3.35/1.72	WORST_CASE(?, O(1))
3.35/1.73	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
3.35/1.73	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.35/1.73	
3.35/1.73	
3.35/1.73	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).
3.35/1.73	
3.35/1.73	(0) CpxIntTrs
3.35/1.73	(1) Koat Proof [FINISHED, 22 ms]
3.35/1.73	(2) BOUNDS(1, 1)
3.35/1.73	
3.35/1.73	
3.35/1.73	----------------------------------------
3.35/1.73	
3.35/1.73	(0)
3.35/1.73	Obligation:
3.35/1.73	Complexity Int TRS consisting of the following rules:
3.35/1.73	f12(A, B, C, D, E, F) -> Com_1(f5(4, 0, 0, D, E, F)) :|: TRUE
3.35/1.73	f5(A, B, C, D, E, F) -> Com_1(f11(A, B, 0, 0, 0, F)) :|: B >= A && C >= 0 && C <= 0
3.35/1.73	f5(A, B, C, D, E, F) -> Com_1(f10(A, B, C, C, C, F)) :|: C >= 1
3.35/1.73	f5(A, B, C, D, E, F) -> Com_1(f10(A, B, C, C, C, F)) :|: 0 >= C + 1
3.35/1.73	f7(A, B, C, D, E, F) -> Com_1(f5(A, 1 + B, G, H, E, G)) :|: TRUE
3.35/1.73	f8(A, B, C, D, E, F) -> Com_1(f5(A, 1 + B, G, H, E, G)) :|: TRUE
3.35/1.73	f5(A, B, C, D, E, F) -> Com_1(f5(A, 1 + B, G, H, E, G)) :|: A >= 1 + B && C >= 0 && C <= 0
3.35/1.73	
3.35/1.73	The start-symbols are:[f12_6]
3.35/1.73	
3.35/1.73	
3.35/1.73	----------------------------------------
3.35/1.73	
3.35/1.73	(1) Koat Proof (FINISHED)
3.35/1.73	YES(?, 8)
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	Initial complexity problem:
3.35/1.73	
3.35/1.73	1:	T:
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f12(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(4, 0, 0, ar_3, ar_4, ar_5))
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f11(ar_0, ar_1, 0, 0, 0, ar_5)) [ ar_1 >= ar_0 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_2, ar_5)) [ ar_2 >= 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_2, ar_5)) [ 0 >= ar_2 + 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f7(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(ar_0, ar_1 + 1, g, h, ar_4, g))
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f8(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(ar_0, ar_1 + 1, g, h, ar_4, g))
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(ar_0, ar_1 + 1, g, h, ar_4, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f12(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ]
3.35/1.73	
3.35/1.73		start location:	koat_start
3.35/1.73	
3.35/1.73		leaf cost:	0
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0, ar_1, ar_2].
3.35/1.73	
3.35/1.73	We thus obtain the following problem:
3.35/1.73	
3.35/1.73	2:	T:
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g))
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g))
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ 0 >= ar_2 + 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ ar_2 >= 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f12(ar_0, ar_1, ar_2) -> Com_1(f5(4, 0, 0))
3.35/1.73	
3.35/1.73		start location:	koat_start
3.35/1.73	
3.35/1.73		leaf cost:	0
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	Testing for reachability in the complexity graph removes the following transitions from problem 2:
3.35/1.73	
3.35/1.73		f8(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g))
3.35/1.73	
3.35/1.73		f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g))
3.35/1.73	
3.35/1.73	We thus obtain the following problem:
3.35/1.73	
3.35/1.73	3:	T:
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ ar_2 >= 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ 0 >= ar_2 + 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f12(ar_0, ar_1, ar_2) -> Com_1(f5(4, 0, 0))
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.35/1.73	
3.35/1.73		start location:	koat_start
3.35/1.73	
3.35/1.73		leaf cost:	0
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	Repeatedly propagating knowledge in problem 3 produces the following problem:
3.35/1.73	
3.35/1.73	4:	T:
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ ar_2 >= 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ 0 >= ar_2 + 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f12(ar_0, ar_1, ar_2) -> Com_1(f5(4, 0, 0))
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.35/1.73	
3.35/1.73		start location:	koat_start
3.35/1.73	
3.35/1.73		leaf cost:	0
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	A polynomial rank function with
3.35/1.73	
3.35/1.73		Pol(f5) = 1
3.35/1.73	
3.35/1.73		Pol(f11) = 0
3.35/1.73	
3.35/1.73		Pol(f10) = 0
3.35/1.73	
3.35/1.73		Pol(f12) = 1
3.35/1.73	
3.35/1.73		Pol(koat_start) = 1
3.35/1.73	
3.35/1.73	orients all transitions weakly and the transitions
3.35/1.73	
3.35/1.73		f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73		f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ ar_2 >= 1 ]
3.35/1.73	
3.35/1.73		f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ 0 >= ar_2 + 1 ]
3.35/1.73	
3.35/1.73	strictly and produces the following problem:
3.35/1.73	
3.35/1.73	5:	T:
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ ar_2 >= 1 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ 0 >= ar_2 + 1 ]
3.35/1.73	
3.35/1.73			(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f12(ar_0, ar_1, ar_2) -> Com_1(f5(4, 0, 0))
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.35/1.73	
3.35/1.73		start location:	koat_start
3.35/1.73	
3.35/1.73		leaf cost:	0
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	A polynomial rank function with
3.35/1.73	
3.35/1.73		Pol(f5) = V_1 - V_2
3.35/1.73	
3.35/1.73		Pol(f11) = V_1 - V_2
3.35/1.73	
3.35/1.73		Pol(f10) = V_1 - V_2
3.35/1.73	
3.35/1.73		Pol(f12) = 4
3.35/1.73	
3.35/1.73		Pol(koat_start) = 4
3.35/1.73	
3.35/1.73	orients all transitions weakly and the transition
3.35/1.73	
3.35/1.73		f5(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73	strictly and produces the following problem:
3.35/1.73	
3.35/1.73	6:	T:
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ ar_2 >= 1 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ 0 >= ar_2 + 1 ]
3.35/1.73	
3.35/1.73			(Comp: 4, Cost: 1)    f5(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ]
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 1)    f12(ar_0, ar_1, ar_2) -> Com_1(f5(4, 0, 0))
3.35/1.73	
3.35/1.73			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.35/1.73	
3.35/1.73		start location:	koat_start
3.35/1.73	
3.35/1.73		leaf cost:	0
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	Complexity upper bound 8
3.35/1.73	
3.35/1.73	
3.35/1.73	
3.35/1.73	Time: 0.058 sec (SMT: 0.051 sec)
3.35/1.73	
3.35/1.73	
3.35/1.73	----------------------------------------
3.35/1.73	
3.35/1.73	(2)
3.35/1.73	BOUNDS(1, 1)
3.35/1.75	EOF