0.64/0.66	MAYBE
0.64/0.66	
0.64/0.66	DP problem for innermost termination.
0.64/0.66	P =
0.64/0.66	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.64/0.66	  f4#(I0, I1) -> f4#(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
0.64/0.66	  f3#(I3, I4) -> f3#(I3 - 2, I5) [1 <= I3 - 1]
0.64/0.66	  f2#(I6, I7) -> f4#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.64/0.66	  f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
0.64/0.66	  f3#(I15, I16) -> f4#(0, I17) [0 = I15]
0.64/0.66	  f2#(I18, I19) -> f3#(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
0.64/0.66	  f1#(I24, I25) -> f2#(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]
0.64/0.66	R = 
0.64/0.66	  init(x1, x2) -> f1(rnd1, rnd2) 
0.64/0.66	  f4(I0, I1) -> f4(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
0.64/0.66	  f3(I3, I4) -> f3(I3 - 2, I5) [1 <= I3 - 1]
0.64/0.66	  f2(I6, I7) -> f4(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.64/0.66	  f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
0.64/0.66	  f3(I15, I16) -> f4(0, I17) [0 = I15]
0.64/0.66	  f2(I18, I19) -> f3(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
0.64/0.66	  f1(I24, I25) -> f2(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]
0.64/0.66	
0.64/0.66	The dependency graph for this problem is:
0.64/0.66	  0 -> 4, 7
0.64/0.66	  1 -> 1
0.64/0.66	  2 -> 2, 5
0.64/0.66	  3 -> 1
0.64/0.66	  4 -> 3, 6
0.64/0.66	  5 -> 
0.64/0.66	  6 -> 2, 5
0.64/0.66	  7 -> 3, 6
0.64/0.66	Where:
0.64/0.66	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.64/0.66	  1) f4#(I0, I1) -> f4#(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
0.64/0.66	  2) f3#(I3, I4) -> f3#(I3 - 2, I5) [1 <= I3 - 1]
0.64/0.66	  3) f2#(I6, I7) -> f4#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.64/0.66	  4) f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
0.64/0.66	  5) f3#(I15, I16) -> f4#(0, I17) [0 = I15]
0.64/0.66	  6) f2#(I18, I19) -> f3#(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
0.64/0.66	  7) f1#(I24, I25) -> f2#(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]
0.64/0.66	
0.64/0.66	We have the following SCCs.
0.64/0.66	  { 2 }
0.64/0.66	  { 1 }
0.64/0.66	
0.64/0.66	DP problem for innermost termination.
0.64/0.66	P =
0.64/0.66	  f4#(I0, I1) -> f4#(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
0.64/0.66	R = 
0.64/0.66	  init(x1, x2) -> f1(rnd1, rnd2) 
0.64/0.66	  f4(I0, I1) -> f4(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
0.64/0.66	  f3(I3, I4) -> f3(I3 - 2, I5) [1 <= I3 - 1]
0.64/0.66	  f2(I6, I7) -> f4(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
0.64/0.66	  f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
0.64/0.66	  f3(I15, I16) -> f4(0, I17) [0 = I15]
0.64/0.66	  f2(I18, I19) -> f3(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
0.64/0.66	  f1(I24, I25) -> f2(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]
0.64/0.66	
0.64/3.64	EOF