	
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(0)
3.52/1.65	Obligation:
3.52/1.65	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.52/1.65	
3.52/1.65	
3.52/1.65	The TRS R consists of the following rules:
3.52/1.65	
3.52/1.65	   p(0) -> 0
3.52/1.65	   p(s(X)) -> X
3.52/1.65	   leq(0, Y) -> true
3.52/1.65	   leq(s(X), 0) -> false
3.52/1.65	   leq(s(X), s(Y)) -> leq(X, Y)
3.52/1.65	   if(true, X, Y) -> activate(X)
3.52/1.65	   if(false, X, Y) -> activate(Y)
3.52/1.65	   diff(X, Y) -> if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
3.52/1.65	   0 -> n__0
3.52/1.65	   s(X) -> n__s(X)
3.52/1.65	   diff(X1, X2) -> n__diff(X1, X2)
3.52/1.65	   p(X) -> n__p(X)
3.52/1.65	   activate(n__0) -> 0
3.52/1.65	   activate(n__s(X)) -> s(activate(X))
3.52/1.65	   activate(n__diff(X1, X2)) -> diff(activate(X1), activate(X2))
3.52/1.65	   activate(n__p(X)) -> p(activate(X))
3.52/1.65	   activate(X) -> X
3.52/1.65	
3.52/1.65	S is empty.
3.52/1.65	Rewrite Strategy: FULL
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.52/1.65	Transformed a relative TRS into a decreasing-loop problem.
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(2)
3.52/1.65	Obligation:
3.52/1.65	Analyzing the following TRS for decreasing loops:
3.52/1.65	
3.52/1.65	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.52/1.65	
3.52/1.65	
3.52/1.65	The TRS R consists of the following rules:
3.52/1.65	
3.52/1.65	   p(0) -> 0
3.52/1.65	   p(s(X)) -> X
3.52/1.65	   leq(0, Y) -> true
3.52/1.65	   leq(s(X), 0) -> false
3.52/1.65	   leq(s(X), s(Y)) -> leq(X, Y)
3.52/1.65	   if(true, X, Y) -> activate(X)
3.52/1.65	   if(false, X, Y) -> activate(Y)
3.52/1.65	   diff(X, Y) -> if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
3.52/1.65	   0 -> n__0
3.52/1.65	   s(X) -> n__s(X)
3.52/1.65	   diff(X1, X2) -> n__diff(X1, X2)
3.52/1.65	   p(X) -> n__p(X)
3.52/1.65	   activate(n__0) -> 0
3.52/1.65	   activate(n__s(X)) -> s(activate(X))
3.52/1.65	   activate(n__diff(X1, X2)) -> diff(activate(X1), activate(X2))
3.52/1.65	   activate(n__p(X)) -> p(activate(X))
3.52/1.65	   activate(X) -> X
3.52/1.65	
3.52/1.65	S is empty.
3.52/1.65	Rewrite Strategy: FULL
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(3) DecreasingLoopProof (LOWER BOUND(ID))
3.52/1.65	The following loop(s) give(s) rise to the lower bound Omega(n^1):
3.52/1.65	
3.52/1.65	The rewrite sequence
3.52/1.65	
3.52/1.65	activate(n__p(X)) ->^+ p(activate(X))
3.52/1.65	
3.52/1.65	gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
3.52/1.65	
3.52/1.65	The pumping substitution is [X / n__p(X)].
3.52/1.65	
3.52/1.65	The result substitution is [ ].
