0.00/0.56	MAYBE
0.00/0.56	
0.00/0.56	DP problem for innermost termination.
0.00/0.56	P =
0.00/0.56	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.56	  f3#(I0, I1) -> f3#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.56	  f3#(I3, I4) -> f3#(I3 - 1, I5) [I3 <= -1]
0.00/0.56	  f2#(I6, I7) -> f3#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.00/0.56	  f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1]
0.00/0.56	  f2#(I15, I16) -> f3#(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20]
0.00/0.56	  f1#(I21, I22) -> f2#(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.56	  f3(I0, I1) -> f3(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.56	  f3(I3, I4) -> f3(I3 - 1, I5) [I3 <= -1]
0.00/0.56	  f2(I6, I7) -> f3(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.00/0.56	  f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1]
0.00/0.56	  f2(I15, I16) -> f3(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20]
0.00/0.56	  f1(I21, I22) -> f2(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1]
0.00/0.56	
0.00/0.56	The dependency graph for this problem is:
0.00/0.56	  0 -> 4, 6
0.00/0.56	  1 -> 1
0.00/0.56	  2 -> 2
0.00/0.56	  3 -> 2
0.00/0.56	  4 -> 3, 5
0.00/0.56	  5 -> 1
0.00/0.56	  6 -> 3, 5
0.00/0.56	Where:
0.00/0.56	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.56	  1) f3#(I0, I1) -> f3#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.56	  2) f3#(I3, I4) -> f3#(I3 - 1, I5) [I3 <= -1]
0.00/0.56	  3) f2#(I6, I7) -> f3#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.00/0.56	  4) f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1]
0.00/0.56	  5) f2#(I15, I16) -> f3#(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20]
0.00/0.56	  6) f1#(I21, I22) -> f2#(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1]
0.00/0.56	
0.00/0.56	We have the following SCCs.
0.00/0.56	  { 1 }
0.00/0.56	  { 2 }
0.00/0.56	
0.00/0.56	DP problem for innermost termination.
0.00/0.56	P =
0.00/0.56	  f3#(I3, I4) -> f3#(I3 - 1, I5) [I3 <= -1]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.56	  f3(I0, I1) -> f3(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.56	  f3(I3, I4) -> f3(I3 - 1, I5) [I3 <= -1]
0.00/0.56	  f2(I6, I7) -> f3(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.00/0.56	  f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1]
0.00/0.56	  f2(I15, I16) -> f3(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20]
0.00/0.56	  f1(I21, I22) -> f2(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1]
0.00/0.56	
0.00/3.54	EOF