2.49/2.55	YES
2.49/2.55	
2.49/2.55	DP problem for innermost termination.
2.49/2.55	P =
2.49/2.55	  init#(x1, x2) -> f1#(rnd1, rnd2) 
2.49/2.55	  f3#(I0, I1) -> f3#(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  f3#(I2, I3) -> f2#(I3, I2) [I2 <= I3 - 1]
2.49/2.55	  f2#(I4, I5) -> f3#(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
2.49/2.55	  f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]
2.49/2.55	R = 
2.49/2.55	  init(x1, x2) -> f1(rnd1, rnd2) 
2.49/2.55	  f3(I0, I1) -> f3(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  f3(I2, I3) -> f2(I3, I2) [I2 <= I3 - 1]
2.49/2.55	  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
2.49/2.55	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]
2.49/2.55	
2.49/2.55	The dependency graph for this problem is:
2.49/2.55	  0 -> 4
2.49/2.55	  1 -> 1, 2
2.49/2.55	  2 -> 3
2.49/2.55	  3 -> 1, 2
2.49/2.55	  4 -> 3
2.49/2.55	Where:
2.49/2.55	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
2.49/2.55	  1) f3#(I0, I1) -> f3#(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  2) f3#(I2, I3) -> f2#(I3, I2) [I2 <= I3 - 1]
2.49/2.55	  3) f2#(I4, I5) -> f3#(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
2.49/2.55	  4) f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]
2.49/2.55	
2.49/2.55	We have the following SCCs.
2.49/2.55	  { 1, 2, 3 }
2.49/2.55	
2.49/2.55	DP problem for innermost termination.
2.49/2.55	P =
2.49/2.55	  f3#(I0, I1) -> f3#(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  f3#(I2, I3) -> f2#(I3, I2) [I2 <= I3 - 1]
2.49/2.55	  f2#(I4, I5) -> f3#(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
2.49/2.55	R = 
2.49/2.55	  init(x1, x2) -> f1(rnd1, rnd2) 
2.49/2.55	  f3(I0, I1) -> f3(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  f3(I2, I3) -> f2(I3, I2) [I2 <= I3 - 1]
2.49/2.55	  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
2.49/2.55	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]
2.49/2.55	
2.49/2.55	We use the reverse value criterion with the projection function NU:
2.49/2.55	  NU[f2#(z1,z2)] = z2
2.49/2.55	  NU[f3#(z1,z2)] = z2 - 1 + -1 * 0
2.49/2.55	
2.49/2.55	This gives the following inequalities:
2.49/2.55	  0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1  ==>  I1 - 1 + -1 * 0 >= I1 - 1 + -1 * 0
2.49/2.55	  I2 <= I3 - 1  ==>  I3 - 1 + -1 * 0 >= I2
2.49/2.55	  0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I5 > I5 - 1 + -1 * 0  with  I5 >= 0
2.49/2.55	
2.49/2.55	We remove all the strictly oriented dependency pairs.
2.49/2.55	
2.49/2.55	DP problem for innermost termination.
2.49/2.55	P =
2.49/2.55	  f3#(I0, I1) -> f3#(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  f3#(I2, I3) -> f2#(I3, I2) [I2 <= I3 - 1]
2.49/2.55	R = 
2.49/2.55	  init(x1, x2) -> f1(rnd1, rnd2) 
2.49/2.55	  f3(I0, I1) -> f3(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  f3(I2, I3) -> f2(I3, I2) [I2 <= I3 - 1]
2.49/2.55	  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
2.49/2.55	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]
2.49/2.55	
2.49/2.55	The dependency graph for this problem is:
2.49/2.55	  1 -> 1, 2
2.49/2.55	  2 -> 
2.49/2.55	Where:
2.49/2.55	  1) f3#(I0, I1) -> f3#(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  2) f3#(I2, I3) -> f2#(I3, I2) [I2 <= I3 - 1]
2.49/2.55	
2.49/2.55	We have the following SCCs.
2.49/2.55	  { 1 }
2.49/2.55	
2.49/2.55	DP problem for innermost termination.
2.49/2.55	P =
2.49/2.55	  f3#(I0, I1) -> f3#(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	R = 
2.49/2.55	  init(x1, x2) -> f1(rnd1, rnd2) 
2.49/2.55	  f3(I0, I1) -> f3(I0 - I1, I1) [0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1]
2.49/2.55	  f3(I2, I3) -> f2(I3, I2) [I2 <= I3 - 1]
2.49/2.55	  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
2.49/2.55	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]
2.49/2.55	
2.49/2.55	We use the basic value criterion with the projection function NU:
2.49/2.55	  NU[f3#(z1,z2)] = z1
2.49/2.55	
2.49/2.55	This gives the following inequalities:
2.49/2.55	  0 <= I1 - 1 /\ I1 <= I0 /\ 0 <= I0 - 1  ==>  I0 >! I0 - I1
2.49/2.55	
2.49/2.55	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.49/5.54	EOF