0.00/0.48	YES
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.00/0.48	  f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
0.00/0.48	  f3#(I4, I5, I6, I7) -> f2#(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
0.00/0.48	  f2#(I9, I10, I11, I12) -> f3#(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
0.00/0.48	  f1#(I13, I14, I15, I16) -> f2#(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]
0.00/0.48	R = 
0.00/0.48	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.00/0.48	  f3(I0, I1, I2, I3) -> f3(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
0.00/0.48	  f3(I4, I5, I6, I7) -> f2(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
0.00/0.48	  f2(I9, I10, I11, I12) -> f3(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
0.00/0.48	  f1(I13, I14, I15, I16) -> f2(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]
0.00/0.48	
0.00/0.48	The dependency graph for this problem is:
0.00/0.48	  0 -> 4
0.00/0.48	  1 -> 1, 2
0.00/0.48	  2 -> 
0.00/0.48	  3 -> 1
0.00/0.48	  4 -> 3
0.00/0.48	Where:
0.00/0.48	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.00/0.48	  1) f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
0.00/0.48	  2) f3#(I4, I5, I6, I7) -> f2#(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
0.00/0.48	  3) f2#(I9, I10, I11, I12) -> f3#(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
0.00/0.48	  4) f1#(I13, I14, I15, I16) -> f2#(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]
0.00/0.48	
0.00/0.48	We have the following SCCs.
0.00/0.48	  { 1 }
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
0.00/0.48	R = 
0.00/0.48	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.00/0.48	  f3(I0, I1, I2, I3) -> f3(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
0.00/0.48	  f3(I4, I5, I6, I7) -> f2(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
0.00/0.48	  f2(I9, I10, I11, I12) -> f3(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
0.00/0.48	  f1(I13, I14, I15, I16) -> f2(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]
0.00/0.48	
0.00/0.48	We use the reverse value criterion with the projection function NU:
0.00/0.48	  NU[f3#(z1,z2,z3,z4)] = z2 - 1 + -1 * z4
0.00/0.48	
0.00/0.48	This gives the following inequalities:
0.00/0.48	  I1 = I2 /\ I3 <= I1 - 1  ==>  I1 - 1 + -1 * I3 > I1 - 1 - 1 + -1 * I3  with  I1 - 1 + -1 * I3 >= 0
0.00/0.48	
0.00/0.48	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.46	EOF