3.14/1.78	WORST_CASE(?, O(1))
3.51/1.79	proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat
3.51/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.51/1.79	
3.51/1.79	
3.51/1.79	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).
3.51/1.79	
3.51/1.79	(0) CpxIntTrs
3.51/1.79	(1) Koat Proof [FINISHED, 10 ms]
3.51/1.79	(2) BOUNDS(1, 1)
3.51/1.79	
3.51/1.79	
3.51/1.79	----------------------------------------
3.51/1.79	
3.51/1.79	(0)
3.51/1.79	Obligation:
3.51/1.79	Complexity Int TRS consisting of the following rules:
3.51/1.79	f11(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f54(A, B, B + A, D, E, F, G, H, I, J, K)) :|: A >= B
3.51/1.79	f11(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f54(A, B, B - A, D, E, F, G, H, I, J, K)) :|: B >= 1 + A
3.51/1.79	f38(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f11(L, B, C, D, E, D, D, H, I, J, K)) :|: D >= 1 + E
3.51/1.79	f38(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f11(L, B, C, D, E, D, D, M, I, J, K)) :|: E >= D
3.51/1.79	f0(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f38(1, 2, C, 1, 10, F, G, H, 10, 2, L)) :|: TRUE
3.51/1.79	
3.51/1.79	The start-symbols are:[f0_11]
3.51/1.79	
3.51/1.79	
3.51/1.79	----------------------------------------
3.51/1.79	
3.51/1.79	(1) Koat Proof (FINISHED)
3.51/1.79	YES(?, 4)
3.51/1.79	
3.51/1.79	
3.51/1.79	
3.51/1.79	Initial complexity problem:
3.51/1.79	
3.51/1.79	1:	T:
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f11(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10) -> Com_1(f54(ar_0, ar_1, ar_1 + ar_0, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10)) [ ar_0 >= ar_1 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f11(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10) -> Com_1(f54(ar_0, ar_1, ar_1 - ar_0, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10)) [ ar_1 >= ar_0 + 1 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f38(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10) -> Com_1(f11(l, ar_1, ar_2, ar_3, ar_4, ar_3, ar_3, ar_7, ar_8, ar_9, ar_10)) [ ar_3 >= ar_4 + 1 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f38(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10) -> Com_1(f11(l, ar_1, ar_2, ar_3, ar_4, ar_3, ar_3, m, ar_8, ar_9, ar_10)) [ ar_4 >= ar_3 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10) -> Com_1(f38(1, 2, ar_2, 1, 10, ar_5, ar_6, ar_7, 10, 2, l))
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10) -> Com_1(f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9, ar_10)) [ 0 <= 0 ]
3.51/1.79	
3.51/1.79		start location:	koat_start
3.51/1.79	
3.51/1.79		leaf cost:	0
3.51/1.79	
3.51/1.79	
3.51/1.79	
3.51/1.79	Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0, ar_1, ar_3, ar_4].
3.51/1.79	
3.51/1.79	We thus obtain the following problem:
3.51/1.79	
3.51/1.79	2:	T:
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_3, ar_4) -> Com_1(f0(ar_0, ar_1, ar_3, ar_4)) [ 0 <= 0 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_3, ar_4) -> Com_1(f38(1, 2, 1, 10))
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f38(ar_0, ar_1, ar_3, ar_4) -> Com_1(f11(l, ar_1, ar_3, ar_4)) [ ar_4 >= ar_3 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f38(ar_0, ar_1, ar_3, ar_4) -> Com_1(f11(l, ar_1, ar_3, ar_4)) [ ar_3 >= ar_4 + 1 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f11(ar_0, ar_1, ar_3, ar_4) -> Com_1(f54(ar_0, ar_1, ar_3, ar_4)) [ ar_1 >= ar_0 + 1 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f11(ar_0, ar_1, ar_3, ar_4) -> Com_1(f54(ar_0, ar_1, ar_3, ar_4)) [ ar_0 >= ar_1 ]
3.51/1.79	
3.51/1.79		start location:	koat_start
3.51/1.79	
3.51/1.79		leaf cost:	0
3.51/1.79	
3.51/1.79	
3.51/1.79	
3.51/1.79	Testing for reachability in the complexity graph removes the following transition from problem 2:
3.51/1.79	
3.51/1.79		f38(ar_0, ar_1, ar_3, ar_4) -> Com_1(f11(l, ar_1, ar_3, ar_4)) [ ar_3 >= ar_4 + 1 ]
3.51/1.79	
3.51/1.79	We thus obtain the following problem:
3.51/1.79	
3.51/1.79	3:	T:
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f11(ar_0, ar_1, ar_3, ar_4) -> Com_1(f54(ar_0, ar_1, ar_3, ar_4)) [ ar_0 >= ar_1 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f11(ar_0, ar_1, ar_3, ar_4) -> Com_1(f54(ar_0, ar_1, ar_3, ar_4)) [ ar_1 >= ar_0 + 1 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f38(ar_0, ar_1, ar_3, ar_4) -> Com_1(f11(l, ar_1, ar_3, ar_4)) [ ar_4 >= ar_3 ]
3.51/1.79	
3.51/1.79			(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_3, ar_4) -> Com_1(f38(1, 2, 1, 10))
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_3, ar_4) -> Com_1(f0(ar_0, ar_1, ar_3, ar_4)) [ 0 <= 0 ]
3.51/1.79	
3.51/1.79		start location:	koat_start
3.51/1.79	
3.51/1.79		leaf cost:	0
3.51/1.79	
3.51/1.79	
3.51/1.79	
3.51/1.79	Repeatedly propagating knowledge in problem 3 produces the following problem:
3.51/1.79	
3.51/1.79	4:	T:
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 1)    f11(ar_0, ar_1, ar_3, ar_4) -> Com_1(f54(ar_0, ar_1, ar_3, ar_4)) [ ar_0 >= ar_1 ]
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 1)    f11(ar_0, ar_1, ar_3, ar_4) -> Com_1(f54(ar_0, ar_1, ar_3, ar_4)) [ ar_1 >= ar_0 + 1 ]
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 1)    f38(ar_0, ar_1, ar_3, ar_4) -> Com_1(f11(l, ar_1, ar_3, ar_4)) [ ar_4 >= ar_3 ]
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_3, ar_4) -> Com_1(f38(1, 2, 1, 10))
3.51/1.79	
3.51/1.79			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_3, ar_4) -> Com_1(f0(ar_0, ar_1, ar_3, ar_4)) [ 0 <= 0 ]
3.51/1.79	
3.51/1.79		start location:	koat_start
3.51/1.79	
3.51/1.79		leaf cost:	0
3.51/1.79	
3.51/1.79	
3.51/1.79	
3.51/1.79	Complexity upper bound 4
3.51/1.79	
3.51/1.79	
3.51/1.79	
3.51/1.79	Time: 0.027 sec (SMT: 0.026 sec)
3.51/1.79	
3.51/1.79	
3.51/1.79	----------------------------------------
3.51/1.79	
3.51/1.79	(2)
3.51/1.79	BOUNDS(1, 1)
3.54/1.82	EOF