/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files


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WORST_CASE(Omega(n^2), O(n^2))
proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, n^2).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 14 ms]
(2) BOUNDS(1, n^2)
(3) Loat Proof [FINISHED, 712 ms]
(4) BOUNDS(n^2, INF)


----------------------------------------

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
evalsipmabubblestart(A, B) -> Com_1(evalsipmabubbleentryin(A, B)) :|: TRUE
evalsipmabubbleentryin(A, B) -> Com_1(evalsipmabubblebb6in(A, B)) :|: TRUE
evalsipmabubblebb6in(A, B) -> Com_1(evalsipmabubblebb4in(A, 0)) :|: A >= 0
evalsipmabubblebb6in(A, B) -> Com_1(evalsipmabubblereturnin(A, B)) :|: 0 >= A + 1
evalsipmabubblebb4in(A, B) -> Com_1(evalsipmabubblebb1in(A, B)) :|: A >= 1 + B
evalsipmabubblebb4in(A, B) -> Com_1(evalsipmabubblebb5in(A, B)) :|: B >= A
evalsipmabubblebb1in(A, B) -> Com_1(evalsipmabubblebb2in(A, B)) :|: C >= D + 1
evalsipmabubblebb1in(A, B) -> Com_1(evalsipmabubblebb3in(A, B)) :|: D >= C
evalsipmabubblebb2in(A, B) -> Com_1(evalsipmabubblebb3in(A, B)) :|: TRUE
evalsipmabubblebb3in(A, B) -> Com_1(evalsipmabubblebb4in(A, B + 1)) :|: TRUE
evalsipmabubblebb5in(A, B) -> Com_1(evalsipmabubblebb6in(A - 1, B)) :|: TRUE
evalsipmabubblereturnin(A, B) -> Com_1(evalsipmabubblestop(A, B)) :|: TRUE

The start-symbols are:[evalsipmabubblestart_2]


----------------------------------------

(1) Koat Proof (FINISHED)
YES(?, 137*Ar_0 + 18*Ar_0^2 + 125)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 1 produces the following problem:

2:	T:

		(Comp: 1, Cost: 1)    evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalsipmabubblestart) = 2

	Pol(evalsipmabubbleentryin) = 2

	Pol(evalsipmabubblebb6in) = 2

	Pol(evalsipmabubblebb4in) = 2

	Pol(evalsipmabubblereturnin) = 1

	Pol(evalsipmabubblebb1in) = 2

	Pol(evalsipmabubblebb5in) = 2

	Pol(evalsipmabubblebb2in) = 2

	Pol(evalsipmabubblebb3in) = 2

	Pol(evalsipmabubblestop) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

	evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

		(Comp: 2, Cost: 1)    evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

		(Comp: ?, Cost: 1)    evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

		(Comp: ?, Cost: 1)    evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

		(Comp: 2, Cost: 1)    evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalsipmabubblestart) = V_1 + 1

	Pol(evalsipmabubbleentryin) = V_1 + 1

	Pol(evalsipmabubblebb6in) = V_1 + 1

	Pol(evalsipmabubblebb4in) = V_1

	Pol(evalsipmabubblereturnin) = V_1

	Pol(evalsipmabubblebb1in) = V_1

	Pol(evalsipmabubblebb5in) = V_1

	Pol(evalsipmabubblebb2in) = V_1

	Pol(evalsipmabubblebb3in) = V_1

	Pol(evalsipmabubblestop) = V_1

	Pol(koat_start) = V_1 + 1

orients all transitions weakly and the transition

	evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)           evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)           evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))

		(Comp: Ar_0 + 1, Cost: 1)    evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

		(Comp: 2, Cost: 1)           evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)           evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)           evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)           evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

		(Comp: ?, Cost: 1)           evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

		(Comp: ?, Cost: 1)           evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)           evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

		(Comp: ?, Cost: 1)           evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

		(Comp: 2, Cost: 1)           evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)           koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalsipmabubblebb5in) = 1

	Pol(evalsipmabubblebb6in) = 0

	Pol(evalsipmabubblebb4in) = 2

	Pol(evalsipmabubblebb1in) = 2

	Pol(evalsipmabubblebb3in) = 2

	Pol(evalsipmabubblebb2in) = 2

and size complexities

	S("koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-0) = Ar_0

	S("koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-1) = Ar_1

	S("evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))", 0-0) = ?

	S("evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))", 0-1) = ?

	S("evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))", 0-0) = ?

	S("evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))", 0-1) = ?

	S("evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))", 0-0) = ?

	S("evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))", 0-1) = ?

	S("evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))", 0-0) = ?

	S("evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))", 0-1) = ?

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]", 0-0) = ?

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]", 0-1) = ?

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]", 0-0) = ?

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]", 0-1) = ?

