4.91/2.38	WORST_CASE(Omega(n^1), O(n^1))
5.11/2.39	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
5.11/2.39	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
5.11/2.39	
5.11/2.39	
5.11/2.39	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).
5.11/2.39	
5.11/2.39	(0) CpxIntTrs
5.11/2.39	(1) Koat Proof [FINISHED, 189 ms]
5.11/2.39	(2) BOUNDS(1, n^1)
5.11/2.39	(3) Loat Proof [FINISHED, 708 ms]
5.11/2.39	(4) BOUNDS(n^1, INF)
5.11/2.39	
5.11/2.39	
5.11/2.39	----------------------------------------
5.11/2.39	
5.11/2.39	(0)
5.11/2.39	Obligation:
5.11/2.39	Complexity Int TRS consisting of the following rules:
5.11/2.39	evalEx6start(A, B, C) -> Com_1(evalEx6entryin(A, B, C)) :|: TRUE
5.11/2.39	evalEx6entryin(A, B, C) -> Com_1(evalEx6bb3in(B, A, C)) :|: TRUE
5.11/2.39	evalEx6bb3in(A, B, C) -> Com_1(evalEx6bbin(A, B, C)) :|: C >= B + 1
5.11/2.39	evalEx6bb3in(A, B, C) -> Com_1(evalEx6returnin(A, B, C)) :|: B >= C
5.11/2.39	evalEx6bbin(A, B, C) -> Com_1(evalEx6bb1in(A, B, C)) :|: A >= B + 1
5.11/2.39	evalEx6bbin(A, B, C) -> Com_1(evalEx6bb2in(A, B, C)) :|: B >= A
5.11/2.39	evalEx6bb1in(A, B, C) -> Com_1(evalEx6bb3in(A, B + 1, C)) :|: TRUE
5.11/2.39	evalEx6bb2in(A, B, C) -> Com_1(evalEx6bb3in(A + 1, B, C)) :|: TRUE
5.11/2.39	evalEx6returnin(A, B, C) -> Com_1(evalEx6stop(A, B, C)) :|: TRUE
5.11/2.39	
5.11/2.39	The start-symbols are:[evalEx6start_3]
5.11/2.39	
5.11/2.39	
5.11/2.39	----------------------------------------
5.11/2.39	
5.11/2.39	(1) Koat Proof (FINISHED)
5.11/2.39	YES(?, 6*ar_1 + 12*ar_2 + 6*ar_0 + 7)
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Initial complexity problem:
5.11/2.39	
5.11/2.39	1:	T:
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
5.11/2.39	
5.11/2.39		start location:	koat_start
5.11/2.39	
5.11/2.39		leaf cost:	0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Repeatedly propagating knowledge in problem 1 produces the following problem:
5.11/2.39	
5.11/2.39	2:	T:
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)    evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)    evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
5.11/2.39	
5.11/2.39		start location:	koat_start
5.11/2.39	
5.11/2.39		leaf cost:	0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	A polynomial rank function with
5.11/2.39	
5.11/2.39		Pol(evalEx6start) = 2
5.11/2.39	
5.11/2.39		Pol(evalEx6entryin) = 2
5.11/2.39	
5.11/2.39		Pol(evalEx6bb3in) = 2
5.11/2.39	
5.11/2.39		Pol(evalEx6bbin) = 2
5.11/2.39	
5.11/2.39		Pol(evalEx6returnin) = 1
5.11/2.39	
5.11/2.39		Pol(evalEx6bb1in) = 2
5.11/2.39	
5.11/2.39		Pol(evalEx6bb2in) = 2
5.11/2.39	
5.11/2.39		Pol(evalEx6stop) = 0
5.11/2.39	
5.11/2.39		Pol(koat_start) = 2
5.11/2.39	
5.11/2.39	orients all transitions weakly and the transitions
5.11/2.39	
5.11/2.39		evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39		evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39	strictly and produces the following problem:
5.11/2.39	
5.11/2.39	3:	T:
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)    evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)    evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)    evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
5.11/2.39	
5.11/2.39		start location:	koat_start
5.11/2.39	
5.11/2.39		leaf cost:	0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Applied AI with 'oct' on problem 3 to obtain the following invariants:
5.11/2.39	
5.11/2.39	  For symbol evalEx6bb1in: -X_2 + X_3 - 1 >= 0 /\ X_1 - X_2 - 1 >= 0
5.11/2.39	
5.11/2.39	  For symbol evalEx6bb2in: -X_2 + X_3 - 1 >= 0 /\ -X_1 + X_3 - 1 >= 0 /\ -X_1 + X_2 >= 0
5.11/2.39	
5.11/2.39	  For symbol evalEx6bbin: -X_2 + X_3 - 1 >= 0
5.11/2.39	
5.11/2.39	  For symbol evalEx6returnin: X_2 - X_3 >= 0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	This yielded the following problem:
5.11/2.39	
5.11/2.39	4:	T:
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)    evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)    evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)    evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39		start location:	koat_start
5.11/2.39	
5.11/2.39		leaf cost:	0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	A polynomial rank function with
5.11/2.39	
