1.63/2.02	YES
1.63/2.02	
1.63/2.02	DP problem for innermost termination.
1.63/2.02	P =
1.63/2.02	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.63/2.02	  f4#(I0, I1, I2) -> f1#(I0, I1, I2) 
1.63/2.02	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/2.02	  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
1.63/2.02	R = 
1.63/2.02	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.63/2.02	  f4(I0, I1, I2) -> f1(I0, I1, I2) 
1.63/2.02	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.63/2.02	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
1.63/2.02	  f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ -1 * I10 + I11 <= 0]
1.63/2.02	
1.63/2.02	The dependency graph for this problem is:
1.63/2.02	  0 -> 1
1.63/2.02	  1 -> 3
1.63/2.02	  2 -> 3
1.63/2.02	  3 -> 2
1.63/2.02	Where:
1.63/2.02	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.63/2.02	  1) f4#(I0, I1, I2) -> f1#(I0, I1, I2) 
1.63/2.02	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/2.02	  3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
1.63/2.02	
1.63/2.02	We have the following SCCs.
1.63/2.02	  { 2, 3 }
1.63/2.02	
1.63/2.02	DP problem for innermost termination.
1.63/2.02	P =
1.63/2.02	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/2.02	  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
1.63/2.02	R = 
1.63/2.02	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.63/2.02	  f4(I0, I1, I2) -> f1(I0, I1, I2) 
1.63/2.02	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.63/2.02	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
1.63/2.02	  f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ -1 * I10 + I11 <= 0]
1.63/2.02	
1.63/2.02	We use the reverse value criterion with the projection function NU:
1.63/2.02	  NU[f1#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0
1.63/2.02	  NU[f3#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0
1.63/2.02	
1.63/2.02	This gives the following inequalities:
1.63/2.02	    ==>  -1 - I4 + I5 + -1 * 0 >= -1 - I4 + I5 + -1 * 0
1.63/2.02	  0 <= -1 - I7 + I8  ==>  -1 - I7 + I8 + -1 * 0 > -1 - (1 + I7) + I8 + -1 * 0  with  -1 - I7 + I8 + -1 * 0 >= 0
1.63/2.02	
1.63/2.02	We remove all the strictly oriented dependency pairs.
1.63/2.02	
1.63/2.02	DP problem for innermost termination.
1.63/2.02	P =
1.63/2.02	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/2.02	R = 
1.63/2.02	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.63/2.02	  f4(I0, I1, I2) -> f1(I0, I1, I2) 
1.63/2.02	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.63/2.02	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
1.63/2.02	  f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ -1 * I10 + I11 <= 0]
1.63/2.02	
1.63/2.02	The dependency graph for this problem is:
1.63/2.02	  2 -> 
1.63/2.02	Where:
1.63/2.02	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/2.02	
1.63/2.02	We have the following SCCs.
1.63/2.02	  
1.63/5.00	EOF