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


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WORST_CASE(?, O(1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 35 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval_nd_loop_start(v_0, v_x_0) -> Com_1(eval_nd_loop_bb0_in(v_0, v_x_0)) :|: TRUE
eval_nd_loop_bb0_in(v_0, v_x_0) -> Com_1(eval_nd_loop_0(v_0, v_x_0)) :|: TRUE
eval_nd_loop_0(v_0, v_x_0) -> Com_1(eval_nd_loop_1(v_0, v_x_0)) :|: TRUE
eval_nd_loop_1(v_0, v_x_0) -> Com_1(eval_nd_loop_2(v_0, v_x_0)) :|: TRUE
eval_nd_loop_2(v_0, v_x_0) -> Com_1(eval_nd_loop_3(v_0, v_x_0)) :|: TRUE
eval_nd_loop_3(v_0, v_x_0) -> Com_1(eval_nd_loop_bb1_in(v_0, 0)) :|: TRUE
eval_nd_loop_bb1_in(v_0, v_x_0) -> Com_1(eval_nd_loop_4(v_0, v_x_0)) :|: TRUE
eval_nd_loop_4(v_0, v_x_0) -> Com_1(eval_nd_loop_5(nondef_0, v_x_0)) :|: TRUE
eval_nd_loop_5(v_0, v_x_0) -> Com_1(eval_nd_loop_bb1_in(v_0, v_0)) :|: v_0 - v_x_0 <= 2 && v_0 - v_x_0 >= 1 && v_0 < 10
eval_nd_loop_5(v_0, v_x_0) -> Com_1(eval_nd_loop_bb2_in(v_0, v_x_0)) :|: v_0 - v_x_0 > 2
eval_nd_loop_5(v_0, v_x_0) -> Com_1(eval_nd_loop_bb2_in(v_0, v_x_0)) :|: v_0 - v_x_0 < 1
eval_nd_loop_5(v_0, v_x_0) -> Com_1(eval_nd_loop_bb2_in(v_0, v_x_0)) :|: v_0 >= 10
eval_nd_loop_bb2_in(v_0, v_x_0) -> Com_1(eval_nd_loop_stop(v_0, v_x_0)) :|: TRUE

The start-symbols are:[eval_nd_loop_start_2]


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(1) Koat Proof (FINISHED)
YES(?, 43)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    evalndloopstart(Ar_0, Ar_1) -> Com_1(evalndloopbb0in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloopbb0in(Ar_0, Ar_1) -> Com_1(evalndloop0(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop0(Ar_0, Ar_1) -> Com_1(evalndloop1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop1(Ar_0, Ar_1) -> Com_1(evalndloop2(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop2(Ar_0, Ar_1) -> Com_1(evalndloop3(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop3(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloopbb1in(Ar_0, Ar_1) -> Com_1(evalndloop4(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop4(Ar_0, Ar_1) -> Com_1(evalndloop5(Ar_0, Fresh_0))

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(Ar_1, Ar_1)) [ Ar_0 + 2 >= Ar_1 /\ Ar_1 >= Ar_0 + 1 /\ 9 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 3 ]

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= 10 ]

		(Comp: ?, Cost: 1)    evalndloopbb2in(Ar_0, Ar_1) -> Com_1(evalndloopstop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalndloopstart(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)    evalndloopstart(Ar_0, Ar_1) -> Com_1(evalndloopbb0in(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloopbb0in(Ar_0, Ar_1) -> Com_1(evalndloop0(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop0(Ar_0, Ar_1) -> Com_1(evalndloop1(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop1(Ar_0, Ar_1) -> Com_1(evalndloop2(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop2(Ar_0, Ar_1) -> Com_1(evalndloop3(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop3(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloopbb1in(Ar_0, Ar_1) -> Com_1(evalndloop4(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop4(Ar_0, Ar_1) -> Com_1(evalndloop5(Ar_0, Fresh_0))

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(Ar_1, Ar_1)) [ Ar_0 + 2 >= Ar_1 /\ Ar_1 >= Ar_0 + 1 /\ 9 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 3 ]

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= 10 ]

		(Comp: ?, Cost: 1)    evalndloopbb2in(Ar_0, Ar_1) -> Com_1(evalndloopstop(Ar_0, Ar_1))

