3.87/1.78	WORST_CASE(?, O(1))
3.87/1.79	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
3.87/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.87/1.79	
3.87/1.79	
3.87/1.79	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).
3.87/1.79	
3.87/1.79	(0) CpxIntTrs
3.87/1.79	(1) Koat Proof [FINISHED, 84 ms]
3.87/1.79	(2) BOUNDS(1, 1)
3.87/1.79	
3.87/1.79	
3.87/1.79	----------------------------------------
3.87/1.79	
3.87/1.79	(0)
3.87/1.79	Obligation:
3.87/1.79	Complexity Int TRS consisting of the following rules:
3.87/1.79	eval(A) -> Com_1(eval(B)) :|: A >= 0 && B + 2 * A >= 10 && 10 >= 2 * A + B
3.87/1.79	start(A) -> Com_1(eval(A)) :|: TRUE
3.87/1.79	
3.87/1.79	The start-symbols are:[start_1]
3.87/1.79	
3.87/1.79	
3.87/1.79	----------------------------------------
3.87/1.79	
3.87/1.79	(1) Koat Proof (FINISHED)
3.87/1.79	YES(?, 5)
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	Initial complexity problem:
3.87/1.79	
3.87/1.79	1:	T:
3.87/1.79	
3.87/1.79			(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: ?, Cost: 1)    start(ar_0) -> Com_1(eval(ar_0))
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]
3.87/1.79	
3.87/1.79		start location:	koat_start
3.87/1.79	
3.87/1.79		leaf cost:	0
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	Repeatedly propagating knowledge in problem 1 produces the following problem:
3.87/1.79	
3.87/1.79	2:	T:
3.87/1.79	
3.87/1.79			(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 1)    start(ar_0) -> Com_1(eval(ar_0))
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]
3.87/1.79	
3.87/1.79		start location:	koat_start
3.87/1.79	
3.87/1.79		leaf cost:	0
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	By chaining the transition start(ar_0) -> Com_1(eval(ar_0)) with all transitions in problem 2, the following new transition is obtained:
3.87/1.79	
3.87/1.79		start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79	We thus obtain the following problem:
3.87/1.79	
3.87/1.79	3:	T:
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 2)    start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]
3.87/1.79	
3.87/1.79		start location:	koat_start
3.87/1.79	
3.87/1.79		leaf cost:	0
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	By chaining the transition start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] with all transitions in problem 3, the following new transition is obtained:
3.87/1.79	
3.87/1.79		start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ]
3.87/1.79	
3.87/1.79	We thus obtain the following problem:
3.87/1.79	
3.87/1.79	4:	T:
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 3)    start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]
3.87/1.79	
3.87/1.79		start location:	koat_start
3.87/1.79	
3.87/1.79		leaf cost:	0
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	By chaining the transition start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ] with all transitions in problem 4, the following new transition is obtained:
3.87/1.79	
3.87/1.79		start(ar_0) -> Com_1(eval(-8*ar_0 + 30)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 ]
3.87/1.79	
3.87/1.79	We thus obtain the following problem:
3.87/1.79	
3.87/1.79	5:	T:
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 4)    start(ar_0) -> Com_1(eval(-8*ar_0 + 30)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]
3.87/1.79	
3.87/1.79		start location:	koat_start
3.87/1.79	
3.87/1.79		leaf cost:	0
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	By chaining the transition start(ar_0) -> Com_1(eval(-8*ar_0 + 30)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 ] with all transitions in problem 5, the following new transition is obtained:
3.87/1.79	
3.87/1.79		start(ar_0) -> Com_1(eval(16*ar_0 - 50)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 /\ -8*ar_0 + 30 >= 0 ]
3.87/1.79	
3.87/1.79	We thus obtain the following problem:
3.87/1.79	
3.87/1.79	6:	T:
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 5)    start(ar_0) -> Com_1(eval(16*ar_0 - 50)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 /\ -8*ar_0 + 30 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]
3.87/1.79	
3.87/1.79		start location:	koat_start
3.87/1.79	
3.87/1.79		leaf cost:	0
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	Testing for reachability in the complexity graph removes the following transition from problem 6:
3.87/1.79	
3.87/1.79		eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]
3.87/1.79	
3.87/1.79	We thus obtain the following problem:
3.87/1.79	
3.87/1.79	7:	T:
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 5)    start(ar_0) -> Com_1(eval(16*ar_0 - 50)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 /\ -8*ar_0 + 30 >= 0 ]
3.87/1.79	
3.87/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]
3.87/1.79	
3.87/1.79		start location:	koat_start
3.87/1.79	
3.87/1.79		leaf cost:	0
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	Complexity upper bound 5
3.87/1.79	
3.87/1.79	
3.87/1.79	
3.87/1.79	Time: 0.196 sec (SMT: 0.188 sec)
3.87/1.79	
3.87/1.79	
3.87/1.79	----------------------------------------
3.87/1.79	
3.87/1.79	(2)
3.87/1.79	BOUNDS(1, 1)
3.87/1.81	EOF