0.81/0.86	YES
0.81/0.86	
0.81/0.86	DP problem for innermost termination.
0.81/0.86	P =
0.81/0.86	  f5#(x1, x2) -> f4#(x1, x2) 
0.81/0.86	  f4#(I0, I1) -> f3#(0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.81/0.86	  f3#(I2, I3) -> f1#(I2, I3) 
0.81/0.86	  f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 50]
0.81/0.86	R = 
0.81/0.86	  f5(x1, x2) -> f4(x1, x2) 
0.81/0.86	  f4(I0, I1) -> f3(0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.81/0.86	  f3(I2, I3) -> f1(I2, I3) 
0.81/0.86	  f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 50]
0.81/0.86	  f1(I6, I7) -> f2(I6, I7) [50 <= I6]
0.81/0.86	
0.81/0.86	The dependency graph for this problem is:
0.81/0.86	  0 -> 1
0.81/0.86	  1 -> 2
0.81/0.86	  2 -> 3
0.81/0.86	  3 -> 2
0.81/0.86	Where:
0.81/0.86	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.81/0.86	  1) f4#(I0, I1) -> f3#(0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.81/0.86	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.81/0.86	  3) f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 50]
0.81/0.86	
0.81/0.86	We have the following SCCs.
0.81/0.86	  { 2, 3 }
0.81/0.86	
0.81/0.86	DP problem for innermost termination.
0.81/0.86	P =
0.81/0.86	  f3#(I2, I3) -> f1#(I2, I3) 
0.81/0.86	  f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 50]
0.81/0.86	R = 
0.81/0.86	  f5(x1, x2) -> f4(x1, x2) 
0.81/0.86	  f4(I0, I1) -> f3(0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.81/0.86	  f3(I2, I3) -> f1(I2, I3) 
0.81/0.86	  f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 50]
0.81/0.86	  f1(I6, I7) -> f2(I6, I7) [50 <= I6]
0.81/0.86	
0.81/0.86	We use the reverse value criterion with the projection function NU:
0.81/0.86	  NU[f1#(z1,z2)] = 50 + -1 * (1 + z1)
0.81/0.86	  NU[f3#(z1,z2)] = 50 + -1 * (1 + z1)
0.81/0.86	
0.81/0.86	This gives the following inequalities:
0.81/0.86	    ==>  50 + -1 * (1 + I2) >= 50 + -1 * (1 + I2)
0.81/0.86	  1 + I4 <= 50  ==>  50 + -1 * (1 + I4) > 50 + -1 * (1 + (1 + I4))  with  50 + -1 * (1 + I4) >= 0
0.81/0.86	
0.81/0.86	We remove all the strictly oriented dependency pairs.
0.81/0.86	
0.81/0.86	DP problem for innermost termination.
0.81/0.86	P =
0.81/0.86	  f3#(I2, I3) -> f1#(I2, I3) 
0.81/0.86	R = 
0.81/0.86	  f5(x1, x2) -> f4(x1, x2) 
0.81/0.86	  f4(I0, I1) -> f3(0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
0.81/0.86	  f3(I2, I3) -> f1(I2, I3) 
0.81/0.86	  f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 50]
0.81/0.86	  f1(I6, I7) -> f2(I6, I7) [50 <= I6]
0.81/0.86	
0.81/0.86	The dependency graph for this problem is:
0.81/0.86	  2 -> 
0.81/0.86	Where:
0.81/0.86	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.81/0.86	
0.81/0.86	We have the following SCCs.
0.81/0.86	  
0.81/3.84	EOF