5.11/2.39		Pol(koat_start) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6start) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6returnin) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6stop) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6bb2in) = -2*V_1 + 2*V_3 - 1
5.11/2.39	
5.11/2.39		Pol(evalEx6bb3in) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6bb1in) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6bbin) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6entryin) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39	orients all transitions weakly and the transitions
5.11/2.39	
5.11/2.39		evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39		evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ]
5.11/2.39	
5.11/2.39	strictly and produces the following problem:
5.11/2.39	
5.11/2.39	5:	T:
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 0)                  koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)                  evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_1 + 2*ar_2, Cost: 1)    evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)                  evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_1 + 2*ar_2, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)                  evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)                  evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)                  evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)                  evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)                  evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39		start location:	koat_start
5.11/2.39	
5.11/2.39		leaf cost:	0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	A polynomial rank function with
5.11/2.39	
5.11/2.39		Pol(koat_start) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6start) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6returnin) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6stop) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6bb2in) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6bb3in) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6bb1in) = -2*V_2 + 2*V_3 - 1
5.11/2.39	
5.11/2.39		Pol(evalEx6bbin) = -2*V_2 + 2*V_3
5.11/2.39	
5.11/2.39		Pol(evalEx6entryin) = -2*V_1 + 2*V_3
5.11/2.39	
5.11/2.39	orients all transitions weakly and the transitions
5.11/2.39	
5.11/2.39		evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39		evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ]
5.11/2.39	
5.11/2.39	strictly and produces the following problem:
5.11/2.39	
5.11/2.39	6:	T:
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 0)                  koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)                  evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_1 + 2*ar_2, Cost: 1)    evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_0 + 2*ar_2, Cost: 1)    evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_1 + 2*ar_2, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_0 + 2*ar_2, Cost: 1)    evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)                  evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39			(Comp: ?, Cost: 1)                  evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)                  evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)                  evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39		start location:	koat_start
5.11/2.39	
5.11/2.39		leaf cost:	0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Repeatedly propagating knowledge in problem 6 produces the following problem:
5.11/2.39	
5.11/2.39	7:	T:
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 0)                               koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)                               evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_1 + 2*ar_2, Cost: 1)                 evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_0 + 2*ar_2, Cost: 1)                 evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_1 + 2*ar_2, Cost: 1)                 evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_0 + 2*ar_2, Cost: 1)                 evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 2, Cost: 1)                               evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ]
