3.87/1.74	WORST_CASE(NON_POLY, ?)
3.87/1.75	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.87/1.75	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.87/1.75	
3.87/1.75	
3.87/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.87/1.75	
3.87/1.75	(0) CpxTRS
3.87/1.75	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.87/1.75	(2) TRS for Loop Detection
3.87/1.75	(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
3.87/1.75	(4) BEST
3.87/1.75	    (5) proven lower bound
3.87/1.75	        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
3.87/1.75	        (7) BOUNDS(n^1, INF)
3.87/1.75	    (8) TRS for Loop Detection
3.87/1.75	        (9) DecreasingLoopProof [FINISHED, 105 ms]
3.87/1.75	        (10) BOUNDS(EXP, INF)
3.87/1.75	
3.87/1.75	
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(0)
3.87/1.75	Obligation:
3.87/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.87/1.75	
3.87/1.75	
3.87/1.75	The TRS R consists of the following rules:
3.87/1.75	
3.87/1.75	   a__minus(0, Y) -> 0
3.87/1.75	   a__minus(s(X), s(Y)) -> a__minus(X, Y)
3.87/1.75	   a__geq(X, 0) -> true
3.87/1.75	   a__geq(0, s(Y)) -> false
3.87/1.75	   a__geq(s(X), s(Y)) -> a__geq(X, Y)
3.87/1.75	   a__div(0, s(Y)) -> 0
3.87/1.75	   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
3.87/1.75	   a__if(true, X, Y) -> mark(X)
3.87/1.75	   a__if(false, X, Y) -> mark(Y)
3.87/1.75	   mark(minus(X1, X2)) -> a__minus(X1, X2)
3.87/1.75	   mark(geq(X1, X2)) -> a__geq(X1, X2)
3.87/1.75	   mark(div(X1, X2)) -> a__div(mark(X1), X2)
3.87/1.75	   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
3.87/1.75	   mark(0) -> 0
3.87/1.75	   mark(s(X)) -> s(mark(X))
3.87/1.75	   mark(true) -> true
3.87/1.75	   mark(false) -> false
3.87/1.75	   a__minus(X1, X2) -> minus(X1, X2)
3.87/1.75	   a__geq(X1, X2) -> geq(X1, X2)
3.87/1.75	   a__div(X1, X2) -> div(X1, X2)
3.87/1.75	   a__if(X1, X2, X3) -> if(X1, X2, X3)
3.87/1.75	
3.87/1.75	S is empty.
3.87/1.75	Rewrite Strategy: FULL
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.87/1.75	Transformed a relative TRS into a decreasing-loop problem.
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(2)
3.87/1.75	Obligation:
3.87/1.75	Analyzing the following TRS for decreasing loops:
3.87/1.75	
3.87/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.87/1.75	
3.87/1.75	
3.87/1.75	The TRS R consists of the following rules:
3.87/1.75	
3.87/1.75	   a__minus(0, Y) -> 0
3.87/1.75	   a__minus(s(X), s(Y)) -> a__minus(X, Y)
3.87/1.75	   a__geq(X, 0) -> true
3.87/1.75	   a__geq(0, s(Y)) -> false
3.87/1.75	   a__geq(s(X), s(Y)) -> a__geq(X, Y)
3.87/1.75	   a__div(0, s(Y)) -> 0
3.87/1.75	   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
3.87/1.75	   a__if(true, X, Y) -> mark(X)
3.87/1.75	   a__if(false, X, Y) -> mark(Y)
3.87/1.75	   mark(minus(X1, X2)) -> a__minus(X1, X2)
3.87/1.75	   mark(geq(X1, X2)) -> a__geq(X1, X2)
3.87/1.75	   mark(div(X1, X2)) -> a__div(mark(X1), X2)
3.87/1.75	   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
3.87/1.75	   mark(0) -> 0
3.87/1.75	   mark(s(X)) -> s(mark(X))
3.87/1.75	   mark(true) -> true
3.87/1.75	   mark(false) -> false
3.87/1.75	   a__minus(X1, X2) -> minus(X1, X2)
3.87/1.75	   a__geq(X1, X2) -> geq(X1, X2)
3.87/1.75	   a__div(X1, X2) -> div(X1, X2)
3.87/1.75	   a__if(X1, X2, X3) -> if(X1, X2, X3)
3.87/1.75	
3.87/1.75	S is empty.
3.87/1.75	Rewrite Strategy: FULL
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(3) DecreasingLoopProof (LOWER BOUND(ID))
3.87/1.75	The following loop(s) give(s) rise to the lower bound Omega(n^1):
3.87/1.75	
3.87/1.75	The rewrite sequence
3.87/1.75	
3.87/1.75	a__geq(s(X), s(Y)) ->^+ a__geq(X, Y)
3.87/1.75	
3.87/1.75	gives rise to a decreasing loop by considering the right hand sides subterm at position [].
3.87/1.75	
3.87/1.75	The pumping substitution is [X / s(X), Y / s(Y)].
3.87/1.75	
3.87/1.75	The result substitution is [ ].
