3.96/1.94	WORST_CASE(Omega(n^1), O(n^1))
4.15/1.95	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
4.15/1.95	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.15/1.95	
4.15/1.95	
4.15/1.95	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).
4.15/1.95	
4.15/1.95	(0) CpxIntTrs
4.15/1.95	(1) Koat Proof [FINISHED, 120 ms]
4.15/1.95	(2) BOUNDS(1, n^1)
4.15/1.95	(3) Loat Proof [FINISHED, 325 ms]
4.15/1.95	(4) BOUNDS(n^1, INF)
4.15/1.95	
4.15/1.95	
4.15/1.95	----------------------------------------
4.15/1.95	
4.15/1.95	(0)
4.15/1.95	Obligation:
4.15/1.95	Complexity Int TRS consisting of the following rules:
4.15/1.95	f0(A, B, C, D) -> Com_1(f6(0, 0, C, D)) :|: TRUE
4.15/1.95	f6(A, B, C, D) -> Com_1(f6(A, B + 1, C, D)) :|: C >= B + 1
4.15/1.95	f6(A, B, C, D) -> Com_1(f6(A + 2, B + 1, C, D)) :|: C >= B + 1
4.15/1.95	f15(A, B, C, D) -> Com_1(f19(C + 1, B, C, 1)) :|: A >= C + 1 && A <= C + 1
4.15/1.95	f15(A, B, C, D) -> Com_1(f19(A, B, C, 0)) :|: C >= A
4.15/1.95	f15(A, B, C, D) -> Com_1(f19(A, B, C, 0)) :|: A >= 2 + C
4.15/1.95	f6(A, B, C, D) -> Com_1(f15(A, B, C, D)) :|: B >= C && C >= A + 1
4.15/1.95	f6(A, B, C, D) -> Com_1(f15(A, B, C, D)) :|: A >= 1 + C && B >= C
4.15/1.95	f6(A, B, C, D) -> Com_1(f19(A, B, A, 1)) :|: B >= C && A >= C && A <= C
4.15/1.95	
4.15/1.95	The start-symbols are:[f0_4]
4.15/1.95	
4.15/1.95	
4.15/1.95	----------------------------------------
4.15/1.95	
4.15/1.95	(1) Koat Proof (FINISHED)
4.15/1.95	YES(?, 2*ar_2 + 13)
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Initial complexity problem:
4.15/1.95	
4.15/1.95	1:	T:
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2, ar_3) -> Com_1(f6(0, 0, ar_2, ar_3))
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2, ar_3) -> Com_1(f6(ar_0, ar_1 + 1, ar_2, ar_3)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2, ar_3) -> Com_1(f6(ar_0 + 2, ar_1 + 1, ar_2, ar_3)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2, ar_3) -> Com_1(f19(ar_2 + 1, ar_1, ar_2, 1)) [ ar_0 = ar_2 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2, ar_3) -> Com_1(f19(ar_0, ar_1, ar_2, 0)) [ ar_2 >= ar_0 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2, ar_3) -> Com_1(f19(ar_0, ar_1, ar_2, 0)) [ ar_0 >= ar_2 + 2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2, ar_3) -> Com_1(f15(ar_0, ar_1, ar_2, ar_3)) [ ar_1 >= ar_2 /\ ar_2 >= ar_0 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2, ar_3) -> Com_1(f15(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_2 + 1 /\ ar_1 >= ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2, ar_3) -> Com_1(f19(ar_0, ar_1, ar_0, 1)) [ ar_1 >= ar_2 /\ ar_0 = ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.15/1.95	
4.15/1.95		start location:	koat_start
4.15/1.95	
4.15/1.95		leaf cost:	0
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0, ar_1, ar_2].
