1.81/1.82	YES
1.81/1.82	
1.81/1.82	DP problem for innermost termination.
1.81/1.82	P =
1.81/1.82	  init#(x1, x2) -> f1#(rnd1, rnd2) 
1.81/1.82	  f3#(I0, I1) -> f3#(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  f3#(I2, I3) -> f2#(I3 - 1, I4) [I3 <= I2]
1.81/1.82	  f2#(I5, I6) -> f3#(0, I5) [0 <= I5 - 1]
1.81/1.82	  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.81/1.82	R = 
1.81/1.82	  init(x1, x2) -> f1(rnd1, rnd2) 
1.81/1.82	  f3(I0, I1) -> f3(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  f3(I2, I3) -> f2(I3 - 1, I4) [I3 <= I2]
1.81/1.82	  f2(I5, I6) -> f3(0, I5) [0 <= I5 - 1]
1.81/1.82	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.81/1.82	
1.81/1.82	The dependency graph for this problem is:
1.81/1.82	  0 -> 4
1.81/1.82	  1 -> 1, 2
1.81/1.82	  2 -> 3
1.81/1.82	  3 -> 1
1.81/1.82	  4 -> 3
1.81/1.82	Where:
1.81/1.82	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
1.81/1.82	  1) f3#(I0, I1) -> f3#(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  2) f3#(I2, I3) -> f2#(I3 - 1, I4) [I3 <= I2]
1.81/1.82	  3) f2#(I5, I6) -> f3#(0, I5) [0 <= I5 - 1]
1.81/1.82	  4) f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.81/1.82	
1.81/1.82	We have the following SCCs.
1.81/1.82	  { 1, 2, 3 }
1.81/1.82	
1.81/1.82	DP problem for innermost termination.
1.81/1.82	P =
1.81/1.82	  f3#(I0, I1) -> f3#(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  f3#(I2, I3) -> f2#(I3 - 1, I4) [I3 <= I2]
1.81/1.82	  f2#(I5, I6) -> f3#(0, I5) [0 <= I5 - 1]
1.81/1.82	R = 
1.81/1.82	  init(x1, x2) -> f1(rnd1, rnd2) 
1.81/1.82	  f3(I0, I1) -> f3(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  f3(I2, I3) -> f2(I3 - 1, I4) [I3 <= I2]
1.81/1.82	  f2(I5, I6) -> f3(0, I5) [0 <= I5 - 1]
1.81/1.82	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.81/1.82	
1.81/1.82	We use the reverse value criterion with the projection function NU:
1.81/1.82	  NU[f2#(z1,z2)] = z1 - 1 + -1 * 0
1.81/1.82	  NU[f3#(z1,z2)] = z2 - 1 - 1 + -1 * 0
1.81/1.82	
1.81/1.82	This gives the following inequalities:
1.81/1.82	  I0 <= I1 - 1  ==>  I1 - 1 - 1 + -1 * 0 >= I1 - 1 - 1 + -1 * 0
1.81/1.82	  I3 <= I2  ==>  I3 - 1 - 1 + -1 * 0 >= I3 - 1 - 1 + -1 * 0
1.81/1.82	  0 <= I5 - 1  ==>  I5 - 1 + -1 * 0 > I5 - 1 - 1 + -1 * 0  with  I5 - 1 + -1 * 0 >= 0
1.81/1.82	
1.81/1.82	We remove all the strictly oriented dependency pairs.
1.81/1.82	
1.81/1.82	DP problem for innermost termination.
1.81/1.82	P =
1.81/1.82	  f3#(I0, I1) -> f3#(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  f3#(I2, I3) -> f2#(I3 - 1, I4) [I3 <= I2]
1.81/1.82	R = 
1.81/1.82	  init(x1, x2) -> f1(rnd1, rnd2) 
1.81/1.82	  f3(I0, I1) -> f3(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  f3(I2, I3) -> f2(I3 - 1, I4) [I3 <= I2]
1.81/1.82	  f2(I5, I6) -> f3(0, I5) [0 <= I5 - 1]
1.81/1.82	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.81/1.82	
1.81/1.82	The dependency graph for this problem is:
1.81/1.82	  1 -> 1, 2
1.81/1.82	  2 -> 
1.81/1.82	Where:
1.81/1.82	  1) f3#(I0, I1) -> f3#(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  2) f3#(I2, I3) -> f2#(I3 - 1, I4) [I3 <= I2]
1.81/1.82	
1.81/1.82	We have the following SCCs.
1.81/1.82	  { 1 }
1.81/1.82	
1.81/1.82	DP problem for innermost termination.
1.81/1.82	P =
1.81/1.82	  f3#(I0, I1) -> f3#(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	R = 
1.81/1.82	  init(x1, x2) -> f1(rnd1, rnd2) 
1.81/1.82	  f3(I0, I1) -> f3(I0 + 1, I1) [I0 <= I1 - 1]
1.81/1.82	  f3(I2, I3) -> f2(I3 - 1, I4) [I3 <= I2]
1.81/1.82	  f2(I5, I6) -> f3(0, I5) [0 <= I5 - 1]
1.81/1.82	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
1.81/1.82	
1.81/1.82	We use the reverse value criterion with the projection function NU:
1.81/1.82	  NU[f3#(z1,z2)] = z2 - 1 + -1 * z1
1.81/1.82	
1.81/1.82	This gives the following inequalities:
1.81/1.82	  I0 <= I1 - 1  ==>  I1 - 1 + -1 * I0 > I1 - 1 + -1 * (I0 + 1)  with  I1 - 1 + -1 * I0 >= 0
1.81/1.82	
1.81/1.82	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.81/4.80	EOF