0.00/0.15	YES
0.00/0.15	
0.00/0.15	DP problem for innermost termination.
0.00/0.15	P =
0.00/0.15	  f6#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 
0.00/0.15	  f5#(I0, I1, I2, I3, I4, I5) -> f4#(I0, I1, I2, rnd4, I4, rnd6) [rnd6 = rnd6 /\ rnd4 = rnd4]
0.00/0.15	  f4#(I6, I7, I8, I9, I10, I11) -> f3#(I6, I7, I8, I9, I10, I11) [1 + I11 <= -1]
0.00/0.15	  f4#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) [0 <= I17]
0.00/0.15	  f4#(I18, I19, I20, I21, I22, I23) -> f1#(rnd1, 0, rnd3, I21, I22, I23) [rnd3 = rnd3 /\ rnd1 = rnd1 /\ -1 <= I23 /\ I23 <= -1]
0.00/0.15	  f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, I27, rnd5, I29) [rnd5 = rnd5]
0.00/0.15	R = 
0.00/0.15	  f6(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) 
0.00/0.15	  f5(I0, I1, I2, I3, I4, I5) -> f4(I0, I1, I2, rnd4, I4, rnd6) [rnd6 = rnd6 /\ rnd4 = rnd4]
0.00/0.15	  f4(I6, I7, I8, I9, I10, I11) -> f3(I6, I7, I8, I9, I10, I11) [1 + I11 <= -1]
0.00/0.15	  f4(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) [0 <= I17]
0.00/0.15	  f4(I18, I19, I20, I21, I22, I23) -> f1(rnd1, 0, rnd3, I21, I22, I23) [rnd3 = rnd3 /\ rnd1 = rnd1 /\ -1 <= I23 /\ I23 <= -1]
0.00/0.15	  f3(I24, I25, I26, I27, I28, I29) -> f1(I24, I25, I26, I27, rnd5, I29) [rnd5 = rnd5]
0.00/0.15	  f1(I30, I31, I32, I33, I34, I35) -> f2(I30, I31, I32, I33, I34, I35) 
0.00/0.15	
0.00/0.15	The dependency graph for this problem is:
0.00/0.15	  0 -> 1
0.00/0.15	  1 -> 2, 3, 4
0.00/0.15	  2 -> 5
0.00/0.15	  3 -> 5
0.00/0.15	  4 -> 
0.00/0.15	  5 -> 
0.00/0.15	Where:
0.00/0.15	  0) f6#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 
0.00/0.15	  1) f5#(I0, I1, I2, I3, I4, I5) -> f4#(I0, I1, I2, rnd4, I4, rnd6) [rnd6 = rnd6 /\ rnd4 = rnd4]
0.00/0.15	  2) f4#(I6, I7, I8, I9, I10, I11) -> f3#(I6, I7, I8, I9, I10, I11) [1 + I11 <= -1]
0.00/0.15	  3) f4#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) [0 <= I17]
0.00/0.15	  4) f4#(I18, I19, I20, I21, I22, I23) -> f1#(rnd1, 0, rnd3, I21, I22, I23) [rnd3 = rnd3 /\ rnd1 = rnd1 /\ -1 <= I23 /\ I23 <= -1]
0.00/0.15	  5) f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, I27, rnd5, I29) [rnd5 = rnd5]
0.00/0.15	
0.00/0.15	We have the following SCCs.
0.00/0.15	  
0.00/3.13	EOF