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, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.00/0.56	  f2#(I0, I1, I2, I3) -> f2#(I0 + 1, I1 + 1, I2, I0 + 1 + I1 + 1) [-2 <= I1 - 1 /\ -2 <= I0 - 1 /\ I3 <= I2 - 1]
0.00/0.56	  f1#(I4, I5, I6, I7) -> f2#(I8, I9, I10, I11) [I8 + I9 = I11 /\ 0 <= I4 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ -1 <= I5 - 1 /\ -1 <= I10 - 1]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.00/0.56	  f2(I0, I1, I2, I3) -> f2(I0 + 1, I1 + 1, I2, I0 + 1 + I1 + 1) [-2 <= I1 - 1 /\ -2 <= I0 - 1 /\ I3 <= I2 - 1]
0.00/0.56	  f1(I4, I5, I6, I7) -> f2(I8, I9, I10, I11) [I8 + I9 = I11 /\ 0 <= I4 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ -1 <= I5 - 1 /\ -1 <= I10 - 1]
0.00/0.56	
0.00/0.56	The dependency graph for this problem is:
0.00/0.56	  0 -> 2
0.00/0.56	  1 -> 1
0.00/0.56	  2 -> 1
0.00/0.56	Where:
0.00/0.56	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.00/0.56	  1) f2#(I0, I1, I2, I3) -> f2#(I0 + 1, I1 + 1, I2, I0 + 1 + I1 + 1) [-2 <= I1 - 1 /\ -2 <= I0 - 1 /\ I3 <= I2 - 1]
0.00/0.56	  2) f1#(I4, I5, I6, I7) -> f2#(I8, I9, I10, I11) [I8 + I9 = I11 /\ 0 <= I4 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ -1 <= I5 - 1 /\ -1 <= I10 - 1]
0.00/0.56	
0.00/0.56	We have the following SCCs.
0.00/0.56	  { 1 }
0.00/0.56	
0.00/0.56	DP problem for innermost termination.
0.00/0.56	P =
0.00/0.56	  f2#(I0, I1, I2, I3) -> f2#(I0 + 1, I1 + 1, I2, I0 + 1 + I1 + 1) [-2 <= I1 - 1 /\ -2 <= I0 - 1 /\ I3 <= I2 - 1]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.00/0.56	  f2(I0, I1, I2, I3) -> f2(I0 + 1, I1 + 1, I2, I0 + 1 + I1 + 1) [-2 <= I1 - 1 /\ -2 <= I0 - 1 /\ I3 <= I2 - 1]
0.00/0.56	  f1(I4, I5, I6, I7) -> f2(I8, I9, I10, I11) [I8 + I9 = I11 /\ 0 <= I4 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ -1 <= I5 - 1 /\ -1 <= I10 - 1]
0.00/0.56	
0.00/3.54	EOF