4.15/1.95	
4.15/1.95	We thus obtain the following problem:
4.15/1.95	
4.15/1.95	2:	T:
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_0)) [ ar_1 >= ar_2 /\ ar_0 = ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 /\ ar_1 >= ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 /\ ar_2 >= ar_0 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_2 + 1, ar_1, ar_2)) [ ar_0 = ar_2 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0 + 2, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f6(0, 0, ar_2))
4.15/1.95	
4.15/1.95		start location:	koat_start
4.15/1.95	
4.15/1.95		leaf cost:	0
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Repeatedly propagating knowledge in problem 2 produces the following problem:
4.15/1.95	
4.15/1.95	3:	T:
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_0)) [ ar_1 >= ar_2 /\ ar_0 = ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 /\ ar_1 >= ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 /\ ar_2 >= ar_0 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 2 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_2 + 1, ar_1, ar_2)) [ ar_0 = ar_2 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0 + 2, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f6(0, 0, ar_2))
4.15/1.95	
4.15/1.95		start location:	koat_start
4.15/1.95	
4.15/1.95		leaf cost:	0
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	A polynomial rank function with
4.15/1.95	
4.15/1.95		Pol(koat_start) = 2
4.15/1.95	
4.15/1.95		Pol(f0) = 2
4.15/1.95	
4.15/1.95		Pol(f6) = 2
4.15/1.95	
4.15/1.95		Pol(f19) = 0
4.15/1.95	
4.15/1.95		Pol(f15) = 1
4.15/1.95	
4.15/1.95	orients all transitions weakly and the transitions
4.15/1.95	
4.15/1.95		f6(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_0)) [ ar_1 >= ar_2 /\ ar_0 = ar_2 ]
4.15/1.95	
4.15/1.95		f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 /\ ar_2 >= ar_0 + 1 ]
4.15/1.95	
4.15/1.95		f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 /\ ar_1 >= ar_2 ]
4.15/1.95	
4.15/1.95		f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_2 + 1, ar_1, ar_2)) [ ar_0 = ar_2 + 1 ]
4.15/1.95	
4.15/1.95		f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ]
4.15/1.95	
4.15/1.95		f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 2 ]
4.15/1.95	
4.15/1.95	strictly and produces the following problem:
4.15/1.95	
4.15/1.95	4:	T:
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_0)) [ ar_1 >= ar_2 /\ ar_0 = ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 /\ ar_1 >= ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 /\ ar_2 >= ar_0 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 2 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)    f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_2 + 1, ar_1, ar_2)) [ ar_0 = ar_2 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0 + 2, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ?, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f6(0, 0, ar_2))
4.15/1.95	
4.15/1.95		start location:	koat_start
4.15/1.95	
4.15/1.95		leaf cost:	0
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	A polynomial rank function with
4.15/1.95	
4.15/1.95		Pol(koat_start) = V_3
4.15/1.95	
4.15/1.95		Pol(f0) = V_3
4.15/1.95	
4.15/1.95		Pol(f6) = -V_2 + V_3
4.15/1.95	
4.15/1.95		Pol(f19) = -V_2 + V_3
4.15/1.95	
4.15/1.95		Pol(f15) = -V_2 + V_3
4.15/1.95	
4.15/1.95	orients all transitions weakly and the transitions
4.15/1.95	
4.15/1.95		f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0 + 2, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95		f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95	strictly and produces the following problem:
4.15/1.95	
4.15/1.95	5:	T:
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 0)       koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)       f6(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_0)) [ ar_1 >= ar_2 /\ ar_0 = ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)       f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 /\ ar_1 >= ar_2 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)       f6(ar_0, ar_1, ar_2) -> Com_1(f15(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 /\ ar_2 >= ar_0 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)       f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 2 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)       f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ]
4.15/1.95	
4.15/1.95			(Comp: 2, Cost: 1)       f15(ar_0, ar_1, ar_2) -> Com_1(f19(ar_2 + 1, ar_1, ar_2)) [ ar_0 = ar_2 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ar_2, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0 + 2, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: ar_2, Cost: 1)    f6(ar_0, ar_1, ar_2) -> Com_1(f6(ar_0, ar_1 + 1, ar_2)) [ ar_2 >= ar_1 + 1 ]
4.15/1.95	
4.15/1.95			(Comp: 1, Cost: 1)       f0(ar_0, ar_1, ar_2) -> Com_1(f6(0, 0, ar_2))
4.15/1.95	
4.15/1.95		start location:	koat_start
4.15/1.95	
4.15/1.95		leaf cost:	0
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Complexity upper bound 2*ar_2 + 13
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Time: 0.116 sec (SMT: 0.106 sec)
4.15/1.95	
4.15/1.95	
4.15/1.95	----------------------------------------
4.15/1.95	
4.15/1.95	(2)
4.15/1.95	BOUNDS(1, n^1)
4.15/1.95	
4.15/1.95	----------------------------------------
4.15/1.95	
4.15/1.95	(3) Loat Proof (FINISHED)
4.15/1.95	
4.15/1.95	
4.15/1.95	### Pre-processing the ITS problem ###
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Initial linear ITS problem
4.15/1.95	
4.15/1.95	   Start location: f0
4.15/1.95	
4.15/1.95	      0: f0 -> f6 : A'=0, B'=0, [], cost: 1
4.15/1.95	
4.15/1.95	      1: f6 -> f6 : B'=1+B, [ C>=1+B ], cost: 1
4.15/1.95	
4.15/1.95	      2: f6 -> f6 : A'=2+A, B'=1+B, [ C>=1+B ], cost: 1
4.15/1.95	
4.15/1.95	      6: f6 -> f15 : [ B>=C && C>=1+A ], cost: 1
4.15/1.95	
4.15/1.95	      7: f6 -> f15 : [ A>=1+C && B>=C ], cost: 1
4.15/1.95	
4.15/1.95	      8: f6 -> f19 : C'=A, D'=1, [ B>=C && A==C ], cost: 1
4.15/1.95	
4.15/1.95	      3: f15 -> f19 : A'=1+C, D'=1, [ A==1+C ], cost: 1
4.15/1.95	
4.15/1.95	      4: f15 -> f19 : D'=0, [ C>=A ], cost: 1
4.15/1.95	
4.15/1.95	      5: f15 -> f19 : D'=0, [ A>=2+C ], cost: 1
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Removed unreachable and leaf rules:
4.15/1.95	
4.15/1.95	   Start location: f0
4.15/1.95	
4.15/1.95	      0: f0 -> f6 : A'=0, B'=0, [], cost: 1
4.15/1.95	
4.15/1.95	      1: f6 -> f6 : B'=1+B, [ C>=1+B ], cost: 1
4.15/1.95	
4.15/1.95	      2: f6 -> f6 : A'=2+A, B'=1+B, [ C>=1+B ], cost: 1
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	### Simplification by acceleration and chaining ###
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Accelerating simple loops of location 1.
