3.49/3.88	YES
3.49/3.88	
3.49/3.88	DP problem for innermost termination.
3.49/3.88	P =
3.49/3.88	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
3.49/3.88	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  f3#(I3, I4, I5) -> f2#(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	  f2#(I6, I7, I8) -> f3#(I7, I6, I8) [I8 <= I7 - 1]
3.49/3.88	  f1#(I9, I10, I11) -> f2#(I12, I13, I14) [0 <= I9 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I10 - 1 /\ -1 <= I13 - 1]
3.49/3.88	R = 
3.49/3.88	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
3.49/3.88	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  f3(I3, I4, I5) -> f2(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	  f2(I6, I7, I8) -> f3(I7, I6, I8) [I8 <= I7 - 1]
3.49/3.88	  f1(I9, I10, I11) -> f2(I12, I13, I14) [0 <= I9 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I10 - 1 /\ -1 <= I13 - 1]
3.49/3.88	
3.49/3.88	The dependency graph for this problem is:
3.49/3.88	  0 -> 4
3.49/3.88	  1 -> 1, 2
3.49/3.88	  2 -> 3
3.49/3.88	  3 -> 1, 2
3.49/3.88	  4 -> 3
3.49/3.88	Where:
3.49/3.88	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
3.49/3.88	  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  2) f3#(I3, I4, I5) -> f2#(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	  3) f2#(I6, I7, I8) -> f3#(I7, I6, I8) [I8 <= I7 - 1]
3.49/3.88	  4) f1#(I9, I10, I11) -> f2#(I12, I13, I14) [0 <= I9 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I10 - 1 /\ -1 <= I13 - 1]
3.49/3.88	
3.49/3.88	We have the following SCCs.
3.49/3.88	  { 1, 2, 3 }
3.49/3.88	
3.49/3.88	DP problem for innermost termination.
3.49/3.88	P =
3.49/3.88	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  f3#(I3, I4, I5) -> f2#(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	  f2#(I6, I7, I8) -> f3#(I7, I6, I8) [I8 <= I7 - 1]
3.49/3.88	R = 
3.49/3.88	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
3.49/3.88	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  f3(I3, I4, I5) -> f2(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	  f2(I6, I7, I8) -> f3(I7, I6, I8) [I8 <= I7 - 1]
3.49/3.88	  f1(I9, I10, I11) -> f2(I12, I13, I14) [0 <= I9 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I10 - 1 /\ -1 <= I13 - 1]
3.49/3.88	
3.49/3.88	We use the reverse value criterion with the projection function NU:
3.49/3.88	  NU[f2#(z1,z2,z3)] = z2 - 1 + -1 * z3
3.49/3.88	  NU[f3#(z1,z2,z3)] = z1 - 1 - 1 + -1 * z3
3.49/3.88	
3.49/3.88	This gives the following inequalities:
3.49/3.88	  I2 <= I1 - 1  ==>  I0 - 1 - 1 + -1 * I2 >= I0 - 1 - 1 + -1 * I2
3.49/3.88	  I4 <= I5  ==>  I3 - 1 - 1 + -1 * I5 >= I3 - 1 - 1 + -1 * I5
3.49/3.88	  I8 <= I7 - 1  ==>  I7 - 1 + -1 * I8 > I7 - 1 - 1 + -1 * I8  with  I7 - 1 + -1 * I8 >= 0
3.49/3.88	
3.49/3.88	We remove all the strictly oriented dependency pairs.
3.49/3.88	
3.49/3.88	DP problem for innermost termination.
3.49/3.88	P =
3.49/3.88	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  f3#(I3, I4, I5) -> f2#(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	R = 
3.49/3.88	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
3.49/3.88	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  f3(I3, I4, I5) -> f2(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	  f2(I6, I7, I8) -> f3(I7, I6, I8) [I8 <= I7 - 1]
3.49/3.88	  f1(I9, I10, I11) -> f2(I12, I13, I14) [0 <= I9 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I10 - 1 /\ -1 <= I13 - 1]
3.49/3.88	
3.49/3.88	The dependency graph for this problem is:
3.49/3.88	  1 -> 1, 2
3.49/3.88	  2 -> 
3.49/3.88	Where:
3.49/3.88	  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  2) f3#(I3, I4, I5) -> f2#(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	
3.49/3.88	We have the following SCCs.
3.49/3.88	  { 1 }
3.49/3.88	
3.49/3.88	DP problem for innermost termination.
3.49/3.88	P =
3.49/3.88	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	R = 
3.49/3.88	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
3.49/3.88	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I2) [I2 <= I1 - 1]
3.49/3.88	  f3(I3, I4, I5) -> f2(I4, I3 - 1, I5) [I4 <= I5]
3.49/3.88	  f2(I6, I7, I8) -> f3(I7, I6, I8) [I8 <= I7 - 1]
3.49/3.88	  f1(I9, I10, I11) -> f2(I12, I13, I14) [0 <= I9 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I10 - 1 /\ -1 <= I13 - 1]
3.49/3.88	
3.49/3.88	We use the reverse value criterion with the projection function NU:
3.49/3.88	  NU[f3#(z1,z2,z3)] = z2 - 1 + -1 * z3
3.49/3.88	
3.49/3.88	This gives the following inequalities:
3.49/3.88	  I2 <= I1 - 1  ==>  I1 - 1 + -1 * I2 > I1 - 1 - 1 + -1 * I2  with  I1 - 1 + -1 * I2 >= 0
3.49/3.88	
3.49/3.88	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
3.49/6.86	EOF