1.93/1.99	YES
1.93/1.99	
1.93/1.99	DP problem for innermost termination.
1.93/1.99	P =
1.93/1.99	  init#(x1, x2) -> f1#(rnd1, rnd2) 
1.93/1.99	  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	  f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9]
1.93/1.99	  f1#(I7, I8) -> f2#(0, I9) 
1.93/1.99	R = 
1.93/1.99	  init(x1, x2) -> f1(rnd1, rnd2) 
1.93/1.99	  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
1.93/1.99	  f1(I7, I8) -> f2(0, I9) 
1.93/1.99	
1.93/1.99	The dependency graph for this problem is:
1.93/1.99	  0 -> 4
1.93/1.99	  1 -> 1, 2
1.93/1.99	  2 -> 3
1.93/1.99	  3 -> 1
1.93/1.99	  4 -> 3
1.93/1.99	Where:
1.93/1.99	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
1.93/1.99	  1) f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  2) f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	  3) f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9]
1.93/1.99	  4) f1#(I7, I8) -> f2#(0, I9) 
1.93/1.99	
1.93/1.99	We have the following SCCs.
1.93/1.99	  { 1, 2, 3 }
1.93/1.99	
1.93/1.99	DP problem for innermost termination.
1.93/1.99	P =
1.93/1.99	  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	  f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9]
1.93/1.99	R = 
1.93/1.99	  init(x1, x2) -> f1(rnd1, rnd2) 
1.93/1.99	  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
1.93/1.99	  f1(I7, I8) -> f2(0, I9) 
1.93/1.99	
1.93/1.99	We use the reverse value criterion with the projection function NU:
1.93/1.99	  NU[f2#(z1,z2)] = 9 + -1 * z1
1.93/1.99	  NU[f3#(z1,z2)] = 9 + -1 * (z1 + 1)
1.93/1.99	
1.93/1.99	This gives the following inequalities:
1.93/1.99	  I1 <= 11  ==>  9 + -1 * (I0 + 1) >= 9 + -1 * (I0 + 1)
1.93/1.99	  11 <= I3 - 1  ==>  9 + -1 * (I2 + 1) >= 9 + -1 * (I2 + 1)
1.93/1.99	  I5 <= 9  ==>  9 + -1 * I5 > 9 + -1 * (I5 + 1)  with  9 + -1 * I5 >= 0
1.93/1.99	
1.93/1.99	We remove all the strictly oriented dependency pairs.
1.93/1.99	
1.93/1.99	DP problem for innermost termination.
1.93/1.99	P =
1.93/1.99	  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	R = 
1.93/1.99	  init(x1, x2) -> f1(rnd1, rnd2) 
1.93/1.99	  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
1.93/1.99	  f1(I7, I8) -> f2(0, I9) 
1.93/1.99	
1.93/1.99	The dependency graph for this problem is:
1.93/1.99	  1 -> 1, 2
1.93/1.99	  2 -> 
1.93/1.99	Where:
1.93/1.99	  1) f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  2) f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	
1.93/1.99	We have the following SCCs.
1.93/1.99	  { 1 }
1.93/1.99	
1.93/1.99	DP problem for innermost termination.
1.93/1.99	P =
1.93/1.99	  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
1.93/1.99	R = 
1.93/1.99	  init(x1, x2) -> f1(rnd1, rnd2) 
1.93/1.99	  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
1.93/1.99	  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
1.93/1.99	  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
1.93/1.99	  f1(I7, I8) -> f2(0, I9) 
1.93/1.99	
1.93/1.99	We use the reverse value criterion with the projection function NU:
1.93/1.99	  NU[f3#(z1,z2)] = 11 + -1 * z2
1.93/1.99	
1.93/1.99	This gives the following inequalities:
1.93/1.99	  I1 <= 11  ==>  11 + -1 * I1 > 11 + -1 * (I1 + 1)  with  11 + -1 * I1 >= 0
1.93/1.99	
1.93/1.99	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.93/4.97	EOF