3.87/1.75	
3.87/1.75	
3.87/1.75	
3.87/1.75	
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(4)
3.87/1.75	Complex Obligation (BEST)
3.87/1.75	
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(5)
3.87/1.75	Obligation:
3.87/1.75	Proved the lower bound n^1 for the following obligation:
3.87/1.75	
3.87/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.87/1.75	
3.87/1.75	
3.87/1.75	The TRS R consists of the following rules:
3.87/1.75	
3.87/1.75	   a__minus(0, Y) -> 0
3.87/1.75	   a__minus(s(X), s(Y)) -> a__minus(X, Y)
3.87/1.75	   a__geq(X, 0) -> true
3.87/1.75	   a__geq(0, s(Y)) -> false
3.87/1.75	   a__geq(s(X), s(Y)) -> a__geq(X, Y)
3.87/1.75	   a__div(0, s(Y)) -> 0
3.87/1.75	   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
3.87/1.75	   a__if(true, X, Y) -> mark(X)
3.87/1.75	   a__if(false, X, Y) -> mark(Y)
3.87/1.75	   mark(minus(X1, X2)) -> a__minus(X1, X2)
3.87/1.75	   mark(geq(X1, X2)) -> a__geq(X1, X2)
3.87/1.75	   mark(div(X1, X2)) -> a__div(mark(X1), X2)
3.87/1.75	   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
3.87/1.75	   mark(0) -> 0
3.87/1.75	   mark(s(X)) -> s(mark(X))
3.87/1.75	   mark(true) -> true
3.87/1.75	   mark(false) -> false
3.87/1.75	   a__minus(X1, X2) -> minus(X1, X2)
3.87/1.75	   a__geq(X1, X2) -> geq(X1, X2)
3.87/1.75	   a__div(X1, X2) -> div(X1, X2)
3.87/1.75	   a__if(X1, X2, X3) -> if(X1, X2, X3)
3.87/1.75	
3.87/1.75	S is empty.
3.87/1.75	Rewrite Strategy: FULL
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(6) LowerBoundPropagationProof (FINISHED)
3.87/1.75	Propagated lower bound.
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(7)
3.87/1.75	BOUNDS(n^1, INF)
3.87/1.75	
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(8)
3.87/1.75	Obligation:
3.87/1.75	Analyzing the following TRS for decreasing loops:
3.87/1.75	
3.87/1.75	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.87/1.75	
3.87/1.75	
3.87/1.75	The TRS R consists of the following rules:
3.87/1.75	
3.87/1.75	   a__minus(0, Y) -> 0
3.87/1.75	   a__minus(s(X), s(Y)) -> a__minus(X, Y)
3.87/1.75	   a__geq(X, 0) -> true
3.87/1.75	   a__geq(0, s(Y)) -> false
3.87/1.75	   a__geq(s(X), s(Y)) -> a__geq(X, Y)
3.87/1.75	   a__div(0, s(Y)) -> 0
3.87/1.75	   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
3.87/1.75	   a__if(true, X, Y) -> mark(X)
3.87/1.75	   a__if(false, X, Y) -> mark(Y)
3.87/1.75	   mark(minus(X1, X2)) -> a__minus(X1, X2)
3.87/1.75	   mark(geq(X1, X2)) -> a__geq(X1, X2)
3.87/1.75	   mark(div(X1, X2)) -> a__div(mark(X1), X2)
3.87/1.75	   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
3.87/1.75	   mark(0) -> 0
3.87/1.75	   mark(s(X)) -> s(mark(X))
3.87/1.75	   mark(true) -> true
3.87/1.75	   mark(false) -> false
3.87/1.75	   a__minus(X1, X2) -> minus(X1, X2)
3.87/1.75	   a__geq(X1, X2) -> geq(X1, X2)
3.87/1.75	   a__div(X1, X2) -> div(X1, X2)
3.87/1.75	   a__if(X1, X2, X3) -> if(X1, X2, X3)
3.87/1.75	
3.87/1.75	S is empty.
3.87/1.75	Rewrite Strategy: FULL
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(9) DecreasingLoopProof (FINISHED)
3.87/1.75	The following loop(s) give(s) rise to the lower bound EXP:
3.87/1.75	
3.87/1.75	The rewrite sequence
3.87/1.75	
3.87/1.75	mark(div(s(X1_0), s(Y2_1))) ->^+ a__if(a__geq(mark(X1_0), Y2_1), s(div(minus(mark(X1_0), Y2_1), s(Y2_1))), 0)
3.87/1.75	
3.87/1.75	gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
3.87/1.75	
3.87/1.75	The pumping substitution is [X1_0 / div(s(X1_0), s(Y2_1))].
3.87/1.75	
3.87/1.75	The result substitution is [ ].
3.87/1.75	
3.87/1.75	
3.87/1.75	
3.87/1.75	The rewrite sequence
3.87/1.75	
3.87/1.75	mark(div(s(X1_0), s(Y2_1))) ->^+ a__if(a__geq(mark(X1_0), Y2_1), s(div(minus(mark(X1_0), Y2_1), s(Y2_1))), 0)
3.87/1.75	
3.87/1.75	gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0,0].
3.87/1.75	
3.87/1.75	The pumping substitution is [X1_0 / div(s(X1_0), s(Y2_1))].
3.87/1.75	
3.87/1.75	The result substitution is [ ].
3.87/1.75	
3.87/1.75	
3.87/1.75	
3.87/1.75	
3.87/1.75	----------------------------------------
3.87/1.75	
3.87/1.75	(10)
3.87/1.75	BOUNDS(EXP, INF)
3.87/1.78	EOF