5.11/2.39	
5.11/2.39			(Comp: 2*ar_0 + 4*ar_2 + 2*ar_1 + 1, Cost: 1)    evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ]
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)                               evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2))
5.11/2.39	
5.11/2.39			(Comp: 1, Cost: 1)                               evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2))
5.11/2.39	
5.11/2.39		start location:	koat_start
5.11/2.39	
5.11/2.39		leaf cost:	0
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Complexity upper bound 6*ar_1 + 12*ar_2 + 6*ar_0 + 7
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Time: 0.192 sec (SMT: 0.172 sec)
5.11/2.39	
5.11/2.39	
5.11/2.39	----------------------------------------
5.11/2.39	
5.11/2.39	(2)
5.11/2.39	BOUNDS(1, n^1)
5.11/2.39	
5.11/2.39	----------------------------------------
5.11/2.39	
5.11/2.39	(3) Loat Proof (FINISHED)
5.11/2.39	
5.11/2.39	
5.11/2.39	### Pre-processing the ITS problem ###
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Initial linear ITS problem
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	      0: evalEx6start -> evalEx6entryin : [], cost: 1
5.11/2.39	
5.11/2.39	      1: evalEx6entryin -> evalEx6bb3in : A'=B, B'=A, [], cost: 1
5.11/2.39	
5.11/2.39	      2: evalEx6bb3in -> evalEx6bbin : [ C>=1+B ], cost: 1
5.11/2.39	
5.11/2.39	      3: evalEx6bb3in -> evalEx6returnin : [ B>=C ], cost: 1
5.11/2.39	
5.11/2.39	      4: evalEx6bbin -> evalEx6bb1in : [ A>=1+B ], cost: 1
5.11/2.39	
5.11/2.39	      5: evalEx6bbin -> evalEx6bb2in : [ B>=A ], cost: 1
5.11/2.39	
5.11/2.39	      6: evalEx6bb1in -> evalEx6bb3in : B'=1+B, [], cost: 1
5.11/2.39	
5.11/2.39	      7: evalEx6bb2in -> evalEx6bb3in : A'=1+A, [], cost: 1
5.11/2.39	
5.11/2.39	      8: evalEx6returnin -> evalEx6stop : [], cost: 1
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Removed unreachable and leaf rules:
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	      0: evalEx6start -> evalEx6entryin : [], cost: 1
5.11/2.39	
5.11/2.39	      1: evalEx6entryin -> evalEx6bb3in : A'=B, B'=A, [], cost: 1
5.11/2.39	
5.11/2.39	      2: evalEx6bb3in -> evalEx6bbin : [ C>=1+B ], cost: 1
5.11/2.39	
5.11/2.39	      4: evalEx6bbin -> evalEx6bb1in : [ A>=1+B ], cost: 1
5.11/2.39	
5.11/2.39	      5: evalEx6bbin -> evalEx6bb2in : [ B>=A ], cost: 1
5.11/2.39	
5.11/2.39	      6: evalEx6bb1in -> evalEx6bb3in : B'=1+B, [], cost: 1
5.11/2.39	
5.11/2.39	      7: evalEx6bb2in -> evalEx6bb3in : A'=1+A, [], cost: 1
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	### Simplification by acceleration and chaining ###
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Eliminated locations (on linear paths):
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	      9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2
5.11/2.39	
5.11/2.39	      2: evalEx6bb3in -> evalEx6bbin : [ C>=1+B ], cost: 1
5.11/2.39	
5.11/2.39	     10: evalEx6bbin -> evalEx6bb3in : B'=1+B, [ A>=1+B ], cost: 2
5.11/2.39	
5.11/2.39	     11: evalEx6bbin -> evalEx6bb3in : A'=1+A, [ B>=A ], cost: 2
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Eliminated locations (on tree-shaped paths):
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	      9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2
5.11/2.39	
5.11/2.39	     12: evalEx6bb3in -> evalEx6bb3in : B'=1+B, [ C>=1+B && A>=1+B ], cost: 3
5.11/2.39	
5.11/2.39	     13: evalEx6bb3in -> evalEx6bb3in : A'=1+A, [ C>=1+B && B>=A ], cost: 3
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Accelerating simple loops of location 2.
5.11/2.39	
5.11/2.39	   Accelerating the following rules:
5.11/2.39	
5.11/2.39	     12: evalEx6bb3in -> evalEx6bb3in : B'=1+B, [ C>=1+B && A>=1+B ], cost: 3
5.11/2.39	
5.11/2.39	     13: evalEx6bb3in -> evalEx6bb3in : A'=1+A, [ C>=1+B && B>=A ], cost: 3
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	   Accelerated rule 12 with backward acceleration, yielding the new rule 14.
5.11/2.39	
5.11/2.39	   Accelerated rule 12 with backward acceleration, yielding the new rule 15.
5.11/2.39	
5.11/2.39	   Accelerated rule 13 with metering function 1-A+B, yielding the new rule 16.
5.11/2.39	
5.11/2.39	   Removing the simple loops: 12 13.