4.15/1.95	
4.15/1.95	   Accelerating the following rules:
4.15/1.95	
4.15/1.95	      1: f6 -> f6 : B'=1+B, [ C>=1+B ], cost: 1
4.15/1.95	
4.15/1.95	      2: f6 -> f6 : A'=2+A, B'=1+B, [ C>=1+B ], cost: 1
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	   Accelerated rule 1 with metering function C-B, yielding the new rule 9.
4.15/1.95	
4.15/1.95	   Accelerated rule 2 with metering function C-B, yielding the new rule 10.
4.15/1.95	
4.15/1.95	   Removing the simple loops: 1 2.
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Accelerated all simple loops using metering functions (where possible):
4.15/1.95	
4.15/1.95	   Start location: f0
4.15/1.95	
4.15/1.95	      0: f0 -> f6 : A'=0, B'=0, [], cost: 1
4.15/1.95	
4.15/1.95	      9: f6 -> f6 : B'=C, [ C>=1+B ], cost: C-B
4.15/1.95	
4.15/1.95	     10: f6 -> f6 : A'=2*C+A-2*B, B'=C, [ C>=1+B ], cost: C-B
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Chained accelerated rules (with incoming rules):
4.15/1.95	
4.15/1.95	   Start location: f0
4.15/1.95	
4.15/1.95	      0: f0 -> f6 : A'=0, B'=0, [], cost: 1
4.15/1.95	
4.15/1.95	     11: f0 -> f6 : A'=0, B'=C, [ C>=1 ], cost: 1+C
4.15/1.95	
4.15/1.95	     12: f0 -> f6 : A'=2*C, B'=C, [ C>=1 ], cost: 1+C
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Removed unreachable locations (and leaf rules with constant cost):
4.15/1.95	
4.15/1.95	   Start location: f0
4.15/1.95	
4.15/1.95	     11: f0 -> f6 : A'=0, B'=C, [ C>=1 ], cost: 1+C
4.15/1.95	
4.15/1.95	     12: f0 -> f6 : A'=2*C, B'=C, [ C>=1 ], cost: 1+C
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	### Computing asymptotic complexity ###
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Fully simplified ITS problem
4.15/1.95	
4.15/1.95	   Start location: f0
4.15/1.95	
4.15/1.95	     12: f0 -> f6 : A'=2*C, B'=C, [ C>=1 ], cost: 1+C
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Computing asymptotic complexity for rule 12
4.15/1.95	
4.15/1.95	   Solved the limit problem by the following transformations:
4.15/1.95	
4.15/1.95	      Created initial limit problem:
4.15/1.95	
4.15/1.95	      1+C (+), C (+/+!) [not solved]
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	      removing all constraints (solved by SMT)
4.15/1.95	
4.15/1.95	      resulting limit problem: [solved]
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	      applying transformation rule (C) using substitution {C==n}
4.15/1.95	
4.15/1.95	      resulting limit problem:
4.15/1.95	
4.15/1.95	      [solved]
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	   Solution:
4.15/1.95	
4.15/1.95	      C / n
4.15/1.95	
4.15/1.95	   Resulting cost 1+n has complexity: Poly(n^1)
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Found new complexity Poly(n^1).
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	Obtained the following overall complexity (w.r.t. the length of the input n):
4.15/1.95	
4.15/1.95	   Complexity:  Poly(n^1)
4.15/1.95	
4.15/1.95	   Cpx degree:  1
4.15/1.95	
4.15/1.95	   Solved cost: 1+n
4.15/1.95	
4.15/1.95	   Rule cost:   1+C
4.15/1.95	
4.15/1.95	   Rule guard:  [ C>=1 ]
4.15/1.95	
4.15/1.95	
4.15/1.95	
4.15/1.95	WORST_CASE(Omega(n^1),?)
4.15/1.95	
4.15/1.95	
4.15/1.95	----------------------------------------
4.15/1.95	
4.15/1.95	(4)
4.15/1.95	BOUNDS(n^1, INF)
4.15/1.97	EOF