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]", 0-0) = ?

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]", 0-1) = ?

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]", 0-0) = ?

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]", 0-1) = ?

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]", 0-0) = ?

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]", 0-1) = ?

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]", 0-0) = ?

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]", 0-1) = 0

	S("evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))", 0-0) = Ar_0

	S("evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))", 0-1) = Ar_1

	S("evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))", 0-0) = Ar_0

	S("evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))", 0-1) = Ar_1

orients the transitions

	evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

	evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

	evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

	evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

	evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

	evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

	evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

weakly and the transitions

	evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

	evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 1)             evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)             evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))

		(Comp: Ar_0 + 1, Cost: 1)      evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

		(Comp: 2, Cost: 1)             evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)             evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: 2*Ar_0 + 2, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)             evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

		(Comp: ?, Cost: 1)             evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

		(Comp: ?, Cost: 1)             evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)             evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

		(Comp: 2*Ar_0 + 2, Cost: 1)    evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

		(Comp: 2, Cost: 1)             evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)             koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalsipmabubblebb4in) = V_1 - V_2 + 1

	Pol(evalsipmabubblebb1in) = V_1 - V_2

	Pol(evalsipmabubblebb3in) = V_1 - V_2

	Pol(evalsipmabubblebb2in) = V_1 - V_2

and size complexities

	S("koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-0) = Ar_0

	S("koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-1) = Ar_1

	S("evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))", 0-0) = 3*Ar_0 + 162

	S("evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))", 0-1) = ?

	S("evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))", 0-1) = ?

	S("evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))", 0-1) = ?

	S("evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))", 0-1) = ?

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]", 0-1) = ?

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]", 0-1) = ?

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]", 0-1) = ?

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]", 0-1) = ?

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]", 0-0) = 3*Ar_0 + 54

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]", 0-1) = ?

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]", 0-0) = 3*Ar_0 + 18

	S("evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]", 0-1) = 0

	S("evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))", 0-0) = Ar_0

	S("evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))", 0-1) = Ar_1

	S("evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))", 0-0) = Ar_0

	S("evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))", 0-1) = Ar_1

orients the transitions

	evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

	evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

	evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

	evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

	evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

weakly and the transition

	evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

strictly and produces the following problem:

6:	T:

		(Comp: 1, Cost: 1)                          evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)                          evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))

		(Comp: Ar_0 + 1, Cost: 1)                   evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

		(Comp: 2, Cost: 1)                          evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

		(Comp: 3*Ar_0^2 + 22*Ar_0 + 19, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: 2*Ar_0 + 2, Cost: 1)                 evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)                          evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

		(Comp: ?, Cost: 1)                          evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

		(Comp: ?, Cost: 1)                          evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)                          evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

		(Comp: 2*Ar_0 + 2, Cost: 1)                 evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

		(Comp: 2, Cost: 1)                          evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)                          koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 6 produces the following problem:

7:	T:

		(Comp: 1, Cost: 1)                          evalsipmabubblestart(Ar_0, Ar_1) -> Com_1(evalsipmabubbleentryin(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)                          evalsipmabubbleentryin(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0, Ar_1))

		(Comp: Ar_0 + 1, Cost: 1)                   evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, 0)) [ Ar_0 >= 0 ]

		(Comp: 2, Cost: 1)                          evalsipmabubblebb6in(Ar_0, Ar_1) -> Com_1(evalsipmabubblereturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]

		(Comp: 3*Ar_0^2 + 22*Ar_0 + 19, Cost: 1)    evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: 2*Ar_0 + 2, Cost: 1)                 evalsipmabubblebb4in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb5in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 ]

		(Comp: 3*Ar_0^2 + 22*Ar_0 + 19, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb2in(Ar_0, Ar_1)) [ C >= D + 1 ]

		(Comp: 3*Ar_0^2 + 22*Ar_0 + 19, Cost: 1)    evalsipmabubblebb1in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1)) [ C >= D ]

		(Comp: 3*Ar_0^2 + 22*Ar_0 + 19, Cost: 1)    evalsipmabubblebb2in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb3in(Ar_0, Ar_1))

		(Comp: 6*Ar_0^2 + 44*Ar_0 + 38, Cost: 1)    evalsipmabubblebb3in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb4in(Ar_0, Ar_1 + 1))

		(Comp: 2*Ar_0 + 2, Cost: 1)                 evalsipmabubblebb5in(Ar_0, Ar_1) -> Com_1(evalsipmabubblebb6in(Ar_0 - 1, Ar_1))

		(Comp: 2, Cost: 1)                          evalsipmabubblereturnin(Ar_0, Ar_1) -> Com_1(evalsipmabubblestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)                          koat_start(Ar_0, Ar_1) -> Com_1(evalsipmabubblestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 137*Ar_0 + 18*Ar_0^2 + 125