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

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalndloopstart) = 2

	Pol(evalndloopbb0in) = 2

	Pol(evalndloop0) = 2

	Pol(evalndloop1) = 2

	Pol(evalndloop2) = 2

	Pol(evalndloop3) = 2

	Pol(evalndloopbb1in) = 2

	Pol(evalndloop4) = 2

	Pol(evalndloop5) = 2

	Pol(evalndloopbb2in) = 1

	Pol(evalndloopstop) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	evalndloopbb2in(Ar_0, Ar_1) -> Com_1(evalndloopstop(Ar_0, Ar_1))

	evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= 10 ]

	evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 3 ]

	evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    evalndloopstart(Ar_0, Ar_1) -> Com_1(evalndloopbb0in(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloopbb0in(Ar_0, Ar_1) -> Com_1(evalndloop0(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop0(Ar_0, Ar_1) -> Com_1(evalndloop1(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop1(Ar_0, Ar_1) -> Com_1(evalndloop2(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop2(Ar_0, Ar_1) -> Com_1(evalndloop3(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop3(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloopbb1in(Ar_0, Ar_1) -> Com_1(evalndloop4(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop4(Ar_0, Ar_1) -> Com_1(evalndloop5(Ar_0, Fresh_0))

		(Comp: ?, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(Ar_1, Ar_1)) [ Ar_0 + 2 >= Ar_1 /\ Ar_1 >= Ar_0 + 1 /\ 9 >= Ar_1 ]

		(Comp: 2, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 3 ]

		(Comp: 2, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: 2, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= 10 ]

		(Comp: 2, Cost: 1)    evalndloopbb2in(Ar_0, Ar_1) -> Com_1(evalndloopstop(Ar_0, Ar_1))

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

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalndloopstart) = 9

	Pol(evalndloopbb0in) = 9

	Pol(evalndloop0) = 9

	Pol(evalndloop1) = 9

	Pol(evalndloop2) = 9

	Pol(evalndloop3) = 9

	Pol(evalndloopbb1in) = -V_1 + 9

	Pol(evalndloop4) = -V_1 + 9

	Pol(evalndloop5) = -V_1 + 9

	Pol(evalndloopbb2in) = -V_1

	Pol(evalndloopstop) = -V_1

	Pol(koat_start) = 9

orients all transitions weakly and the transition

	evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(Ar_1, Ar_1)) [ Ar_0 + 2 >= Ar_1 /\ Ar_1 >= Ar_0 + 1 /\ 9 >= Ar_1 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)    evalndloopstart(Ar_0, Ar_1) -> Com_1(evalndloopbb0in(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloopbb0in(Ar_0, Ar_1) -> Com_1(evalndloop0(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop0(Ar_0, Ar_1) -> Com_1(evalndloop1(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop1(Ar_0, Ar_1) -> Com_1(evalndloop2(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop2(Ar_0, Ar_1) -> Com_1(evalndloop3(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalndloop3(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloopbb1in(Ar_0, Ar_1) -> Com_1(evalndloop4(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalndloop4(Ar_0, Ar_1) -> Com_1(evalndloop5(Ar_0, Fresh_0))

		(Comp: 9, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(Ar_1, Ar_1)) [ Ar_0 + 2 >= Ar_1 /\ Ar_1 >= Ar_0 + 1 /\ 9 >= Ar_1 ]

		(Comp: 2, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 3 ]

		(Comp: 2, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: 2, Cost: 1)    evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= 10 ]

		(Comp: 2, Cost: 1)    evalndloopbb2in(Ar_0, Ar_1) -> Com_1(evalndloopstop(Ar_0, Ar_1))

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

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 4 produces the following problem:

5:	T:

		(Comp: 1, Cost: 1)     evalndloopstart(Ar_0, Ar_1) -> Com_1(evalndloopbb0in(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)     evalndloopbb0in(Ar_0, Ar_1) -> Com_1(evalndloop0(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)     evalndloop0(Ar_0, Ar_1) -> Com_1(evalndloop1(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)     evalndloop1(Ar_0, Ar_1) -> Com_1(evalndloop2(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)     evalndloop2(Ar_0, Ar_1) -> Com_1(evalndloop3(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)     evalndloop3(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(0, Ar_1))

		(Comp: 10, Cost: 1)    evalndloopbb1in(Ar_0, Ar_1) -> Com_1(evalndloop4(Ar_0, Ar_1))

		(Comp: 10, Cost: 1)    evalndloop4(Ar_0, Ar_1) -> Com_1(evalndloop5(Ar_0, Fresh_0))

		(Comp: 9, Cost: 1)     evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb1in(Ar_1, Ar_1)) [ Ar_0 + 2 >= Ar_1 /\ Ar_1 >= Ar_0 + 1 /\ 9 >= Ar_1 ]

		(Comp: 2, Cost: 1)     evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 3 ]

		(Comp: 2, Cost: 1)     evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: 2, Cost: 1)     evalndloop5(Ar_0, Ar_1) -> Com_1(evalndloopbb2in(Ar_0, Ar_1)) [ Ar_1 >= 10 ]

		(Comp: 2, Cost: 1)     evalndloopbb2in(Ar_0, Ar_1) -> Com_1(evalndloopstop(Ar_0, Ar_1))

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

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 43



Time: 0.058 sec (SMT: 0.044 sec)


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(2)
BOUNDS(1, 1)