0.94/1.03	YES
0.94/1.03	
0.94/1.03	DP problem for innermost termination.
0.94/1.03	P =
0.94/1.03	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.94/1.03	  f3#(I0, I1) -> f2#(I2, I3) [I0 - 2 * y1 = 0 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2]
0.94/1.03	  f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  f3#(I8, I9) -> f2#(I8 - 1, I10) [0 <= I8 - 1 /\ I8 - 2 * I11 = 1 /\ I8 - 2 * I11 <= 1 /\ 0 <= I8 - 2 * I11]
0.94/1.03	  f2#(I12, I13) -> f3#(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	  f1#(I16, I17) -> f2#(I18, I19) [0 <= I16 - 1 /\ -1 <= I17 - 1 /\ 0 <= I18 - 1]
0.94/1.03	R = 
0.94/1.03	  init(x1, x2) -> f1(rnd1, rnd2) 
0.94/1.03	  f3(I0, I1) -> f2(I2, I3) [I0 - 2 * y1 = 0 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2]
0.94/1.03	  f2(I4, I5) -> f3(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  f3(I8, I9) -> f2(I8 - 1, I10) [0 <= I8 - 1 /\ I8 - 2 * I11 = 1 /\ I8 - 2 * I11 <= 1 /\ 0 <= I8 - 2 * I11]
0.94/1.03	  f2(I12, I13) -> f3(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	  f1(I16, I17) -> f2(I18, I19) [0 <= I16 - 1 /\ -1 <= I17 - 1 /\ 0 <= I18 - 1]
0.94/1.03	
0.94/1.03	The dependency graph for this problem is:
0.94/1.03	  0 -> 5
0.94/1.03	  1 -> 2, 4
0.94/1.03	  2 -> 1
0.94/1.03	  3 -> 2
0.94/1.03	  4 -> 3
0.94/1.03	  5 -> 2, 4
0.94/1.03	Where:
0.94/1.03	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.94/1.03	  1) f3#(I0, I1) -> f2#(I2, I3) [I0 - 2 * y1 = 0 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2]
0.94/1.03	  2) f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  3) f3#(I8, I9) -> f2#(I8 - 1, I10) [0 <= I8 - 1 /\ I8 - 2 * I11 = 1 /\ I8 - 2 * I11 <= 1 /\ 0 <= I8 - 2 * I11]
0.94/1.03	  4) f2#(I12, I13) -> f3#(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	  5) f1#(I16, I17) -> f2#(I18, I19) [0 <= I16 - 1 /\ -1 <= I17 - 1 /\ 0 <= I18 - 1]
0.94/1.03	
0.94/1.03	We have the following SCCs.
0.94/1.03	  { 1, 2, 3, 4 }
0.94/1.03	
0.94/1.03	DP problem for innermost termination.
0.94/1.03	P =
0.94/1.03	  f3#(I0, I1) -> f2#(I2, I3) [I0 - 2 * y1 = 0 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2]
0.94/1.03	  f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  f3#(I8, I9) -> f2#(I8 - 1, I10) [0 <= I8 - 1 /\ I8 - 2 * I11 = 1 /\ I8 - 2 * I11 <= 1 /\ 0 <= I8 - 2 * I11]
0.94/1.03	  f2#(I12, I13) -> f3#(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	R = 
0.94/1.03	  init(x1, x2) -> f1(rnd1, rnd2) 
0.94/1.03	  f3(I0, I1) -> f2(I2, I3) [I0 - 2 * y1 = 0 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2]
0.94/1.03	  f2(I4, I5) -> f3(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  f3(I8, I9) -> f2(I8 - 1, I10) [0 <= I8 - 1 /\ I8 - 2 * I11 = 1 /\ I8 - 2 * I11 <= 1 /\ 0 <= I8 - 2 * I11]
0.94/1.03	  f2(I12, I13) -> f3(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	  f1(I16, I17) -> f2(I18, I19) [0 <= I16 - 1 /\ -1 <= I17 - 1 /\ 0 <= I18 - 1]
0.94/1.03	
0.94/1.03	We use the basic value criterion with the projection function NU:
0.94/1.03	  NU[f2#(z1,z2)] = z1
0.94/1.03	  NU[f3#(z1,z2)] = z1
0.94/1.03	
0.94/1.03	This gives the following inequalities:
0.94/1.03	  I0 - 2 * y1 = 0 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2  ==>  I0 >! I2
0.94/1.03	  I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1  ==>  I4 (>! \union =) I4
0.94/1.03	  0 <= I8 - 1 /\ I8 - 2 * I11 = 1 /\ I8 - 2 * I11 <= 1 /\ 0 <= I8 - 2 * I11  ==>  I8 >! I8 - 1
0.94/1.03	  I12 - 2 * I15 = 1 /\ 0 <= I12 - 1  ==>  I12 (>! \union =) I12
0.94/1.03	
0.94/1.03	We remove all the strictly oriented dependency pairs.
0.94/1.03	
0.94/1.03	DP problem for innermost termination.
0.94/1.03	P =
0.94/1.03	  f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  f2#(I12, I13) -> f3#(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	R = 
0.94/1.03	  init(x1, x2) -> f1(rnd1, rnd2) 
0.94/1.03	  f3(I0, I1) -> f2(I2, I3) [I0 - 2 * y1 = 0 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2]
0.94/1.03	  f2(I4, I5) -> f3(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  f3(I8, I9) -> f2(I8 - 1, I10) [0 <= I8 - 1 /\ I8 - 2 * I11 = 1 /\ I8 - 2 * I11 <= 1 /\ 0 <= I8 - 2 * I11]
0.94/1.03	  f2(I12, I13) -> f3(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	  f1(I16, I17) -> f2(I18, I19) [0 <= I16 - 1 /\ -1 <= I17 - 1 /\ 0 <= I18 - 1]
0.94/1.03	
0.94/1.03	The dependency graph for this problem is:
0.94/1.03	  2 -> 
0.94/1.03	  4 -> 
0.94/1.03	Where:
0.94/1.03	  2) f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 0 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1]
0.94/1.03	  4) f2#(I12, I13) -> f3#(I12, I14) [I12 - 2 * I15 = 1 /\ 0 <= I12 - 1]
0.94/1.03	
0.94/1.03	We have the following SCCs.
0.94/1.03	  
0.94/4.01	EOF