0.00/1.04	MAYBE
0.00/1.04	
0.00/1.04	DP problem for innermost termination.
0.00/1.04	P =
0.00/1.04	  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.00/1.04	  f2#(I0, I1, I2, I3) -> f2#(I4, I5, I6, I3 + 2) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ -1 <= y1 - 1 /\ -1 <= y2 - 1 /\ -1 <= I1 - 1 /\ 0 <= I1 + y2 + 1 /\ I0 - y1 = I4 /\ I1 + y2 + 1 = I5 /\ I0 - y1 - (I1 + y2 + 1) = I6]
0.00/1.04	  f1#(I7, I8, I9, I10) -> f2#(I11, I12, I13, 2) [I11 - I12 = I13 /\ 0 <= I7 - 1 /\ -1 <= I12 - 1 /\ -1 <= I8 - 1 /\ -1 <= I11 - 1]
0.00/1.04	R = 
0.00/1.04	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.00/1.04	  f2(I0, I1, I2, I3) -> f2(I4, I5, I6, I3 + 2) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ -1 <= y1 - 1 /\ -1 <= y2 - 1 /\ -1 <= I1 - 1 /\ 0 <= I1 + y2 + 1 /\ I0 - y1 = I4 /\ I1 + y2 + 1 = I5 /\ I0 - y1 - (I1 + y2 + 1) = I6]
0.00/1.04	  f1(I7, I8, I9, I10) -> f2(I11, I12, I13, 2) [I11 - I12 = I13 /\ 0 <= I7 - 1 /\ -1 <= I12 - 1 /\ -1 <= I8 - 1 /\ -1 <= I11 - 1]
0.00/1.04	
0.00/1.04	The dependency graph for this problem is:
0.00/1.04	  0 -> 2
0.00/1.04	  1 -> 1
0.00/1.04	  2 -> 1
0.00/1.04	Where:
0.00/1.04	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.00/1.04	  1) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I6, I3 + 2) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ -1 <= y1 - 1 /\ -1 <= y2 - 1 /\ -1 <= I1 - 1 /\ 0 <= I1 + y2 + 1 /\ I0 - y1 = I4 /\ I1 + y2 + 1 = I5 /\ I0 - y1 - (I1 + y2 + 1) = I6]
0.00/1.04	  2) f1#(I7, I8, I9, I10) -> f2#(I11, I12, I13, 2) [I11 - I12 = I13 /\ 0 <= I7 - 1 /\ -1 <= I12 - 1 /\ -1 <= I8 - 1 /\ -1 <= I11 - 1]
0.00/1.04	
0.00/1.04	We have the following SCCs.
0.00/1.04	  { 1 }
0.00/1.04	
0.00/1.04	DP problem for innermost termination.
0.00/1.04	P =
0.00/1.04	  f2#(I0, I1, I2, I3) -> f2#(I4, I5, I6, I3 + 2) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ -1 <= y1 - 1 /\ -1 <= y2 - 1 /\ -1 <= I1 - 1 /\ 0 <= I1 + y2 + 1 /\ I0 - y1 = I4 /\ I1 + y2 + 1 = I5 /\ I0 - y1 - (I1 + y2 + 1) = I6]
0.00/1.04	R = 
0.00/1.04	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.00/1.04	  f2(I0, I1, I2, I3) -> f2(I4, I5, I6, I3 + 2) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ -1 <= y1 - 1 /\ -1 <= y2 - 1 /\ -1 <= I1 - 1 /\ 0 <= I1 + y2 + 1 /\ I0 - y1 = I4 /\ I1 + y2 + 1 = I5 /\ I0 - y1 - (I1 + y2 + 1) = I6]
0.00/1.04	  f1(I7, I8, I9, I10) -> f2(I11, I12, I13, 2) [I11 - I12 = I13 /\ 0 <= I7 - 1 /\ -1 <= I12 - 1 /\ -1 <= I8 - 1 /\ -1 <= I11 - 1]
0.00/1.04	
0.61/1.04	EOF