0.66/0.69	MAYBE
0.66/0.69	
0.66/0.69	DP problem for innermost termination.
0.66/0.69	P =
0.66/0.69	  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.66/0.69	  f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
0.66/0.69	  f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
0.66/0.69	  f1#(I10, I11, I12, I13) -> f2#(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
0.66/0.69	  f1#(I16, I17, I18, I19) -> f2#(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
0.66/0.69	  f1#(I21, I22, I23, I24) -> f2#(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]
0.66/0.69	R = 
0.66/0.69	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.66/0.69	  f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
0.66/0.69	  f2(I6, I7, I8, I9) -> f2(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
0.66/0.69	  f1(I10, I11, I12, I13) -> f2(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
0.66/0.69	  f1(I16, I17, I18, I19) -> f2(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
0.66/0.69	  f1(I21, I22, I23, I24) -> f2(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]
0.66/0.69	
0.66/0.69	The dependency graph for this problem is:
0.66/0.69	  0 -> 3, 4, 5
0.66/0.69	  1 -> 1, 2
0.66/0.69	  2 -> 2
0.66/0.69	  3 -> 1, 2
0.66/0.69	  4 -> 
0.66/0.69	  5 -> 
0.66/0.69	Where:
0.66/0.69	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.66/0.69	  1) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
0.66/0.69	  2) f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
0.66/0.69	  3) f1#(I10, I11, I12, I13) -> f2#(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
0.66/0.69	  4) f1#(I16, I17, I18, I19) -> f2#(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
0.66/0.69	  5) f1#(I21, I22, I23, I24) -> f2#(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]
0.66/0.69	
0.66/0.69	We have the following SCCs.
0.66/0.69	  { 1 }
0.66/0.69	  { 2 }
0.66/0.69	
0.66/0.69	DP problem for innermost termination.
0.66/0.69	P =
0.66/0.69	  f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
0.66/0.69	R = 
0.66/0.69	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.66/0.69	  f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
0.66/0.69	  f2(I6, I7, I8, I9) -> f2(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
0.66/0.69	  f1(I10, I11, I12, I13) -> f2(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
0.66/0.69	  f1(I16, I17, I18, I19) -> f2(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
0.66/0.69	  f1(I21, I22, I23, I24) -> f2(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]
0.66/0.69	
0.66/3.67	EOF