0.00/0.40	MAYBE
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.40	  f2#(I0, I1) -> f2#(I0 - 1, I2) [5 <= I0 - 1]
0.00/0.40	  f2#(I3, I4) -> f2#(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 4]
0.00/0.40	  f2#(I6, I7) -> f2#(5, I8) [5 = I6]
0.00/0.40	  f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.40	  f2(I0, I1) -> f2(I0 - 1, I2) [5 <= I0 - 1]
0.00/0.40	  f2(I3, I4) -> f2(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 4]
0.00/0.40	  f2(I6, I7) -> f2(5, I8) [5 = I6]
0.00/0.40	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.40	
0.00/0.40	The dependency graph for this problem is:
0.00/0.40	  0 -> 4
0.00/0.40	  1 -> 1, 3
0.00/0.40	  2 -> 2
0.00/0.40	  3 -> 3
0.00/0.40	  4 -> 1, 2, 3
0.00/0.40	Where:
0.00/0.40	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.40	  1) f2#(I0, I1) -> f2#(I0 - 1, I2) [5 <= I0 - 1]
0.00/0.40	  2) f2#(I3, I4) -> f2#(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 4]
0.00/0.40	  3) f2#(I6, I7) -> f2#(5, I8) [5 = I6]
0.00/0.40	  4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.40	
0.00/0.40	We have the following SCCs.
0.00/0.40	  { 2 }
0.00/0.40	  { 1 }
0.00/0.40	  { 3 }
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  f2#(I6, I7) -> f2#(5, I8) [5 = I6]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.40	  f2(I0, I1) -> f2(I0 - 1, I2) [5 <= I0 - 1]
0.00/0.40	  f2(I3, I4) -> f2(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 4]
0.00/0.40	  f2(I6, I7) -> f2(5, I8) [5 = I6]
0.00/0.40	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.40	
0.00/3.38	EOF