0.00/0.54	MAYBE
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.54	  f2#(I0, I1) -> f2#(1, I1) [0 = I0 /\ 0 <= I1 - 1]
0.00/0.54	  f2#(I2, I3) -> f2#(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
0.00/0.54	  f2#(I4, I5) -> f2#(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
0.00/0.54	  f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6]
0.00/0.54	  f1#(I8, I9) -> f2#(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
0.00/0.54	R = 
0.00/0.54	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.54	  f2(I0, I1) -> f2(1, I1) [0 = I0 /\ 0 <= I1 - 1]
0.00/0.54	  f2(I2, I3) -> f2(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
0.00/0.54	  f2(I4, I5) -> f2(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
0.00/0.54	  f2(I6, I7) -> f2(0, 0) [0 = I7 /\ 0 = I6]
0.00/0.54	  f1(I8, I9) -> f2(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
0.00/0.54	
0.00/0.54	The dependency graph for this problem is:
0.00/0.54	  0 -> 5
0.00/0.54	  1 -> 2, 3
0.00/0.54	  2 -> 1, 4
0.00/0.54	  3 -> 2, 3
0.00/0.54	  4 -> 4
0.00/0.54	  5 -> 1, 2, 3
0.00/0.54	Where:
0.00/0.54	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.54	  1) f2#(I0, I1) -> f2#(1, I1) [0 = I0 /\ 0 <= I1 - 1]
0.00/0.54	  2) f2#(I2, I3) -> f2#(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
0.00/0.54	  3) f2#(I4, I5) -> f2#(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
0.00/0.54	  4) f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6]
0.00/0.54	  5) f1#(I8, I9) -> f2#(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
0.00/0.54	
0.00/0.54	We have the following SCCs.
0.00/0.54	  { 1, 2, 3 }
0.00/0.54	  { 4 }
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6]
0.00/0.54	R = 
0.00/0.54	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.54	  f2(I0, I1) -> f2(1, I1) [0 = I0 /\ 0 <= I1 - 1]
0.00/0.54	  f2(I2, I3) -> f2(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
0.00/0.54	  f2(I4, I5) -> f2(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
0.00/0.54	  f2(I6, I7) -> f2(0, 0) [0 = I7 /\ 0 = I6]
0.00/0.54	  f1(I8, I9) -> f2(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
0.00/0.54	
0.00/3.52	EOF