Time: 0.078 sec (SMT: 0.066 sec)


----------------------------------------

(2)
BOUNDS(1, n^2)

----------------------------------------

(3) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: evalsipmabubblestart

      0: evalsipmabubblestart -> evalsipmabubbleentryin : [], cost: 1

      1: evalsipmabubbleentryin -> evalsipmabubblebb6in : [], cost: 1

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

      3: evalsipmabubblebb6in -> evalsipmabubblereturnin : [ 0>=1+A ], cost: 1

      4: evalsipmabubblebb4in -> evalsipmabubblebb1in : [ A>=1+B ], cost: 1

      5: evalsipmabubblebb4in -> evalsipmabubblebb5in : [ B>=A ], cost: 1

      6: evalsipmabubblebb1in -> evalsipmabubblebb2in : [ free>=1+free_1 ], cost: 1

      7: evalsipmabubblebb1in -> evalsipmabubblebb3in : [ free_2>=free_3 ], cost: 1

      8: evalsipmabubblebb2in -> evalsipmabubblebb3in : [], cost: 1

      9: evalsipmabubblebb3in -> evalsipmabubblebb4in : B'=1+B, [], cost: 1

     10: evalsipmabubblebb5in -> evalsipmabubblebb6in : A'=-1+A, [], cost: 1

     11: evalsipmabubblereturnin -> evalsipmabubblestop : [], cost: 1



Checking for constant complexity:

   The following rule is satisfiable with cost >= 1, yielding constant complexity:

      0: evalsipmabubblestart -> evalsipmabubbleentryin : [], cost: 1



Removed unreachable and leaf rules:

   Start location: evalsipmabubblestart

      0: evalsipmabubblestart -> evalsipmabubbleentryin : [], cost: 1

      1: evalsipmabubbleentryin -> evalsipmabubblebb6in : [], cost: 1

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

      4: evalsipmabubblebb4in -> evalsipmabubblebb1in : [ A>=1+B ], cost: 1

      5: evalsipmabubblebb4in -> evalsipmabubblebb5in : [ B>=A ], cost: 1

      6: evalsipmabubblebb1in -> evalsipmabubblebb2in : [ free>=1+free_1 ], cost: 1

      7: evalsipmabubblebb1in -> evalsipmabubblebb3in : [ free_2>=free_3 ], cost: 1

      8: evalsipmabubblebb2in -> evalsipmabubblebb3in : [], cost: 1

      9: evalsipmabubblebb3in -> evalsipmabubblebb4in : B'=1+B, [], cost: 1

     10: evalsipmabubblebb5in -> evalsipmabubblebb6in : A'=-1+A, [], cost: 1



Simplified all rules, resulting in:

   Start location: evalsipmabubblestart

      0: evalsipmabubblestart -> evalsipmabubbleentryin : [], cost: 1

      1: evalsipmabubbleentryin -> evalsipmabubblebb6in : [], cost: 1

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

      4: evalsipmabubblebb4in -> evalsipmabubblebb1in : [ A>=1+B ], cost: 1

      5: evalsipmabubblebb4in -> evalsipmabubblebb5in : [ B>=A ], cost: 1

      6: evalsipmabubblebb1in -> evalsipmabubblebb2in : [], cost: 1

      7: evalsipmabubblebb1in -> evalsipmabubblebb3in : [], cost: 1

      8: evalsipmabubblebb2in -> evalsipmabubblebb3in : [], cost: 1

      9: evalsipmabubblebb3in -> evalsipmabubblebb4in : B'=1+B, [], cost: 1

     10: evalsipmabubblebb5in -> evalsipmabubblebb6in : A'=-1+A, [], cost: 1



### Simplification by acceleration and chaining ###



Eliminated locations (on linear paths):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

      4: evalsipmabubblebb4in -> evalsipmabubblebb1in : [ A>=1+B ], cost: 1

     13: evalsipmabubblebb4in -> evalsipmabubblebb6in : A'=-1+A, [ B>=A ], cost: 2

      7: evalsipmabubblebb1in -> evalsipmabubblebb3in : [], cost: 1

     14: evalsipmabubblebb1in -> evalsipmabubblebb3in : [], cost: 2

      9: evalsipmabubblebb3in -> evalsipmabubblebb4in : B'=1+B, [], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

     13: evalsipmabubblebb4in -> evalsipmabubblebb6in : A'=-1+A, [ B>=A ], cost: 2

     15: evalsipmabubblebb4in -> evalsipmabubblebb3in : [ A>=1+B ], cost: 2

     16: evalsipmabubblebb4in -> evalsipmabubblebb3in : [ A>=1+B ], cost: 3

      9: evalsipmabubblebb3in -> evalsipmabubblebb4in : B'=1+B, [], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

     13: evalsipmabubblebb4in -> evalsipmabubblebb6in : A'=-1+A, [ B>=A ], cost: 2

     17: evalsipmabubblebb4in -> evalsipmabubblebb4in : B'=1+B, [ A>=1+B ], cost: 3

     18: evalsipmabubblebb4in -> evalsipmabubblebb4in : B'=1+B, [ A>=1+B ], cost: 4



Accelerating simple loops of location 3.

