0.00/0.44	MAYBE
0.00/0.44	
0.00/0.44	DP problem for innermost termination.
0.00/0.44	P =
0.00/0.44	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.44	  f2#(I0, I1) -> f2#(35, I2) [I0 <= 35 /\ 30 <= I0 - 1]
0.00/0.44	  f2#(I3, I4) -> f2#(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30]
0.00/0.44	  f2#(I6, I7) -> f2#(0, I8) [35 <= I6 - 1]
0.00/0.44	  f1#(I9, I10) -> f2#(I10 + 28, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ I10 <= I10 + 28 - 1]
0.00/0.44	R = 
0.00/0.44	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.44	  f2(I0, I1) -> f2(35, I2) [I0 <= 35 /\ 30 <= I0 - 1]
0.00/0.44	  f2(I3, I4) -> f2(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30]
0.00/0.44	  f2(I6, I7) -> f2(0, I8) [35 <= I6 - 1]
0.00/0.44	  f1(I9, I10) -> f2(I10 + 28, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ I10 <= I10 + 28 - 1]
0.00/0.44	
0.00/0.44	The dependency graph for this problem is:
0.00/0.44	  0 -> 4
0.00/0.44	  1 -> 1
0.00/0.44	  2 -> 2
0.00/0.44	  3 -> 
0.00/0.44	  4 -> 1, 2, 3
0.00/0.44	Where:
0.00/0.44	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.44	  1) f2#(I0, I1) -> f2#(35, I2) [I0 <= 35 /\ 30 <= I0 - 1]
0.00/0.44	  2) f2#(I3, I4) -> f2#(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30]
0.00/0.44	  3) f2#(I6, I7) -> f2#(0, I8) [35 <= I6 - 1]
0.00/0.44	  4) f1#(I9, I10) -> f2#(I10 + 28, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ I10 <= I10 + 28 - 1]
0.00/0.44	
0.00/0.44	We have the following SCCs.
0.00/0.44	  { 2 }
0.00/0.44	  { 1 }
0.00/0.44	
0.00/0.44	DP problem for innermost termination.
0.00/0.44	P =
0.00/0.44	  f2#(I0, I1) -> f2#(35, I2) [I0 <= 35 /\ 30 <= I0 - 1]
0.00/0.44	R = 
0.00/0.44	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.44	  f2(I0, I1) -> f2(35, I2) [I0 <= 35 /\ 30 <= I0 - 1]
0.00/0.44	  f2(I3, I4) -> f2(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30]
0.00/0.44	  f2(I6, I7) -> f2(0, I8) [35 <= I6 - 1]
0.00/0.44	  f1(I9, I10) -> f2(I10 + 28, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ I10 <= I10 + 28 - 1]
0.00/0.44	
0.00/3.42	EOF