1.68/1.70	YES
1.68/1.70	
1.68/1.70	DP problem for innermost termination.
1.68/1.70	P =
1.68/1.70	  f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
1.68/1.70	  f5#(I0, I1, I2) -> f4#(0, I1, I2) 
1.68/1.70	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.68/1.70	  f4#(I9, I10, I11) -> f3#(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
1.68/1.70	  f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
1.68/1.70	R = 
1.68/1.70	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
1.68/1.70	  f5(I0, I1, I2) -> f4(0, I1, I2) 
1.68/1.70	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.68/1.70	  f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7]
1.68/1.70	  f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
1.68/1.70	  f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
1.68/1.70	  f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15]
1.68/1.70	
1.68/1.70	The dependency graph for this problem is:
1.68/1.70	  0 -> 1
1.68/1.70	  1 -> 3
1.68/1.70	  2 -> 4
1.68/1.70	  3 -> 2
1.68/1.70	  4 -> 2
1.68/1.70	Where:
1.68/1.70	  0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
1.68/1.70	  1) f5#(I0, I1, I2) -> f4#(0, I1, I2) 
1.68/1.70	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.68/1.70	  3) f4#(I9, I10, I11) -> f3#(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
1.68/1.70	  4) f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
1.68/1.70	
1.68/1.70	We have the following SCCs.
1.68/1.70	  { 2, 4 }
1.68/1.70	
1.68/1.70	DP problem for innermost termination.
1.68/1.70	P =
1.68/1.70	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.68/1.70	  f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
1.68/1.70	R = 
1.68/1.70	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
1.68/1.70	  f5(I0, I1, I2) -> f4(0, I1, I2) 
1.68/1.70	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.68/1.70	  f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7]
1.68/1.70	  f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
1.68/1.70	  f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
1.68/1.70	  f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15]
1.68/1.70	
1.68/1.70	We use the reverse value criterion with the projection function NU:
1.68/1.70	  NU[f1#(z1,z2,z3)] = 36 + -1 * (1 + z1)
1.68/1.70	  NU[f3#(z1,z2,z3)] = 36 + -1 * (1 + z1)
1.68/1.70	
1.68/1.70	This gives the following inequalities:
1.68/1.70	    ==>  36 + -1 * (1 + I3) >= 36 + -1 * (1 + I3)
1.68/1.70	  1 + I12 <= 36  ==>  36 + -1 * (1 + I12) > 36 + -1 * (1 + (1 + I12))  with  36 + -1 * (1 + I12) >= 0
1.68/1.70	
1.68/1.70	We remove all the strictly oriented dependency pairs.
1.68/1.70	
1.68/1.70	DP problem for innermost termination.
1.68/1.70	P =
1.68/1.70	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.68/1.70	R = 
1.68/1.70	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
1.68/1.70	  f5(I0, I1, I2) -> f4(0, I1, I2) 
1.68/1.70	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.68/1.70	  f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7]
1.68/1.70	  f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
1.68/1.70	  f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
1.68/1.70	  f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15]
1.68/1.70	
1.68/1.70	The dependency graph for this problem is:
1.68/1.70	  2 -> 
1.68/1.70	Where:
1.68/1.70	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.68/1.70	
1.68/1.70	We have the following SCCs.
1.68/1.70	  
1.68/4.68	EOF