   Simplified some of the simple loops (and removed duplicate rules).

   Accelerating the following rules:

     18: evalsipmabubblebb4in -> evalsipmabubblebb4in : B'=1+B, [ A>=1+B ], cost: 4



   Accelerated rule 18 with metering function A-B, yielding the new rule 19.

   Removing the simple loops: 18.



Accelerated all simple loops using metering functions (where possible):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

     13: evalsipmabubblebb4in -> evalsipmabubblebb6in : A'=-1+A, [ B>=A ], cost: 2

     19: evalsipmabubblebb4in -> evalsipmabubblebb4in : B'=A, [ A>=1+B ], cost: 4*A-4*B



Chained accelerated rules (with incoming rules):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

      2: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=0, [ A>=0 ], cost: 1

     20: evalsipmabubblebb6in -> evalsipmabubblebb4in : B'=A, [ A>=1 ], cost: 1+4*A

     13: evalsipmabubblebb4in -> evalsipmabubblebb6in : A'=-1+A, [ B>=A ], cost: 2



Eliminated locations (on tree-shaped paths):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

     21: evalsipmabubblebb6in -> evalsipmabubblebb6in : A'=-1+A, B'=0, [ A>=0 && 0>=A ], cost: 3

     22: evalsipmabubblebb6in -> evalsipmabubblebb6in : A'=-1+A, B'=A, [ A>=1 ], cost: 3+4*A



Accelerating simple loops of location 2.

   Simplified some of the simple loops (and removed duplicate rules).

   Accelerating the following rules:

     21: evalsipmabubblebb6in -> evalsipmabubblebb6in : A'=-1+A, B'=0, [ -A==0 ], cost: 3

     22: evalsipmabubblebb6in -> evalsipmabubblebb6in : A'=-1+A, B'=A, [ A>=1 ], cost: 3+4*A



   Accelerated rule 21 with metering function 1+A, yielding the new rule 23.

   Accelerated rule 22 with metering function A, yielding the new rule 24.

   Removing the simple loops: 21 22.



Accelerated all simple loops using metering functions (where possible):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

     23: evalsipmabubblebb6in -> evalsipmabubblebb6in : A'=-1, B'=0, [ -A==0 ], cost: 3+3*A

     24: evalsipmabubblebb6in -> evalsipmabubblebb6in : A'=0, B'=1, [ A>=1 ], cost: 5*A+2*A^2



Chained accelerated rules (with incoming rules):

   Start location: evalsipmabubblestart

     12: evalsipmabubblestart -> evalsipmabubblebb6in : [], cost: 2

     25: evalsipmabubblestart -> evalsipmabubblebb6in : A'=-1, B'=0, [ -A==0 ], cost: 5+3*A

     26: evalsipmabubblestart -> evalsipmabubblebb6in : A'=0, B'=1, [ A>=1 ], cost: 2+5*A+2*A^2



Removed unreachable locations (and leaf rules with constant cost):

   Start location: evalsipmabubblestart

     25: evalsipmabubblestart -> evalsipmabubblebb6in : A'=-1, B'=0, [ -A==0 ], cost: 5+3*A

     26: evalsipmabubblestart -> evalsipmabubblebb6in : A'=0, B'=1, [ A>=1 ], cost: 2+5*A+2*A^2



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: evalsipmabubblestart

     25: evalsipmabubblestart -> evalsipmabubblebb6in : A'=-1, B'=0, [ -A==0 ], cost: 5+3*A

     26: evalsipmabubblestart -> evalsipmabubblebb6in : A'=0, B'=1, [ A>=1 ], cost: 2+5*A+2*A^2



Computing asymptotic complexity for rule 25

   Could not solve the limit problem.

   Resulting cost 0 has complexity: Unknown



Computing asymptotic complexity for rule 26

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      A (+/+!), 2+5*A+2*A^2 (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {A==n}

      resulting limit problem:

      [solved]



   Solution:

      A / n

   Resulting cost 2+2*n^2+5*n has complexity: Poly(n^2)



Found new complexity Poly(n^2).



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Poly(n^2)

   Cpx degree:  2

   Solved cost: 2+2*n^2+5*n

   Rule cost:   2+5*A+2*A^2

   Rule guard:  [ A>=1 ]



WORST_CASE(Omega(n^2),?)


----------------------------------------

(4)
BOUNDS(n^2, INF)