3.52/1.65	
3.52/1.65	
3.52/1.65	
3.52/1.65	
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(4)
3.52/1.65	Complex Obligation (BEST)
3.52/1.65	
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(5)
3.52/1.65	Obligation:
3.52/1.65	Proved the lower bound n^1 for the following obligation:
3.52/1.65	
3.52/1.65	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.52/1.65	
3.52/1.65	
3.52/1.65	The TRS R consists of the following rules:
3.52/1.65	
3.52/1.65	   p(0) -> 0
3.52/1.65	   p(s(X)) -> X
3.52/1.65	   leq(0, Y) -> true
3.52/1.65	   leq(s(X), 0) -> false
3.52/1.65	   leq(s(X), s(Y)) -> leq(X, Y)
3.52/1.65	   if(true, X, Y) -> activate(X)
3.52/1.65	   if(false, X, Y) -> activate(Y)
3.52/1.65	   diff(X, Y) -> if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
3.52/1.65	   0 -> n__0
3.52/1.65	   s(X) -> n__s(X)
3.52/1.65	   diff(X1, X2) -> n__diff(X1, X2)
3.52/1.65	   p(X) -> n__p(X)
3.52/1.65	   activate(n__0) -> 0
3.52/1.65	   activate(n__s(X)) -> s(activate(X))
3.52/1.65	   activate(n__diff(X1, X2)) -> diff(activate(X1), activate(X2))
3.52/1.65	   activate(n__p(X)) -> p(activate(X))
3.52/1.65	   activate(X) -> X
3.52/1.65	
3.52/1.65	S is empty.
3.52/1.65	Rewrite Strategy: FULL
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(6) LowerBoundPropagationProof (FINISHED)
3.52/1.65	Propagated lower bound.
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(7)
3.52/1.65	BOUNDS(n^1, INF)
3.52/1.65	
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(8)
3.52/1.65	Obligation:
3.52/1.65	Analyzing the following TRS for decreasing loops:
3.52/1.65	
3.52/1.65	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.52/1.65	
3.52/1.65	
3.52/1.65	The TRS R consists of the following rules:
3.52/1.65	
3.52/1.65	   p(0) -> 0
3.52/1.65	   p(s(X)) -> X
3.52/1.65	   leq(0, Y) -> true
3.52/1.65	   leq(s(X), 0) -> false
3.52/1.65	   leq(s(X), s(Y)) -> leq(X, Y)
3.52/1.65	   if(true, X, Y) -> activate(X)
3.52/1.65	   if(false, X, Y) -> activate(Y)
3.52/1.65	   diff(X, Y) -> if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
3.52/1.65	   0 -> n__0
3.52/1.65	   s(X) -> n__s(X)
3.52/1.65	   diff(X1, X2) -> n__diff(X1, X2)
3.52/1.65	   p(X) -> n__p(X)
3.52/1.65	   activate(n__0) -> 0
3.52/1.65	   activate(n__s(X)) -> s(activate(X))
3.52/1.65	   activate(n__diff(X1, X2)) -> diff(activate(X1), activate(X2))
3.52/1.65	   activate(n__p(X)) -> p(activate(X))
3.52/1.65	   activate(X) -> X
3.52/1.65	
3.52/1.65	S is empty.
3.52/1.65	Rewrite Strategy: FULL
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(9) DecreasingLoopProof (FINISHED)
3.52/1.65	The following loop(s) give(s) rise to the lower bound EXP:
3.52/1.65	
3.52/1.65	The rewrite sequence
3.52/1.65	
3.52/1.65	activate(n__diff(X1, X2)) ->^+ if(leq(activate(X1), activate(X2)), n__0, n__s(n__diff(n__p(activate(X1)), activate(X2))))
3.52/1.65	
3.52/1.65	gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
3.52/1.65	
3.52/1.65	The pumping substitution is [X1 / n__diff(X1, X2)].
3.52/1.65	
3.52/1.65	The result substitution is [ ].
3.52/1.65	
3.52/1.65	
3.52/1.65	
3.52/1.65	The rewrite sequence
3.52/1.65	
3.52/1.65	activate(n__diff(X1, X2)) ->^+ if(leq(activate(X1), activate(X2)), n__0, n__s(n__diff(n__p(activate(X1)), activate(X2))))
3.52/1.65	
3.52/1.65	gives rise to a decreasing loop by considering the right hand sides subterm at position [2,0,0,0].
3.52/1.65	
3.52/1.65	The pumping substitution is [X1 / n__diff(X1, X2)].
3.52/1.65	
3.52/1.65	The result substitution is [ ].
3.52/1.65	
3.52/1.65	
3.52/1.65	
3.52/1.65	
3.52/1.65	----------------------------------------
3.52/1.65	
3.52/1.65	(10)
3.52/1.65	BOUNDS(EXP, INF)
3.61/1.69	EOF