1.29/1.32	YES
1.29/1.32	
1.29/1.32	DP problem for innermost termination.
1.29/1.32	P =
1.29/1.32	  init#(x1, x2) -> f1#(rnd1, rnd2) 
1.29/1.32	  f3#(I0, I1) -> f3#(I0, I1 - 1) [0 <= I1 - 1]
1.29/1.32	  f3#(I2, I3) -> f2#(0, I2 - 1) [0 = I3]
1.29/1.32	  f2#(I4, I5) -> f3#(I5, I4) [0 <= I5 - 1]
1.29/1.32	  f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I8 - 1]
1.29/1.32	R = 
1.29/1.32	  init(x1, x2) -> f1(rnd1, rnd2) 
1.29/1.32	  f3(I0, I1) -> f3(I0, I1 - 1) [0 <= I1 - 1]
1.29/1.32	  f3(I2, I3) -> f2(0, I2 - 1) [0 = I3]
1.29/1.32	  f2(I4, I5) -> f3(I5, I4) [0 <= I5 - 1]
1.29/1.32	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I8 - 1]
1.29/1.32	
1.29/1.32	The dependency graph for this problem is:
1.29/1.32	  0 -> 4
1.29/1.32	  1 -> 1, 2
1.29/1.32	  2 -> 3
1.29/1.32	  3 -> 1, 2
1.29/1.32	  4 -> 3
1.29/1.32	Where:
1.29/1.32	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
1.29/1.32	  1) f3#(I0, I1) -> f3#(I0, I1 - 1) [0 <= I1 - 1]
1.29/1.32	  2) f3#(I2, I3) -> f2#(0, I2 - 1) [0 = I3]
1.29/1.32	  3) f2#(I4, I5) -> f3#(I5, I4) [0 <= I5 - 1]
1.29/1.32	  4) f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I8 - 1]
1.29/1.32	
1.29/1.32	We have the following SCCs.
1.29/1.32	  { 1, 2, 3 }
1.29/1.32	
1.29/1.32	DP problem for innermost termination.
1.29/1.32	P =
1.29/1.32	  f3#(I0, I1) -> f3#(I0, I1 - 1) [0 <= I1 - 1]
1.29/1.32	  f3#(I2, I3) -> f2#(0, I2 - 1) [0 = I3]
1.29/1.32	  f2#(I4, I5) -> f3#(I5, I4) [0 <= I5 - 1]
1.29/1.32	R = 
1.29/1.32	  init(x1, x2) -> f1(rnd1, rnd2) 
1.29/1.32	  f3(I0, I1) -> f3(I0, I1 - 1) [0 <= I1 - 1]
1.29/1.32	  f3(I2, I3) -> f2(0, I2 - 1) [0 = I3]
1.29/1.32	  f2(I4, I5) -> f3(I5, I4) [0 <= I5 - 1]
1.29/1.32	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I8 - 1]
1.29/1.32	
1.29/1.32	We use the basic value criterion with the projection function NU:
1.29/1.32	  NU[f2#(z1,z2)] = z1
1.29/1.32	  NU[f3#(z1,z2)] = z2
1.29/1.32	
1.29/1.32	This gives the following inequalities:
1.29/1.32	  0 <= I1 - 1  ==>  I1 >! I1 - 1
1.29/1.32	  0 = I3  ==>  I3 (>! \union =) 0
1.29/1.32	  0 <= I5 - 1  ==>  I4 (>! \union =) I4
1.29/1.32	
1.29/1.32	We remove all the strictly oriented dependency pairs.
1.29/1.32	
1.29/1.32	DP problem for innermost termination.
1.29/1.32	P =
1.29/1.32	  f3#(I2, I3) -> f2#(0, I2 - 1) [0 = I3]
1.29/1.32	  f2#(I4, I5) -> f3#(I5, I4) [0 <= I5 - 1]
1.29/1.32	R = 
1.29/1.32	  init(x1, x2) -> f1(rnd1, rnd2) 
1.29/1.32	  f3(I0, I1) -> f3(I0, I1 - 1) [0 <= I1 - 1]
1.29/1.32	  f3(I2, I3) -> f2(0, I2 - 1) [0 = I3]
1.29/1.32	  f2(I4, I5) -> f3(I5, I4) [0 <= I5 - 1]
1.29/1.32	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I8 - 1]
1.29/1.32	
1.29/1.32	We use the reverse value criterion with the projection function NU:
1.29/1.32	  NU[f2#(z1,z2)] = z2 - 1 + -1 * 0
1.29/1.32	  NU[f3#(z1,z2)] = z1 - 1 - 1 + -1 * 0
1.29/1.32	
1.29/1.32	This gives the following inequalities:
1.29/1.32	  0 = I3  ==>  I2 - 1 - 1 + -1 * 0 >= I2 - 1 - 1 + -1 * 0
1.29/1.32	  0 <= I5 - 1  ==>  I5 - 1 + -1 * 0 > I5 - 1 - 1 + -1 * 0  with  I5 - 1 + -1 * 0 >= 0
1.29/1.32	
1.29/1.32	We remove all the strictly oriented dependency pairs.
1.29/1.32	
1.29/1.32	DP problem for innermost termination.
1.29/1.32	P =
1.29/1.32	  f3#(I2, I3) -> f2#(0, I2 - 1) [0 = I3]
1.29/1.32	R = 
1.29/1.32	  init(x1, x2) -> f1(rnd1, rnd2) 
1.29/1.32	  f3(I0, I1) -> f3(I0, I1 - 1) [0 <= I1 - 1]
1.29/1.32	  f3(I2, I3) -> f2(0, I2 - 1) [0 = I3]
1.29/1.32	  f2(I4, I5) -> f3(I5, I4) [0 <= I5 - 1]
1.29/1.32	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I8 - 1]
1.29/1.32	
1.29/1.32	The dependency graph for this problem is:
1.29/1.32	  2 -> 
1.29/1.32	Where:
1.29/1.32	  2) f3#(I2, I3) -> f2#(0, I2 - 1) [0 = I3]
1.29/1.32	
1.29/1.32	We have the following SCCs.
1.29/1.32	  
1.29/4.30	EOF