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Accelerated all simple loops using metering functions (where possible):
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	      9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2
5.11/2.39	
5.11/2.39	     14: evalEx6bb3in -> evalEx6bb3in : B'=C, [ C>=1+B && A>=1+B && A>=C ], cost: 3*C-3*B
5.11/2.39	
5.11/2.39	     15: evalEx6bb3in -> evalEx6bb3in : B'=A, [ C>=1+B && A>=1+B && C>=A ], cost: 3*A-3*B
5.11/2.39	
5.11/2.39	     16: evalEx6bb3in -> evalEx6bb3in : A'=1+B, [ C>=1+B && B>=A ], cost: 3-3*A+3*B
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Chained accelerated rules (with incoming rules):
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	      9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2
5.11/2.39	
5.11/2.39	     17: evalEx6start -> evalEx6bb3in : A'=B, B'=C, [ C>=1+A && B>=1+A && B>=C ], cost: 2+3*C-3*A
5.11/2.39	
5.11/2.39	     18: evalEx6start -> evalEx6bb3in : A'=B, [ C>=1+A && B>=1+A && C>=B ], cost: 2-3*A+3*B
5.11/2.39	
5.11/2.39	     19: evalEx6start -> evalEx6bb3in : A'=1+A, B'=A, [ C>=1+A && A>=B ], cost: 5+3*A-3*B
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Removed unreachable locations (and leaf rules with constant cost):
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	     17: evalEx6start -> evalEx6bb3in : A'=B, B'=C, [ C>=1+A && B>=1+A && B>=C ], cost: 2+3*C-3*A
5.11/2.39	
5.11/2.39	     18: evalEx6start -> evalEx6bb3in : A'=B, [ C>=1+A && B>=1+A && C>=B ], cost: 2-3*A+3*B
5.11/2.39	
5.11/2.39	     19: evalEx6start -> evalEx6bb3in : A'=1+A, B'=A, [ C>=1+A && A>=B ], cost: 5+3*A-3*B
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	### Computing asymptotic complexity ###
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Fully simplified ITS problem
5.11/2.39	
5.11/2.39	   Start location: evalEx6start
5.11/2.39	
5.11/2.39	     17: evalEx6start -> evalEx6bb3in : A'=B, B'=C, [ C>=1+A && B>=1+A && B>=C ], cost: 2+3*C-3*A
5.11/2.39	
5.11/2.39	     18: evalEx6start -> evalEx6bb3in : A'=B, [ C>=1+A && B>=1+A && C>=B ], cost: 2-3*A+3*B
5.11/2.39	
5.11/2.39	     19: evalEx6start -> evalEx6bb3in : A'=1+A, B'=A, [ C>=1+A && A>=B ], cost: 5+3*A-3*B
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Computing asymptotic complexity for rule 17
5.11/2.39	
5.11/2.39	   Solved the limit problem by the following transformations:
5.11/2.39	
5.11/2.39	      Created initial limit problem:
5.11/2.39	
5.11/2.39	      1-C+B (+/+!), -A+B (+/+!), 2+3*C-3*A (+), C-A (+/+!) [not solved]
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	      removing all constraints (solved by SMT)
5.11/2.39	
5.11/2.39	      resulting limit problem: [solved]
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	      applying transformation rule (C) using substitution {C==0,A==-n,B==0}
5.11/2.39	
5.11/2.39	      resulting limit problem:
5.11/2.39	
5.11/2.39	      [solved]
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	   Solution:
5.11/2.39	
5.11/2.39	      C / 0
5.11/2.39	
5.11/2.39	      A / -n
5.11/2.39	
5.11/2.39	      B / 0
5.11/2.39	
5.11/2.39	   Resulting cost 2+3*n has complexity: Poly(n^1)
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Found new complexity Poly(n^1).
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	Obtained the following overall complexity (w.r.t. the length of the input n):
5.11/2.39	
5.11/2.39	   Complexity:  Poly(n^1)
5.11/2.39	
5.11/2.39	   Cpx degree:  1
5.11/2.39	
5.11/2.39	   Solved cost: 2+3*n
5.11/2.39	
5.11/2.39	   Rule cost:   2+3*C-3*A
5.11/2.39	
5.11/2.39	   Rule guard:  [ C>=1+A && B>=1+A && B>=C ]
5.11/2.39	
5.11/2.39	
5.11/2.39	
5.11/2.39	WORST_CASE(Omega(n^1),?)
5.11/2.39	
5.11/2.39	
5.11/2.39	----------------------------------------
5.11/2.39	
5.11/2.39	(4)
5.11/2.39	BOUNDS(n^1, INF)
5.11/2.41	EOF